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					<description><![CDATA[<p>1. Price Randomness 2. Price Factors 3. Technical Analysis 3.1. Chart Patterns 3.2. Market Indicators 4. Mathematical Expectancy 5. The&#8230;</p>
<p>The post <a href="https://www.math4all.es/maths-of-trading/">The Mathematics of Trading</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="&#x1f522; The MATHS of TRADING &#x1f4c8;&#x1f680;" width="1170" height="658" src="https://www.youtube.com/embed/_XZG_z-xQI0?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe></p>
<p><a href="#aleatoriedad">1. Price Randomness</a><br />
<a href="#factores">2. Price Factors</a><br />
<a href="#analisis">3. Technical Analysis</a><br />
<a href="#analisis">3.1. Chart Patterns</a><br />
<a href="#analisis">3.2. Market Indicators</a><br />
<a href="#esperanza">4. Mathematical Expectancy</a><br />
<a href="#datos">5. The Importance of Data in Trading</a><br />
<a href="#riesgo">6. Risk Management</a><br />
<a href="#ror">6.1. Risk of Ruin (ROR)</a><br />
<a href="#optimo">6.2. Optimal Risk</a><br />
<a href="#posicion">6.3. Calculating Position Size</a><br />
<a href="#variable">6.4. Variable Risk</a><br />
<a href="#parametros">7. Parameters of a Trading System</a><br />
<a href="#curso">8. TRADING COURSE</a></p>
<p>What do you know about trading? Can you beat the markets over time, or is it something impossible?</p>
<p>Do you believe mathematics can help you design a winning system, or is bankruptcy inevitable?<br />
<a id="aleatoriedad" name="Ancla"></a></p>
<h2>1. Price Randomness</h2>
<p>They say that 90% of traders lose 90% of their money in the first 90 days, which isn&#8217;t very encouraging. But if many people lose, others must be winning, although&#8230; does that mean it&#8217;s possible to predict the price?</p>
<p>There are theories claiming that the price is completely random (<em><strong>Random walk</strong></em>) and the certainty that it doesn&#8217;t follow any known distribution. There is even the hypothesis that it&#8217;s impossible to anticipate it (<em><strong>efficient market</strong></em>), as markets are efficient and reflect all available advantageous information too quickly. This theory, which sounds quite good, was accepted and taught for many years.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-mercados-eficientes.jpg"><img fetchpriority="high" decoding="async" class="aligncenter size-large wp-image-1340" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-mercados-eficientes-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-mercados-eficientes-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-mercados-eficientes-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-mercados-eficientes-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-mercados-eficientes.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>But&#8230; since when do markets act only efficiently? Aren&#8217;t our emotions part of the game?</p>
<p>If we analyze what happens in a <strong>financial bubble</strong>, we can see a strong sense of euphoria is created, driving the price irrationally and ultimately leaving stocks overvalued. Something that cannot be explained by the efficient market model, because the price is always perfectly valued.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-burbuja-financiera.jpg"><img decoding="async" class="aligncenter wp-image-1384 size-large" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-burbuja-financiera-1024x649.jpg" alt="" width="1024" height="649" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-burbuja-financiera-1024x649.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-burbuja-financiera-300x190.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-burbuja-financiera-768x486.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-burbuja-financiera.jpg 1102w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>If we accept the emotional factor, we then have the <strong><em>Adaptive Market Hypothesis</em></strong>, which combines the best of both and better explains price behavior.</p>
<p>Although&#8230; how many factors actually influence the price?<br />
<a id="factores" name="Ancla"></a></p>
<h2>2. Price Factors</h2>
<p>Here is a <strong>list</strong> where you can see most of them:</p>
<div id="attachment_1606" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/financial-market-factors.jpg"><img decoding="async" aria-describedby="caption-attachment-1606" class="wp-image-1606 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/financial-market-factors-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/financial-market-factors-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/financial-market-factors-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/financial-market-factors-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/financial-market-factors.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1606" class="wp-caption-text">Screenshot</p></div>
<p>Some are highly <strong>psychological</strong>, such as news, social media posts, and even predictions themselves. It may seem strange, but the price itself influences the price—curious, right?</p>
<p>But one of the most influential factors is <strong>manipulation</strong>. The movements of large &#8220;whales&#8221; or groups of many investors can cause sharp rises and falls in a short time. By suddenly removing many sellers or buyers from the auction, the price takes on a very different value, which drags more investors to follow the movement, further expanding its effects.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-manipulaciones.jpg"><img decoding="async" class="aligncenter size-large wp-image-1343" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-manipulaciones-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-manipulaciones-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-manipulaciones-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-manipulaciones-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-manipulaciones.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>The fact is, some factors are more influential than others.</p>
<p>But&#8230; how can mathematics help you in trading?<br />
<a id="analisis" name="Ancla"></a></p>
<h2>3. Technical Analysis</h2>
<p>To analyze a market, you can focus on its peculiarities through <strong>fundamental analysis</strong>. For example, if you wanted to invest in Apple long-term, you would study its sales data, economic reports, and information relative to the company to develop a prediction of the true value of its shares.</p>
<p>In contrast, a trader focuses mainly on a technical analysis of the price because, according to <em><strong>Dow Theory</strong></em>, the history already displays all factors influencing it.</p>
<p>And here again, mathematics helps us&#8230; but what are those factors?<br />
<a id="patrones" name="Ancla"></a></p>
<h3>3.1 Chart Patterns</h3>
<p>Technically analyzing a market involves analyzing price trends and their corrections to take advantage of them.</p>
<p>To do this, we draw <strong>trend lines</strong>, <strong>support and resistance</strong>, <strong>Fibonacci levels</strong>, or <strong>Elliott Waves</strong>, which are nothing more than graphic elements that mark patterns in the price and have a certain <strong>statistical success</strong> rate.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-patrones-graficos.jpg"><img decoding="async" class="aligncenter size-large wp-image-1421" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-patrones-graficos-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-patrones-graficos-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-patrones-graficos-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-patrones-graficos-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>For example, any trader knows that when a very strong resistance level, like an <strong>all-time high</strong>, is broken, it&#8217;s very likely the price will gain more momentum. This is just one example, but these tools allow you to easily design a system with a high percentage of hits.<br />
<a id="indicadores" name="Ancla"></a></p>
<h3>3.2. Market Indicators</h3>
<p>The problem is that there is no rigorous consensus on how to use them because every analyst draws these elements their own way. So, if you want more unified buy and sell signals, there are <strong>market indicators</strong>. These are simply charts plotted with mathematical calculations that give the same signals to both you and me.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-indicadores-bursatiles.jpg"><img decoding="async" class="aligncenter wp-image-1445 size-large" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-indicadores-bursatiles-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-indicadores-bursatiles-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-indicadores-bursatiles-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-indicadores-bursatiles-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-indicadores-bursatiles.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>One example is the <strong>moving average</strong>, which is simply a line where each point is calculated using the average value of the periods we choose. For example, if we choose a 9-period moving average, each point is obtained from the average of the 9 previous prices. That is, a line of averages is drawn to smooth out the noise caused by short-term factors like news or manipulation. The moving average summarizes the general information better and is visually useful.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-media-movil.jpg"><img decoding="async" class="aligncenter wp-image-1406 size-large" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-media-movil-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-media-movil-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-media-movil-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-media-movil-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-media-movil.jpg 1238w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<div id="attachment_1607" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-formula.jpg"><img decoding="async" aria-describedby="caption-attachment-1607" class="wp-image-1607 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-formula-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-formula-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-formula-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-formula-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-formula.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1607" class="wp-caption-text">Screenshot</p></div>
<p>There are different <strong>types of moving averages</strong>, and they differ mainly in the priority they give to the values of those averages. Most indicators use <strong>exponential moving averages</strong> because they give more importance to the latest price values.</p>
<div id="attachment_1608" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-types.jpg"><img decoding="async" aria-describedby="caption-attachment-1608" class="wp-image-1608 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-types-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-types-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-types-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-types-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-types.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1608" class="wp-caption-text">Screenshot</p></div>
<p>But getting back to the topic, would you like to see an indicator?</p>
<div id="attachment_1609" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-cross.jpg"><img decoding="async" aria-describedby="caption-attachment-1609" class="wp-image-1609 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-cross-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/moving-average-cross-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-cross-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-cross-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/moving-average-cross.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1609" class="wp-caption-text">Screenshot</p></div>
<p>The simplest one is a <strong>moving average crossover</strong>, where two moving averages of different periods are superimposed—one faster and more sensitive line with fewer periods, against one slower and more stable line with more periods—and the relationship is studied. Can you imagine how?</p>
<p>The <strong>crossover points</strong> represent moments where the short term diverges from the long term, thus indicating a possible buy or sell.</p>
<p>Note that when they cross, there is usually a change in price. However, notice there is also a <strong>lag</strong>.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-macd.jpg"><img decoding="async" class="aligncenter size-large wp-image-1353" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-macd-1024x512.jpg" alt="" width="1024" height="512" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-macd-1024x512.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-macd-300x150.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-macd-768x384.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-macd.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Another common technical indicator is the <strong>MACD</strong>, which looks for another line crossover—in this case, a very fast signal achieved by subtracting a 26-period average from a 12-period average, with a slower signal achieved by smoothing this with a 9-period average.</p>
<div id="attachment_1610" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/rsi-overbought-oversold.jpg"><img decoding="async" aria-describedby="caption-attachment-1610" class="wp-image-1610 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/rsi-overbought-oversold-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/rsi-overbought-oversold-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/rsi-overbought-oversold-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/rsi-overbought-oversold-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/rsi-overbought-oversold.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1610" class="wp-caption-text">Screenshot</p></div>
<p>Then there is the <strong>RSI</strong> or Relative Strength Index, which indicates if an asset is considered overbought or oversold when it reaches levels above 70 or below 30.</p>
<p>Nevertheless, there are also indicators created by mathematicians, such as <b>BOLLINGER BANDS</b>, which represent market volatility with their width.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-bandas-de-bollinger.jpg"><img decoding="async" class="aligncenter size-large wp-image-1355" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-bandas-de-bollinger-1024x384.jpg" alt="" width="1024" height="384" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-bandas-de-bollinger-1024x384.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-bandas-de-bollinger-300x113.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-bandas-de-bollinger-768x288.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-bandas-de-bollinger.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>And others like <b>VOLUME</b>, widely used by some traders.</p>
<div id="attachment_1611" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/volume-indicator.jpg"><img decoding="async" aria-describedby="caption-attachment-1611" class="wp-image-1611 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/volume-indicator-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/volume-indicator-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/volume-indicator-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/volume-indicator-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/volume-indicator.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1611" class="wp-caption-text">Screenshot</p></div>
<p>All of these also serve to design systems with a good <strong>probability of success</strong>, and since many analysts use them, they ultimately end up influencing the price. A bit paradoxical, isn&#8217;t it?</p>
<p>The point is that indicators can have some effectiveness, but that doesn&#8217;t mean they are always right. Note that in the MACD, when a large trend change occurs, there is a crossover; however, not every time a crossover occurs does the momentum have the same strength. You must be very clear that indicators are calculated from the price and not the other way around, and if the price changes, the indicator changes too because it is drawn with <strong>past data</strong>. That&#8217;s why you must be careful, as price momentum can blow right through any indicator. You&#8217;ve been warned!</p>
<p>It&#8217;s important that before using them, you know all their strengths and weaknesses well (so if you&#8217;d like to learn them, check out the course I have prepared below).<br />
<a id="esperanza" name="Ancla"></a></p>
<h2>4. Mathematical Expectancy</h2>
<p>Note that although we don&#8217;t have a crystal ball to predict the price, we have introduced a concept that can indeed make you a winner: <strong>Statistical Success</strong>.</p>
<p>It is not necessary to know what the price is going to do at all times to have a profitable system.</p>
<p>But how do we know if it&#8217;s profitable? Do you think you can know?</p>
<p>Actually, it&#8217;s very simple: a system is <strong>profitable</strong> when the profit from all positive trades exceeds the loss from all negative trades. In other words, the balance between winning and losing trades is positive.</p>
<p>Simple as that. What were you expecting?</p>
<p>If you summarize that <strong>average balance for a single trade</strong>, you get the <strong>mathematical expectancy</strong> of the system. This is how much money you earn on average per trade.</p>
<div id="attachment_1612" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/mathematical-expectation-trading.jpg"><img decoding="async" aria-describedby="caption-attachment-1612" class="wp-image-1612 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/mathematical-expectation-trading-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/mathematical-expectation-trading-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/mathematical-expectation-trading-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/mathematical-expectation-trading-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/mathematical-expectation-trading.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1612" class="wp-caption-text">Screenshot</p></div>
<p>It is calculated using the <strong>win rate</strong> (percentage of winning trades) and the <strong>risk-reward ratio</strong> of all of them.</p>
<p>Let&#8217;s look at an example:</p>
<div id="attachment_1613" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/heads-tails.jpg"><img decoding="async" aria-describedby="caption-attachment-1613" class="wp-image-1613 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/heads-tails-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/heads-tails-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/heads-tails-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/heads-tails-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/heads-tails.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1613" class="wp-caption-text">Screenshot</p></div>
<p>Imagine your system was like <strong>flipping a coin</strong> with <strong>2 rules</strong>. If it&#8217;s heads, they pay you another coin, but if it&#8217;s tails, you lose your coin. This means your risk-reward ratio is 1 to 1 (1 coin you can win for every 1 coin you can lose). Since the probability of getting heads is 50%, the <em><strong>law of large numbers</strong></em> ensures that after many flips, you would have a win rate close to that value, and since the prize when you win is the same as what you lose when you lose, you would have a system that neither wins nor loses. Mathematical expectancy: 0.</p>
<p>Now look at this other one that makes <strong>5 trades</strong>. 3 of them are positive (which is a 60% success rate), and the risk-reward ratio is 2 to 1.</p>
<div id="attachment_1614" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-probability.jpg"><img decoding="async" aria-describedby="caption-attachment-1614" class="wp-image-1614 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-probability-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-probability-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-probability-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-probability-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-probability.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1614" class="wp-caption-text">Screenshot</p></div>
<p>If we use the formula, it has an average profit per trade of 1.6. This means that so far, its expectancy is clearly a winner.</p>
<div id="attachment_1615" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-win-loss-ratio.jpg"><img decoding="async" aria-describedby="caption-attachment-1615" class="wp-image-1615 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-win-loss-ratio-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-win-loss-ratio-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-win-loss-ratio-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-win-loss-ratio-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/profitable-trading-system-high-win-loss-ratio.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1615" class="wp-caption-text">Screenshot</p></div>
<p>However, if I show you this other one, you might think it isn&#8217;t profitable because it has a <strong>win rate lower than 50%</strong>, right? This is just a perception, because in this case, the ratio compensates for the other variable.</p>
<div id="attachment_1616" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/loser-trading-system-low-win-ratio.jpg"><img decoding="async" aria-describedby="caption-attachment-1616" class="wp-image-1616 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/loser-trading-system-low-win-ratio-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/loser-trading-system-low-win-ratio-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/loser-trading-system-low-win-ratio-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/loser-trading-system-low-win-ratio-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/loser-trading-system-low-win-ratio.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1616" class="wp-caption-text">Screenshot</p></div>
<p>Though the opposite can also happen. Systems with a high win rate may not be profitable if the risk-reward ratio doesn&#8217;t ultimately compensate. I tell you this because some people take advantage of this perception to sell you <strong>trading bots</strong> that have a high success rate but, when they lose, they lose such large amounts that it stops being profitable. Unfortunately, all the profit up until the catastrophe is just a <strong>statistical illusion</strong>, so be very careful.</p>
<div id="attachment_1617" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-profitable-table-probability-win-loss-ratio.jpg"><img decoding="async" aria-describedby="caption-attachment-1617" class="wp-image-1617 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-profitable-table-probability-win-loss-ratio-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-profitable-table-probability-win-loss-ratio-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-profitable-table-probability-win-loss-ratio-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-profitable-table-probability-win-loss-ratio-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-profitable-table-probability-win-loss-ratio.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1617" class="wp-caption-text">Screenshot</p></div>
<p>Always remember that <strong>a system is a winner only when its mathematical expectancy is positive</strong>, and not just because of one of the two variables separately.<br />
<a id="datos" name="Ancla"></a></p>
<h2>5. The importance of data in trading</h2>
<p>But if you are a good observer, you would tell me that 5 trades are not enough to guarantee a certain <strong>confidence level</strong>, because for that you need a much larger <strong>sample of trades</strong>. And the truth is you wouldn&#8217;t be wrong; to obtain decisive conclusions, it is necessary to test any system first in a <strong>simulator</strong> with plenty of data.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-simulador.jpg"><img decoding="async" class="aligncenter size-large wp-image-1363" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-simulador-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-simulador-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-simulador-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-simulador-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-simulador.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>So, don&#8217;t even think about trading live without knowing your mathematical expectancy well!</p>
<p>But don&#8217;t go overboard with the <strong>historical data size</strong> either. Keep in mind that markets are adaptive and change over time. This means neither the win rate nor the risk-reward ratio will always tend toward the same value, and if you include data that is too old, you may get <strong>obsolete conclusions</strong>. Remember that statistics show us how the system has behaved up until the <strong>current moment</strong>, but not how it will do in the future. That&#8217;s why it&#8217;s more interesting to see how they evolve than to draw conclusions from useless data.</p>
<p>Not bad, right?<br />
<a id="riesgo" name="Ancla"></a></p>
<h2>6. Risk Management</h2>
<p>Now you have a way of knowing when a system is profitable, but&#8230; is it enough to win?</p>
<p>The answer is no, and <strong>mathematics</strong> will prove it to you.<br />
<a id="ror" name="Ancla"></a></p>
<h3>6.1. Risk of Ruin (RoR)</h3>
<p>Even if your system is very good, if you risk too much per trade, you will likely lose.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-resultados.jpg"><img decoding="async" class="aligncenter size-large wp-image-1364" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-resultados-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-resultados-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-resultados-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-resultados-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-resultados.jpg 1240w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Returning to the previous example, even if you have a 60% win rate, if you risk a lot of capital on each trade, a <strong>negative streak of trades</strong> is enough to wipe out your account. Because 60% isn&#8217;t that decisive, and it could happen even if you had a 99% rate.</p>
<p>The key is knowing that probability of going broke, which is precisely the <strong>probability or Risk of Ruin (RoR)</strong>. This depends on the system parameters you already know, but also on the risk you want to assume. The more your system risks, the greater the possibility that a series of negative trades will ruin you. (In fact, this relationship is exponential).</p>
<div id="attachment_1618" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/risk-of-ruin.jpg"><img decoding="async" aria-describedby="caption-attachment-1618" class="wp-image-1618 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/risk-of-ruin-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/risk-of-ruin-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/risk-of-ruin-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/risk-of-ruin-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/risk-of-ruin.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1618" class="wp-caption-text">Screenshot</p></div>
<p>So, having a positive mathematical expectancy isn&#8217;t enough!</p>
<p>To fully convince you, I&#8217;ve created a <strong>trading simulator</strong> where you can test all these concepts. I&#8217;m going to input a system that is profitable on paper, but with a quite high <strong>risk per trade</strong>, for example, <strong>40%</strong>. You can see that what seemed like a winner turns out not to be. Surprise!</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-loser.jpg"><img decoding="async" class="aligncenter wp-image-1625 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-loser-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-loser-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-loser-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-loser-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-loser.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>The truth is many systems with positive expectancy can still ruin you if you don&#8217;t have proper <strong>risk control</strong>. Because even if they are winners in the long run, expectancy doesn&#8217;t take into account that <strong>your capital is limited</strong>, and if you go broke, you won&#8217;t be able to prove it&#8217;s profitable because you&#8217;ll have no balance left to recover.</p>
<p>Keep in mind that trading success is based on trading many times, so if you want a future as a trader, you are going to have to control this parameter no matter what.</p>
<p>That&#8217;s why we&#8217;re going to try reducing the risk in the simulator, so you can see that this time the system can make you invincible in the long run.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-system.jpg"><img decoding="async" class="aligncenter wp-image-1619 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-system-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-system-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-system-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-system-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-simulator-system.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>&nbsp;</p>
<p style="text-align: center;"><a href="http://math4all.es/trading/risk-simulator.xlsx">DOWNLOAD RISK SIMULATOR</a></p>
<p>You&#8217;ve just discovered the holy grail of trading. How about giving me a like or sharing this?<br />
<a id="optimo" name="Ancla"></a></p>
<h3>6.2. Optimal Risk</h3>
<p>Now you just need to know which numbers are appropriate, so the question you must ask yourself is&#8230; exactly how much risk should be assumed?</p>
<p>Years ago, the mathematician <strong>Ralph Vince</strong> discovered that an optimal risk (Optimal f) exists to obtain the greatest possible benefit, but unfortunately, as you will see, it requires assuming a percentage that is too high.</p>
<p>The idea for not having drawdowns or large drops is to assume a <strong>small risk percentage</strong>. But which one exactly?</p>
<p>For this, I want you to see a chart where you can see how many consecutive negative trades you need to go broke with some common risks. Notice that a low risk allows you a much larger <strong>margin</strong> for error.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-risk-percentaje.jpg"><img decoding="async" class="aligncenter wp-image-1626 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-risk-percentaje-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-risk-percentaje-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-risk-percentaje-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-risk-percentaje-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-risk-percentaje.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>And while any of these risks could work for you, <strong>there is no optimal risk</strong> for everyone, because every system is different and every person is different. Everyone tolerates risk in their own way and isn&#8217;t willing to risk the same amount for the same benefit. Right?</p>
<p>Higher risk means growing faster, but also more chances of ending up badly. So depending on your system and your <strong>tolerance</strong>, you can assume higher or lower values, but what you can never do if you care about your capital is assume a percentage that puts your account at risk.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-tolerancia-riesgo.jpg"><img decoding="async" class="aligncenter wp-image-1418 size-large" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-tolerancia-riesgo-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-tolerancia-riesgo-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-tolerancia-riesgo-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-tolerancia-riesgo-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-tolerancia-riesgo.jpg 1239w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Now you know what math is for, but&#8230; how do you calculate your entry size for that percentage if you also have a loss limit?<br />
<a id="posicion" name="Ancla"></a></p>
<h3>6.3. Calculating Position Size</h3>
<p>To calculate the position size for a specific <strong>risk</strong> and <strong>stop loss</strong>, we use the following formula:</p>
<div id="attachment_1621" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-posizion-sizing-formula.jpg"><img decoding="async" aria-describedby="caption-attachment-1621" class="wp-image-1621 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-posizion-sizing-formula-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-posizion-sizing-formula-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-posizion-sizing-formula-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-posizion-sizing-formula-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-posizion-sizing-formula.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1621" class="wp-caption-text">Screenshot</p></div>
<p>Where we input both parameters along with the capital, and it calculates the exact amount to invest.</p>
<p>In practice, this is adjusted <strong>manually</strong> by setting the stop loss and adjusting the entry to get the <strong>risk percentage</strong> we want. However, it&#8217;s important that you know how it&#8217;s calculated.</p>
<p>Now you&#8217;ve learned how to calculate the risk for each trade. But&#8230; what if you don&#8217;t want to assume a fixed percentage? In that case, don&#8217;t worry, because <strong>variable risk</strong> systems also exist.<br />
<a id="variable" name="Ancla"></a></p>
<h3><strong>6.4. Variable Risk</strong></h3>
<p>Imagine you start with a <strong>$10,000</strong> account and a <strong>1% risk</strong> (meaning you risk $100 per trade). After some time trading, you manage to increase your account to $15,000. If you keep the same percentage, on the next trade you will risk $150, because it is now calculated as 1% of $15,000. With a <strong>fixed percentage</strong>, you also risk more when your account is larger, and that means your <strong>probability of ruin</strong> remains the same because 1% will always be <strong>proportional</strong> to your account size. What I&#8217;m saying is that if what interests you is for your system to become less vulnerable as you grow, then you must use a <strong>variable risk</strong> system.</p>
<div id="attachment_1622" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-fixed-risk.jpg"><img decoding="async" aria-describedby="caption-attachment-1622" class="wp-image-1622 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-fixed-risk-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-fixed-risk-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-fixed-risk-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-fixed-risk-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-fixed-risk.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1622" class="wp-caption-text">Screenshot</p></div>
<p>Back to the example, what do you think would happen if in every trade you always assumed $100 instead of 1%? The answer is logical: you would risk a smaller percentage as you grow, and that&#8217;s great. But you wouldn&#8217;t grow as fast either, because with the <strong>same risk</strong>, you would always opt for the <strong>same profit</strong>. So you should only do it if your profit seems sufficient to you.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2026/02/trading-variable-risk.jpg"><img decoding="async" class="aligncenter wp-image-1623 size-large" src="https://www.math4all.es/wp-content/uploads/2026/02/trading-variable-risk-1024x597.jpg" alt="" width="1024" height="597" srcset="https://www.math4all.es/wp-content/uploads/2026/02/trading-variable-risk-1024x597.jpg 1024w, https://www.math4all.es/wp-content/uploads/2026/02/trading-variable-risk-300x175.jpg 300w, https://www.math4all.es/wp-content/uploads/2026/02/trading-variable-risk-768x448.jpg 768w, https://www.math4all.es/wp-content/uploads/2026/02/trading-variable-risk.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Assuming the same amount isn&#8217;t perfect either, but&#8230; do you think it would be possible to <strong>reduce risk</strong> and <strong>increase profit</strong> at the same time? Let me know how you would do it in the comments!</p>
<p>In case you&#8217;re interested, there are multiple fixed and variable risk systems like <em><strong>fixed risk</strong></em> or <em><strong>fixed ratio</strong></em> with their pros and cons, which I explain in more detail in my trading course.<br />
<a id="parametros" name="Ancla"></a></p>
<h2>7. Parameters of a System</h2>
<p>The important thing is that you can now control the most essential <strong>variables</strong>, although these are not all of them. Mathematical expectancy or risk of ruin are just some of the many <strong>parameters</strong> used to analyze a trading system.</p>
<p>Within a <strong>backtest</strong>, many others can be measured, such as the <strong>Sharpe Ratio</strong> to measure profitability based on risk, or the <strong>Z-Score</strong> to know the randomness between trades. There are many more, and it&#8217;s important to know them in detail if you want to measure your system correctly.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-sistema.jpg"><img decoding="async" class="aligncenter wp-image-1402 size-large" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-sistema-1024x575.jpg" alt="" width="1024" height="575" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-sistema-1024x575.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-sistema-300x168.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-sistema-768x431.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/trading-sistema.jpg 1238w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Deciding whether an <strong>expert advisor</strong> or any system is appropriate or not comes down to evaluating a series of numbers that tell you if it is <strong>suitable for your operation</strong> or your tolerance. That&#8217;s why knowing its mathematics well is so important.<br />
<a id="curso" name="Ancla"></a></p>
<h2>8. TRADING COURSE</h2>
<p>If you&#8217;ve been left wanting to <strong>expand</strong> or <strong>delve deeper</strong> into technical patterns and indicators, <strong>discover</strong> the different risk systems that exist, or <strong>get to know</strong> these parameters well, I am developing a <strong>trading course</strong> where I explain all these concepts <strong>thoroughly</strong> and <strong>with examples</strong>. So, if you&#8217;re interested in learning <strong>trading from scratch</strong> with a mathematician, don&#8217;t miss this opportunity. I will leave you a link once it&#8217;s available.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/trading-curso.jpg"><img decoding="async" class="aligncenter wp-image-1443 size-large" src="https://www.math4all.es/wp-content/uploads/2021/09/trading-curso-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2021/09/trading-curso-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/trading-curso-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/trading-curso-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>If you are interested, <strong>leave your email</strong> in the comments.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2021/09/math4all.jpg"><img decoding="async" class="aligncenter wp-image-1435 size-medium" src="https://www.math4all.es/wp-content/uploads/2021/09/math4all-300x273.jpg" alt="" width="300" height="273" srcset="https://www.math4all.es/wp-content/uploads/2021/09/math4all-300x273.jpg 300w, https://www.math4all.es/wp-content/uploads/2021/09/math4all-768x698.jpg 768w, https://www.math4all.es/wp-content/uploads/2021/09/math4all-1024x931.jpg 1024w, https://www.math4all.es/wp-content/uploads/2021/09/math4all.jpg 1100w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p style="text-align: center;">See you in the course!</p>
<p>&nbsp;</p>
<img src="https://www.math4all.es/wp-content/uploads/2021/09/matematicas-del-trading.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/maths-of-trading/" data-pin-media="https://www.math4all.es/wp-content/uploads/2021/09/matematicas-del-trading.jpg" data-pin-description="The Mathematics of Trading"/><p>The post <a href="https://www.math4all.es/maths-of-trading/">The Mathematics of Trading</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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		<title>The Mathematics of Ludo &#8211; Rules, Probabilities and Strategy</title>
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		<pubDate>Mon, 02 Nov 2020 18:13:32 +0000</pubDate>
				<category><![CDATA[The Mathematics of games]]></category>
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					<description><![CDATA[<p>Parcheesi or Ludo is a more complicated game than it seems, and in this article I&#8217;m going to show you&#8230;</p>
<p>The post <a href="https://www.math4all.es/the-mathematics-of-ludo-rules-probabilities-and-strategy/">The Mathematics of Ludo – Rules, Probabilities and Strategy</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="Las MATEMÁTICAS del PARCHÍS &#x1f3b2;&#x1f3b2;" width="1170" height="658" src="https://www.youtube.com/embed/1p1pKsQ3Y7A?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe><br />
Parcheesi or Ludo is a more <strong>complicated</strong> game than it seems, and in this article I&#8217;m going to show you how the probabilities can help you <strong>improve your strategy</strong> so you don&#8217;t depend exclusively on luck.</p>
<p class="p3">Welcome then to the mathematics of Ludo.</p>
<p><a href="#reglas">1.Ludo Rules</a><br />
<a href="#ventaja">2.The advantage in Ludo</a><br />
<a href="#gestionarventaja"> 2.1.Manage the advantage</a><br />
<a href="#prioridad">3.Priority</a><br />
<a href="#probabilidad">4.Ludo Probabilities</a><br />
<a href="#probabilidadcomer"> 4.1.Probability of Eating/be eaten</a><br />
<a href="#probabilidadsalvarnos"> 4.2.Probability to Save us</a><br />
<a href="#probabilidadataque"> 4.3.Probability in the Attack</a><br />
<a href="#jugar"> 5.Online Ludo</a><br />
<a href="#tutorial"> 6.Practical Ludo Tutorial</a></p>
<p><a id="reglas" name="Ancla"></a></p>
<h2 class="p3">1.Ludo Rules</h2>
<p>Ludo is a board game of up to <strong>4 players</strong>, where we can play 1 against 1, in pairs, or 4 individually. It can be played with <strong>1 die</strong> in the classic version, although normally <strong>2 dice</strong> are used to speed up the games.</p>
<ul>
<li>Each player has <strong>4 chips</strong> to move them in their turn with the result of the roll of the dice, moving the desired checker with each die.</li>
<li>The objective is to <strong>reach the goal with all the chips</strong> before the opponent, turning around with all of them until the entry of our color, introducing them in the goal with the exact value. And if the game is by pairs, they will have to be all the pieces of the pair.</li>
<li>Before reaching the finish line with each tile, the player can <strong>eat the opponent&#8217;s chips</strong> that are not safe, <strong>build bridges</strong> or barriers with two tiles or <strong>put their chips saf</strong>e in the <strong>safe boxes</strong>. These locks are distributed every 5 or 7 boxes.</li>
<li>The rules for leaving the house are to <strong>roll a 5</strong> with one of the dice or with the sum of the two.</li>
<li>If you <strong>roll a double</strong> with both dice, you roll again, and if you roll a double <strong>3 times in a row</strong>, your chip goes home as long as it is not already in the goal zone. There are other rules for different modalities but these are the official ones.</li>
</ul>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/tablero-de-parchis.jpg"><img decoding="async" class="aligncenter wp-image-763 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/tablero-de-parchis.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/07/tablero-de-parchis.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/tablero-de-parchis-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/tablero-de-parchis-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Our task then is to manage the <strong>movement of the chips</strong> in order to have the best chances of <strong>winning the game</strong>.</p>
<p>I am a specialist in <strong>1-on-1</strong> Ludo, and I will focus on this modality, but most of the theory I will teach is easily applicable in pairs or 4 individual.</p>
<p><a id="ventaja" name="Ancla"></a></p>
<h2 class="p1"><strong>2.The advantage in Ludo</strong></h2>
<p>As you have seen The rules are very simple, but before we know how to beat our opponent, <strong>how do we know who is winning?</strong></p>
<p>Knowing which player has the <strong>advantage</strong> in Ludo is important and to find out this we will first analyze the board.</p>
<p>In Ludo, each piece goes through exactly <strong>71 boxes</strong> from the moment it leaves the house until it enters the goal. If we multiply that by each checker we need a total of <strong>284 moves</strong> between the 4 checkers to finish our game.</p>
<p>Suppose you have your tiles distributed with this configuration:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_MOVES.jpg"><img decoding="async" class="aligncenter wp-image-1112 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_MOVES-1024x609.jpg" alt="" width="1024" height="609" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_MOVES-1024x609.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_MOVES-300x178.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_MOVES-768x457.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_MOVES.jpg 1580w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In this one, the chip closest to the goal needs <strong>8 moves</strong> to enter the goal, the next closest <strong>25 moves</strong>, the next closest <strong>42 moves</strong>, and the furthest <strong>59 moves</strong>. That is to say that between all of them they add up to <strong>134 movements</strong> to finish our game or what is the same as saying that you have made <strong>150 movements</strong> of 284 (it&#8217;s simply to make the subtraction).</p>
<p>If at the same moment your opponent has <strong>130 moves</strong>, you know for sure that a priori you have more probability of winning the game, because you have less moves left to finish. It&#8217;s simple, isn&#8217;t it?</p>
<p>We have an a priori way of knowing who has the advantage, and although the advantage is not exactly the <strong>probabilities of winning</strong> as we will see, but it gives us an idea of how we are doing.</p>
<p>If you are able to see who has this <strong>advantage</strong> even if it is in a visual way, you have a very valuable information to <strong>control the game</strong>. Can you imagine what for?</p>
<p>The <strong>Ludo strategy</strong> is very simple, when you have an advantage you have to <strong>keep it</strong>, and if you lose it you have to <strong>recover it</strong>, that&#8217;s why it&#8217;s important to know if you are really winning or losing, to know if there is an aggressive or more conservative strategy to be applied. And this depends on the risk!</p>
<p><a id="gestionarventaja" name="Ancla"></a></p>
<h3>2.1.Manage the advantage</h3>
<p>Imagine that you already have a big advantage and your opponent has all his chips on the left side of the board. In addition, you have a lagging chip that acts as a plug for his more advanced pieces. A common technique when your opponent sees that you have an advantage is to <strong>catch our chip</strong> with a bridge by bringing two chips closer together to catch it. But if you are aware of your ample advantage, you will see that it makes little sense to block their passage if for that we need to risk a checker. So the most intelligent thing is to <strong>move it</strong> to the other part of the board as soon as possible where it is safe, to continue with the <strong>same advantage</strong> but this time <strong>without risk</strong>.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-1.jpg"><img decoding="async" class="aligncenter wp-image-793 size-large" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-1-1024x613.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-1.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-1-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-1-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Imagine another situation where you have enough of an advantage and can choose to move <strong>the most advanced chip</strong> by priority or <strong>the chip that is unsafe</strong> to put to safety. If you analyze and see that your advantage is <strong>enough</strong> to win, what is the point of leaving a chip at risk despite being a considerable distance away?</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-2.jpg"><img decoding="async" class="aligncenter wp-image-794 size-large" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-2-1024x613.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-2-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-caso-2-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>If your opponent rolls a <strong>double</strong>, he can bring the chip closer to yours and even kill it on the next roll, but what is the probability of rolling a double?</p>
<p>Drawing a double is probably easier than you think, since with <strong>6 cases out of 36</strong> we have approximately a <strong>16.6%</strong> chance of getting it, or 1 sixth, that is to say that <strong>1 out of every 6</strong> times we make this decision we would be risking a game that we have practically won. This shows us the importance of having all the <strong>chips safe</strong>, especially in an advantageous situation. Any miscalculated detail can put the entire game at risk. So if you have an advantage, <strong>don&#8217;t risk</strong> it if it&#8217;s not necessary.</p>
<p class="p1"><span class="s2"><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DOUBLE.jpg"><img decoding="async" class="aligncenter wp-image-1113 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DOUBLE-1024x609.jpg" alt="" width="1024" height="609" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DOUBLE-1024x609.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DOUBLE-300x178.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DOUBLE-768x457.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DOUBLE.jpg 1580w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></span></p>
<p>Another risk situation to avoid is to secure our chip in the opponent&#8217;s<strong> starting box</strong> when he has one chip at the start and another at home, or both at home. Because having a chip outside <strong>is worth a 5 to kill us</strong> and that will happen almost half of the times he throws (<strong>41.67%</strong>), which we are not interested in.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-ficha-en-casa.jpg"><img decoding="async" class="aligncenter wp-image-767 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-ficha-en-casa.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-ficha-en-casa.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-ficha-en-casa-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-ficha-en-casa-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In conclusion, if you have an advantage big enough to win, you should take a <strong>conservative</strong> stance, that is, avoid risk or confrontation as much as possible by moving the necessary piece.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_STRATEGY.jpg"><img decoding="async" class="aligncenter wp-image-1115 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_STRATEGY-1024x567.jpg" alt="" width="1024" height="567" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_STRATEGY-1024x567.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_STRATEGY-300x166.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_STRATEGY-768x425.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_STRATEGY.jpg 1352w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><br />
<a id="prioridad" name="Ancla"></a></p>
<h2 class="p1"><b>3.Priority in Ludo</b></h2>
<p>As you have seen, there are situations where putting our chips in safety is more important than moving chips further ahead, so <strong>it doesn&#8217;t matter which chip we move?</strong></p>
<p>I imagine that you think that if you move a backward or a <strong>forward chip</strong> you are moving the same chips in global computation, then choosing one or the other does not seem to give us more or less advantage.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-que-ficha.jpg"><img decoding="async" class="aligncenter wp-image-772 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-que-ficha.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-que-ficha.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-que-ficha-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-que-ficha-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>But even if you don&#8217;t believe it, moving the <strong>most forward</strong> can give you more chances to win. Keep in mind that <strong>each goal adds up to 10</strong>, and if you always move the most advanced piece, you will arrive earlier with the same rolls, since you win that advantage earlier. Therefore, if you only have to reach the goal, always do it by <strong>priority</strong> and don&#8217;t move chips further back in a grouped way.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-tablero.jpg"><img decoding="async" class="aligncenter wp-image-773 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-tablero.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-tablero.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-tablero-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-tablero-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Moving all the <strong>chips in a group</strong> without a concept of priority is a <strong>mistake</strong> if it is not justified. Group movement involves moving the most backward chips and that moves your defense forward, allowing your opponent to move forward more easily and to <strong>pressure you</strong> with the advanced chips. Besides, when you move the chips in group, you advance very slowly with the chip in front of you and this does not allow you to reach the opponent quickly to <strong>press as soon as possible</strong>. In short, moving in a group does not make you lose the advantage as such, but it does have a strategic reason <strong>it takes away your possibilities of winning</strong>.</p>
<p>I guess you wonder if I am saying that the <strong>probability of winning</strong> and the advantage are not exactly the same, and indeed they are not.</p>
<p>In this other example you can see that our opponent is further away in <strong>moves</strong>, however our probability of winning is lower. You wonder why? If you look at it, you still have to put in <strong>4 chips</strong> and he is practically <strong>1 throw away</strong> from being able to reach the goal with just one. You need at least 2 rolls to put in all 4 chips in, but if you don&#8217;t <strong>roll a 1</strong>, you still don&#8217;t have a choice because you still have <strong>2 blocked chips</strong>.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-fichas-acumuladas.jpg"><img decoding="async" class="aligncenter wp-image-774 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-fichas-acumuladas.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-fichas-acumuladas.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-fichas-acumuladas-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-fichas-acumuladas-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Now you can understand that the probability of winning depends also on how <strong>we manage and position</strong> the chips apart from the advantage, and in this case it would have been better to put the chips further ahead in goal by priority before waiting so long.</p>
<p>Another technique to avoid accumulating <strong>chips in the finish</strong> line is to leave your chips when they reach a <strong>distance of 10</strong>, because if you manage to put one chip in the finish line zone, the others will go immediately. You simply need <strong>1 hit</strong> to place several chips, and that makes them less likely to pile up.</p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-distancia-10-meta.jpg"><img decoding="async" class="aligncenter wp-image-775 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-distancia-10-meta.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-distancia-10-meta.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-distancia-10-meta-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-distancia-10-meta-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In short, when there is no risk, we always move the <strong>most advanced chip</strong>, but what happens when there is?</p>
<p>In this example we have already played <strong>2 doubles in a row</strong> and we run the risk of roll a third one. If you do not have a chip to move in the goal zone that overrides the 3-double rule, you must choose which chip you put at risk in case the <strong>third double</strong> arrives. In this situation the decision that best minimizes the risk without taking into account the strategy is to move the <strong>most delayed</strong> chip, because in case of being eliminated it is the one that makes us lose less advantage. In this case we are choosing by priority the most delayed chip.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-3-dobles.jpg"><img decoding="async" class="aligncenter wp-image-770 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-3-dobles.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-3-dobles.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-3-dobles-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-3-dobles-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Although you may think that getting <strong>3 doubles</strong> is complicated, and you are not wrong, in fact you have <strong>0.46%</strong> chance of pulling 3 doubles before it happens, but when you have already pulled 2, getting a third one is exactly <strong>as easy as</strong> pulling the second one.</p>
<p>In short, the priority when there is <strong>risk</strong> works the other way around, and it is better to risk the most <strong>delayed chip</strong> whenever possible and strategically correct.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_CHIPS_STRATEGY.jpg"><img decoding="async" class="aligncenter wp-image-1116 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_CHIPS_STRATEGY-1024x555.jpg" alt="" width="1024" height="555" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_CHIPS_STRATEGY-1024x555.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_CHIPS_STRATEGY-300x163.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_CHIPS_STRATEGY-768x416.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_CHIPS_STRATEGY.jpg 1380w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>But if there <strong>is no risk</strong>, nor strategy, you will move and enter the goal with the <strong>most advanced chips</strong>, to avoid guys like minime to end up winning because you are absent-minded <img src="https://s.w.org/images/core/emoji/17.0.2/72x72/1f609.png" alt="😉" class="wp-smiley" style="height: 1em; max-height: 1em;" /></p>
<p>However, if your opponent is good, he will normally prevent you from doing this by <strong>forcing you to move</strong> the backward chips, so that he can move the front pieces to <strong>control the game</strong>. So be very careful!</p>
<p><a id="probabilidad" name="Ancla"></a></p>
<h2 class="p5">4.Ludo Probabilities</h2>
<p>You just saw how risk changes the priority to choose which chip to move, but how far should you move a tile so you don&#8217;t get killed? Is it better to be 2 boxes away, 4 or the same? What do you think?</p>
<p>&#8211;<strong>Out of shot</strong></p>
<p>As you know the <strong>maximum value</strong> of a die is 6, and that makes the sum of two dice a <strong>maximum of 12</strong>. So I recommend you to be at a distance greater than 12 if possible, because more than 12 boxes can not kill you directly.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-suma-maxima-2-dados.jpg"><img decoding="async" class="aligncenter wp-image-776 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-suma-maxima-2-dados.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-suma-maxima-2-dados.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-suma-maxima-2-dados-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-suma-maxima-2-dados-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>And what happens <strong>below 12</strong>, all distances are equally likely?</p>
<p>&#8211;<strong>On within range (distance less than 12)</strong></p>
<p>The truth is that it is not, because there are boxes that are within direct shot and others that are not. You&#8217;ll agree that at <strong>6 boxes</strong> or less, either die can kill you, since a die reaches 6.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-probabilidad-distancia-6.jpg"><img decoding="async" class="wp-image-782 size-full aligncenter" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-probabilidad-distancia-6.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-probabilidad-distancia-6.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-probabilidad-distancia-6-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-probabilidad-distancia-6-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In contrast to more than 6 the opponent needs the <strong>sum of the two dice</strong> to kill you, which is more difficult by probability. Therefore it is also better to be 7 boxes or more away from each other whenever possible.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/probabilidad-distancia-8.jpg"><img decoding="async" class="wp-image-784 size-full aligncenter" src="https://www.math4all.es/wp-content/uploads/2020/07/probabilidad-distancia-8.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/07/probabilidad-distancia-8.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/probabilidad-distancia-8-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/probabilidad-distancia-8-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<h4>&#8211;<strong>On direct within range (distance less than 7)</strong></h4>
<p>And what happens to less than 7?</p>
<p>Even at a distance <strong>less than 7</strong>, the probability is not the same either, because besides killing you <strong>directly</strong>, it can do so with the <strong>sum</strong> and not all sums are equally probable.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_5.jpg"><img decoding="async" class="aligncenter wp-image-1118 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_5-1024x609.jpg" alt="" width="1024" height="609" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_5-1024x609.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_5-300x178.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_5-768x457.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_5.jpg 1580w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Do it <strong>directly</strong> if it has the same probability, because taking any value off the die is just as likely, it is the same to take a 3 as a 6. But do you think it is just as likely to add a 3 as a 6?</p>
<p>As you can deduce we can <strong>add 3</strong> only in 2 different ways, 2-1 and 1-2, but we can add 6 in <strong>4 different ways</strong>, so the probability of adding changes and in this case is higher for 6.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_6.jpg"><img decoding="async" class="aligncenter wp-image-1119 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_6-1024x609.jpg" alt="" width="1024" height="609" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_6-1024x609.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_6-300x178.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_6-768x457.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_6.jpg 1580w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>But <strong>how do we calculate all the probability</strong>?</p>
<p><a id="probabilidadcomer" name="Ancla"></a></p>
<h3>4.1.Probability of Eating/be eaten</h3>
<p>Knowing the probability of being killed both with the sum, as with the first die or the second one in a direct way is the probability of the <strong>union of the 3 events</strong>:</p>
<ul>
<li class="p5"><em><em>A: &#8220;Let the value come out with dice 1</em></em><em>&#8220;</em></li>
<li class="p5"><em>B: &#8220;Let the value come out with dice 2&#8221;</em></li>
<li class="p5"><em>C: &#8220;Let the value come out with the sum of dice 1 and 2&#8221;</em></li>
</ul>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_UNION_PROBABILITY.jpg"><img decoding="async" class="aligncenter wp-image-1120 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_UNION_PROBABILITY-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_UNION_PROBABILITY.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_UNION_PROBABILITY-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_UNION_PROBABILITY-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a>We apply the <strong>probability for the union</strong> of 3 events::</p>
<p class="p5"><strong>P(AUBUC)=P(A)+P(B)+P(C)-P(A∩B)-P(A∩C)-P(B∩C)+P(A∩B∩C)</strong></p>
<p>and this is simplified as <strong>P(A)+P(B)+P(C)-P(A∩B)</strong> because P(A∩C), P(B∩C) y P(A∩B∩C) are 0 since the sum of two dice can never give the same value as one of the two. (when talking about events A and B, A and C mutually exclusive)</p>
<p>But since I know that it doesn&#8217;t have to interest you, I&#8217;ll show you the <strong>results</strong> directly:</p>
<h4>4.1.1.Probability regardless of the next die roll<a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE.jpg"><img decoding="async" class="aligncenter wp-image-1123 size-large" title="P(ser comido a distancia 0)=0% P(ser comido a distancia 1)=30,56% P(ser comido a distancia 2)=33,33% p(ser comido a distancia 3)=36,11% p(ser comido a distancia 4)=38,89% p(ser comido a distancia 5)=41,67% p(ser comido a distancia 6)=44,44% p(ser comido a distancia 7)=16,67% p(ser comido a distancia 8)=13,89% p(ser comido a distancia 9)=11,11% p(ser comido a distancia 10)=8,33% p(ser comido a distancia 11)=5,56% p(ser comido a distancia 12)=2,78%" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE-1024x577.jpg" alt="" width="1024" height="577" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE-1024x577.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE-768x433.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE.jpg 1820w" sizes="(max-width: 1024px) 100vw, 1024px" /></a>Which can be represented by the following graph:</h4>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART_2.jpg"><img decoding="async" class="aligncenter wp-image-1129 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART_2-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART_2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART_2-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART_2-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>As you can see, being <strong>6 boxes</strong> away is where you have the highest probability of being killed, thanks to the fact that, as you have already seen, it is easier to obtain with the sum, and the further you go from 6 the <strong>more difficult</strong> it becomes, as we deduced.</p>
<p>On the other hand, the closer you get to 1, the <strong>more difficult</strong> it is to be killed. This is the best distance at first, because if you think about it, 1 cannot be obtained as the sum of two dice. Was it what you imagined?</p>
<p>The problem is that at such a close distance there is more risk of staying<strong> </strong>within range on the next roll and this calculation is not taking it into account. For that reason we must also add the probability that the opponent has of killing us in <strong>the next roll</strong> if we are in that position.</p>
<h4>4.1.2.Probability taking into account the next die roll</h4>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE_2.jpg"><img decoding="async" class="aligncenter wp-image-1125 size-large" title="p(ser comido a distancia 0)=0% + 23,98% = 23,98% p(ser comido a distancia 1)=30,56% + 19,28% = 49,84% P(ser comido a distancia 2)=33,33% + 14,73% = 48,06% P(ser comido a distancia 3)=36,11% + 10,56% = 46,67% P(ser comido a distancia 4)=38,89% + 7,02% = 45,91% P(ser comido a distancia 5)=41,67% + 4,32% = 45,99% P(ser comido a distancia 6)=44,44% + 2,7% = 47,14% p(ser comido a distancia 7)=16,67% + 1,54% = 18,21% p(ser comido a distancia 8)=13,89% + 0,77% = 14,66% p(ser comido a distancia 9)=11,11% + 0,31% = 11,42% p(ser comido a distancia 10)=8,33% + 0,08% = 8,41% p(ser comido a distancia 11)=5,56% + 0% = 5,56% p(ser comido a distancia 12)=2,78% + 0% = 2,78%" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE_2-1024x577.jpg" alt="" width="1024" height="577" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE_2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE_2-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_DISTANCE_2-768x433.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p5">Updating also the graph:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART.jpg"><img decoding="async" class="aligncenter wp-image-1128 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART-1024x604.jpg" alt="" width="1024" height="604" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART-300x177.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_DICE_CHART-768x453.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>And as you can see, things have changed a lot! The first positions have much <strong>more risk</strong> because of the next roll, and in the end the worst is the <strong>first position</strong>, because the probability of being killed in the next roll decompensates the balance of the other probability. And the best option is to put the chip at a <strong>distance of 4 or 5</strong>, which positions you further away and without as much probability of dying as at 6.</p>
<p>Did you never think that the Ludo could give for so much truth?</p>
<p><a id="probabilidadsalvarnos" name="Ancla"></a></p>
<h3 class="p5">4.2.Probability to Save us</h3>
<p>Now you know where it&#8217;s best to stand so you don&#8217;t get killed. But when you&#8217;re already <strong>far away</strong>, you have to think more about where to place your unprotected chip in order to <strong>reach an insurance</strong> with greater chances.</p>
<p>&nbsp;</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-seguro-lejos.jpg"><img decoding="async" class="aligncenter wp-image-789 size-large" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-seguro-lejos-1024x575.jpg" alt="" width="1024" height="575" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-seguro-lejos.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-seguro-lejos-300x168.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-seguro-lejos-768x431.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In this example we see that by placing the chip in <strong>box 15</strong> you choose to save it with a 1 or by adding 7 with the two dice to get to 22, which is <strong>less likely</strong>. On the other hand, if you have the possibility of placing it <strong>in the box 16</strong>, you save the chip with a 1 and a 6, so you have a much <strong>greater chances</strong> of placing your chip in a safe box on the next roll.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-zonas-para-alcanzar-2-seguros.jpg"><img decoding="async" class="aligncenter wp-image-788 size-full" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-zonas-para-alcanzar-2-seguros.jpg" alt="" width="1024" height="613" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-zonas-para-alcanzar-2-seguros.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-zonas-para-alcanzar-2-seguros-300x180.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-zonas-para-alcanzar-2-seguros-768x460.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>and that&#8217;s because these <strong>zones</strong> are always better for getting you to safety.</p>
<p><a id="probabilidadataque" name="Ancla"></a></p>
<h3 class="p5">4.3.Probability in the Attack</h3>
<p>As you have seen, you can <strong>avoid unnecessary risks</strong> to avoid being eaten, but how do you manage an attack?</p>
<p>Thinking about attacking is using the same information from the table but this time in reverse, to position your chip behind, but this time looking for the <strong>highest probability</strong> and not the lowest. This is to place you from behind at <strong>6 distance</strong> if you are going to repeat a turn or know that your opponent cannot move that piece. And in case you have the ability to escape then <strong>as close as possible</strong>, which is 1 if you are behind or 0 if you are in safety, because that is where you have more probability of eating him in the next roll if he decides to move it, as the table indicates.</p>
<p>But be careful not to run out of room for manoeuvre with the other chips, because being so close you run the risk of <strong>going over</strong> with the dice and being the one who ends up shooting, either with that checker or with the one you decide to move. So if that risk exists, the best according to mathematics is to be at a prudential distance of <strong>6 boxes</strong> where you get an optimal <strong>benefit-risk ratio</strong> if your intention is to kill.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_OPTIMAL_DISTANCE.jpg"><img decoding="async" class="aligncenter wp-image-1126 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_OPTIMAL_DISTANCE-1024x609.jpg" alt="" width="1024" height="609" srcset="https://www.math4all.es/wp-content/uploads/2020/11/LUDO_OPTIMAL_DISTANCE-1024x609.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_OPTIMAL_DISTANCE-300x178.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_OPTIMAL_DISTANCE-768x457.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/LUDO_OPTIMAL_DISTANCE.jpg 1896w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<h2>5.Online Ludo</h2>
<p>In the next video you will be able to see the <strong>attack strategies</strong> that exist in Ludo, and how to manage the barriers, but in order not to make the wait too long you can try these strategies by playing Ludo online on different<strong> ludo games:</strong> <strong>Ludo Star</strong>, <strong>Ludo mundijuegos</strong> or <strong>Ludo Playspace</strong>.</p>
<h4><strong>Ludo Star</strong></h4>
<h4><strong><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-star.jpg"><img decoding="async" class="aligncenter wp-image-840" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-star-300x169.jpg" alt="" width="684" height="385" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-star-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-star-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-star-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-star.jpg 1280w" sizes="(max-width: 684px) 100vw, 684px" /></a></strong></h4>
<p><strong>Pros</strong>: Very big community, it&#8217;s free.</p>
<p><strong>Cons</strong>: Unobtrusive design, no turbo game. Few variants.</p>
<h4><strong>Ludo Mundijuegos</strong></h4>
<p>&nbsp;</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-mundijuegos.jpg"><img decoding="async" class="aligncenter wp-image-839" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-mundijuegos-300x169.jpg" alt="" width="684" height="385" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-mundijuegos-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-mundijuegos-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-mundijuegos-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-mundijuegos.jpg 1280w" sizes="(max-width: 684px) 100vw, 684px" /></a></p>
<p><strong>Pros</strong>: You can also play on the website, it&#8217;s free.</p>
<p><strong>Cons</strong>: Need flash player for web, small community.</p>
<p><a id="jugar" name="Ancla"></a></p>
<h4><strong>Ludo Playspace</strong></h4>
<div id="attachment_841" style="width: 694px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2020/07/parchis-playspace.jpg"><img decoding="async" aria-describedby="caption-attachment-841" class="wp-image-841" src="https://www.math4all.es/wp-content/uploads/2020/07/parchis-playspace-300x169.jpg" alt="" width="684" height="385" srcset="https://www.math4all.es/wp-content/uploads/2020/07/parchis-playspace-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-playspace-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-playspace-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/07/parchis-playspace.jpg 1280w" sizes="(max-width: 684px) 100vw, 684px" /></a><p id="caption-attachment-841" class="wp-caption-text">Playspace Ludo (for me the best)</p></div>
<p><center><a class="btn btn-success btn-lg active" style="font-size: 20px; font-family: Luckiest Guy,helvetica neue;" href="http://www.math4all.es/parchis/jugarparchis.php"><i class="glyphicon glyphicon-play"></i> ¡play LUDO!</a></center><br />
<strong>Pros</strong>: Puedes jugar en facebook, muchas modalidades, gran comunidad, gran personalización, es gratis.</p>
<p><strong>Cons</strong>: Some customizations are chargeable.</p>
<p>I usually play <strong>playspace</strong> ludo because I like its design better, but above all because it allows to <strong>play turbo games</strong>, which is what I like, so I leave you a <strong>link</strong> in case you want to play it</p>
<p>I hope you find it practical and if you want more videos like this one of live play, leave a comment at the end of the article. Greetings.</p>
<img src="https://www.math4all.es/wp-content/uploads/2020/11/ludo-maths.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/the-mathematics-of-ludo-rules-probabilities-and-strategy/" data-pin-media="https://www.math4all.es/wp-content/uploads/2020/11/ludo-maths.jpg" data-pin-description="The Mathematics of Ludo &#8211; Rules, Probabilities and Strategy"/><p>The post <a href="https://www.math4all.es/the-mathematics-of-ludo-rules-probabilities-and-strategy/">The Mathematics of Ludo – Rules, Probabilities and Strategy</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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		<title>The Mathematics of Poker &#8211; Range, Position and EV</title>
		<link>https://www.math4all.es/the-mathematics-of-poker-range-position-and-ev/</link>
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		<dc:creator><![CDATA[math4all]]></dc:creator>
		<pubDate>Mon, 02 Nov 2020 16:06:05 +0000</pubDate>
				<category><![CDATA[The Mathematics of games]]></category>
		<category><![CDATA[The mathematics of the casino]]></category>
		<category><![CDATA[casino games]]></category>
		<category><![CDATA[ev allin]]></category>
		<category><![CDATA[ev calculator]]></category>
		<category><![CDATA[ev call]]></category>
		<category><![CDATA[game]]></category>
		<category><![CDATA[nash equilibrium]]></category>
		<category><![CDATA[poker]]></category>
		<category><![CDATA[poker ev]]></category>
		<category><![CDATA[poker heads up]]></category>
		<category><![CDATA[poker math]]></category>
		<category><![CDATA[poker position]]></category>
		<category><![CDATA[poker range]]></category>
		<category><![CDATA[poker strategy]]></category>
		<category><![CDATA[poker tracker]]></category>
		<category><![CDATA[ranges chart]]></category>
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					<description><![CDATA[<p>In this video I analyze other more technical aspects of Poker such as range, position or EV calculation. 1.Range 2.Poker&#8230;</p>
<p>The post <a href="https://www.math4all.es/the-mathematics-of-poker-range-position-and-ev/">The Mathematics of Poker – Range, Position and EV</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="Las MATEMÁTICAS del PÓKER &#x2663;&#x2665; (Rangos y Posición)" width="1170" height="658" src="https://www.youtube.com/embed/-ZP2eJB-GAE?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe></p>
<p>In this video I analyze other more technical aspects of Poker such as range, position or EV calculation.</p>
<p><a href="#rango">1.Range</a><br />
<a href="#factores">2.Poker Factors</a><br />
<a href="#posicion">3.Position</a><br />
<a href="#ev">4.Poker EV</a><br />
<a href="#evcall"> 4.1.Call EV</a><br />
<a href="#calculadora"> 4.1.1.Call EV calculator</a><br />
<a href="#evallin"> 4.2.All-in EV</a><br />
<a href="#otros">5.Other Factors</a><br />
<a href="#headsup">6.Heads up Strategy</a><br />
<a href="#equilibrio">7.Nash Equilibrium</a></p>
<p>In the <strong>last poker math video</strong> we saw a summary of the <strong>rules of the game</strong>, and what the probability of linking each hand is.</p>
<p>We then looked at the <strong>outs</strong> and <strong>odds</strong>, which are the probabilities used in poker, and learned how to calculate them and refer to our <strong>table of odds and outs</strong>.</p>
<p>We also saw how we can approximate these values mentally without the need for tables, and finally we learned how to use a <strong>poker calculator</strong>, and saw how the odds we calculate for our projects relate to our real probabilities.</p>
<p>In any case, <strong>I didn&#8217;t answer</strong> how to calculate our odds when we don&#8217;t know the cards of our opponents, and neither because a hand that has little chance of being tied can be profitable. That&#8217;s why if you were left with the doubt or are simply curious to answer these questions we go with the second part of the poker math.</p>
<p><a id="rango" name="Ancla"></a></p>
<h2 class="p1">1.Range</h2>
<p>As we saw in the previous video, Poker is a game of <strong>incomplete information</strong>, since we do not know the cards of our rivals. Consequently what we can calculate comes from an estimation of what hands they might have, based on the projects they can complete with the community cards. But also knowing <strong>which hands they usually play frequently</strong> or which ones they prefer to discard in certain phases of the game.</p>
<p>Some players tend to play a very small number of hands, because they feel safer playing like this. Others, more <strong>recreational</strong>, prefer to use a larger number. Have you ever wondered what kind of player you are?</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_PLAYING_STYLES.jpg"><img decoding="async" class="aligncenter wp-image-1093 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_PLAYING_STYLES-1024x597.jpg" alt="" width="1024" height="597" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_PLAYING_STYLES-1024x597.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_PLAYING_STYLES-300x175.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_PLAYING_STYLES-768x448.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_PLAYING_STYLES.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>The reality is that we all have a <strong>filter</strong> to choose which hands we play and which we don&#8217;t, which is more or less big depending on our <strong>style of play</strong>.</p>
<p>Well, the group of cards that make up that estimate is known as <strong>range</strong> and is represented at the software level with a <strong>matrix</strong> where you can select which cards make it up.</p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2020/01/Poker-rango.jpg"><img decoding="async" class="aligncenter wp-image-710 size-large" src="https://www.math4all.es/wp-content/uploads/2020/01/Poker-rango-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/01/Poker-rango.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-rango-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-rango-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>If you want to calculate your hands <strong>p</strong><b>robability</b> taking it into account, what you should do is calculate the probabilities of your cards against that range you estimate, eliminating all those hands that are impossible or very unlikely that the opponent is playing, given his profile or the game situation. This way you will get a calculation as <strong>close as possible to the real</strong> one with the information you have. That&#8217;s why the better the estimation you make, the more accurate will be the calculation of your probability, and that&#8217;s why it&#8217;s important to acquire a technical knowledge about what ranks our opponents may have. But it is also important to know <strong>our own rank</strong>.</p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER-RANGES.jpg"><img decoding="async" class="aligncenter wp-image-1094 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER-RANGES-1024x558.jpg" alt="" width="1024" height="558" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER-RANGES-1024x558.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER-RANGES-300x164.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER-RANGES-768x419.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER-RANGES.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>On a practical level, I recommend you a <strong>Tracker Software</strong> to do this, because besides calculating probabilities with ranks it is able to <strong>analyze the game</strong> of your opponents and offer you <strong>personalized statistics</strong>. It does this by saving the data of the games you are playing and in this way you can know what number of hands you play voluntarily, the percentage of times you have increased in the preflop, or any variable you consider important. A player&#8217;s statistics can always give us a <strong>long term advantage</strong> in making decisions. Although keep in mind that this is only for <strong>Online Poker</strong>.</p>
<p>&nbsp;</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Tracker.png"><img decoding="async" class="aligncenter wp-image-743 size-large" src="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Tracker-1024x576.png" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Tracker-1024x576.png 1024w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Tracker-300x169.png 300w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Tracker-768x432.png 768w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Tracker.png 1280w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p style="text-align: center;"><a href="https://pokertracker.com/?a_aid=math4all" rel="nofollow">DOWNLOAD TRACKER SOFTWARE</a></p>
<p>The truth is that this software is very, very cool. And now that you&#8217;ve seen how to calculate ranks, what else do you think you should take into account to make a decision?</p>
<p><a id="factores" name="Ancla"></a></p>
<h2 class="p1">2.Poker Factors</h2>
<p>Deciding when to match a bet, raise it with a certain amount or throw away our cards is what determines how <strong>good players</strong> we are in the long run, but choosing one or the other depends on many <strong>factors</strong>, so I have made a <strong>list</strong> for you to look at:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_FACTORS.jpg"><img decoding="async" class="aligncenter wp-image-1096 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_FACTORS-1024x597.jpg" alt="" width="1024" height="597" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_FACTORS-1024x597.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_FACTORS-300x175.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_FACTORS-768x448.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_FACTORS.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Some rely solely on <strong>our information</strong>, and others rely on information provided to us by <strong>our competitors</strong>. But then there are others that depend solely on the game such as the number of players.  The more players there are at a table, the more likely it is that one of them will have a stronger hand than ours, as we face more players on average per round. That is why we must adapt our game with a more <strong>exclusive range</strong>, and also reduce the range we estimate of <strong>our opponents</strong>, because it is less likely to win with the same hand when there are more possible better hands.</p>
<p><a id="posicion" name="Ancla"></a></p>
<h2>3.Position</h2>
<p>As we see there are <strong>a lot of factors</strong>, but not all of them have the same importance.</p>
<p>One of the factors considered <strong>most important</strong> in poker is the <strong>position</strong>, as it can be used to get <strong>extra information</strong> from your opponents, or also to <strong>lie</strong>. If you play in an <strong>early position</strong>, you have no information about what your opponents are going to do, and since you can&#8217;t discard any player you are forced to play a <strong>shorter range</strong> of hands. On the other hand, if you play in a <strong>late position</strong>, you get information about the bets that are made or the players that withdraw, so you have certain information that can give you an advantage and <strong>more chances to win</strong>. That makes each player adopt a different <strong>game and range depending on their position</strong> and that logically offers different results. But <strong>how much of different?</strong> Well, <strong>9 players</strong> in over <strong>1000 hands</strong> have been analyzed to see which position has the most advantage, and this is what has been observed:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS.jpg"><img decoding="async" class="aligncenter wp-image-1097 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS-1024x597.jpg" alt="" width="1024" height="597" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS-1024x597.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS-300x175.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS-768x448.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p1">As we had already deduced the <strong>last positions</strong> are usually the ones that offer us more profit, but curiously and against what I had said, the <strong>first position</strong> also has a quite high percentage of victories. This is explained by the fact that <strong>many players use early positions to bet or raise</strong> as if they had a strong hand, whether they have it or not.</p>
<p>If we make a <strong>summary</strong>, we can see that <strong>a late position is the one that benefits you the most</strong>, but the early position can also serve to bluff you.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS_2.jpg"><img decoding="async" class="aligncenter wp-image-1098 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS_2-1024x597.jpg" alt="" width="1024" height="597" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS_2-1024x597.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS_2-300x175.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS_2-768x448.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_POSITION_STATS_2.jpg 1200w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Playing good poker doesn&#8217;t seem to be easy, you have to be good enough to overcome the <strong>losses of the game</strong> and even losses on commissions. And with all these factors our decisions depend on many things. So, <strong>is there any mathematical formula to decide what action we take?</strong></p>
<p><a id="ev" name="Ancla"></a></p>
<h2>4.Poker EV</h2>
<p>Although <strong>position</strong> is a very important factor, its influence decreases as <strong>fewer players are left</strong>, since by reducing the number of rivals we can consequently steal fewer chips. So when there are only <strong>2 players</strong> left at the table, which is usually the case, the position is no longer a factor to take into account and we can establish a <strong>mathematical method</strong> to know what decision we should take, based only on the <strong>size of the pot</strong>, the <strong>size of the bets</strong> and our <strong>probabilities</strong>.</p>
<p>For that a new concept is defined called <strong>EV</strong> that comes from (Expected value) and is based on the concept of <strong>long term gain</strong>. To know if a play suits you or not, the most intelligent thing is to calculate what <strong>average gain</strong> we would obtain if we repeated many times that same play. And that is precisely what calculates the EV, subtracting the probability of winning by the profit less the probability of losing by the loss. In this way you can know that <strong>profitability</strong> has for example to <strong>equal a bet</strong>.</p>
<p><a id="evcall" name="Ancla"></a></p>
<h3>4.1.Call EV</h3>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_CALL.jpg"><img decoding="async" class="aligncenter wp-image-1103 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_CALL-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_CALL-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_CALL-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_CALL-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_CALL.jpg 1919w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>When the <strong>EV is negative</strong>, equalizing or making a call <strong>is not profitable</strong>, but if it is <strong>positive</strong>, it is a priori <strong>profitable in the long term</strong> as it is higher than the EV of folding, which is 0. Note that it <strong>is also possible to have a positive EV even though you are more likely to lose</strong>, if the benefit outweighs the risk. Since it can be more profitable to win few times with much money, than many times with little money. This answers the question of why it can be convenient to bet with a <strong>small probability</strong>. I have included this <strong>EV call calculator</strong> in case you are interested in calculating the expected value of your possible call.</p>
<p><a id="calculadora" name="Ancla"></a><br />
<a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALL_EV_CALCULATOR.jpg"><img decoding="async" class="aligncenter wp-image-1105 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALL_EV_CALCULATOR-1024x592.jpg" alt="" width="1024" height="592" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALL_EV_CALCULATOR.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALL_EV_CALCULATOR-300x173.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALL_EV_CALCULATOR-768x444.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p style="text-align: center;"><a href="https://www.math4all.es/pokerev/ev-calculator.html">CALL EV CALCULATOR</a></p>
<p>To calculate the EV mentally you can use the known <strong>pot odds</strong>, which are no more than the proportion between what we win against the bet we have to make, and which are compared with the odds (which are the times we lose for each time we win). As both are calculated with the same proportion, you only have to <strong>compare odds and pot odds to know if it is profitable to go to a raise or not</strong>, so this is a practical way to calculate the EV for the <strong>live game</strong>.</p>
<p><a id="evallin" name="Ancla"></a></p>
<h3>4.2.All-In EV</h3>
<p>To know instead if doing an <strong>all-in</strong> is profitable, the thing changes, since it is important to take into account the <strong>response</strong> of our rival, which can be to <strong>go or to fold</strong> based on the amount that we bet, so the EV in this case also includes the <strong>estimated probability</strong> that one action or the other happens, and can be calculated in this way.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_ALLIN.jpg"><img decoding="async" class="aligncenter wp-image-1099 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_ALLIN-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_ALLIN-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_ALLIN-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_ALLIN-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_EV_ALLIN.jpg 1919w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Which is basically the <strong>Fold EV</strong> plus the <strong>Call EV</strong></p>
<p>There are <strong>many formulas</strong> to determine our actions, such as the minimum bet you have to make to remove the odds necessary to the rival to pursue their project and many others that you can search on the Internet:</p>
<p class="p1"><strong>AM = (Outs x P)/(50 – 2 x Outs)</strong></p>
<p><a id="otros" name="Ancla"></a></p>
<h2>5.Other factors</h2>
<p>The truth is that although these formulas are useful, there are <strong>other factors</strong> that can also influence when making a decision:</p>
<p>-One of them is the <strong>confrontation cost</strong>. If you have a profitable move but few chips and you don&#8217;t have the possibility of re-buying or you simply play in the short term, it may not be smart to match or to do all-in if doing so may have a confrontation cost of <strong>losing all or a large amount of chips</strong>, compared to not confronting (especially when our probability of winning is low), so in modalities such as <strong>tournament</strong>, it is also important to evaluate our decisions based on the chips we have and those of our rivals. A mathematical model that takes this into account is the <strong>ICM</strong> model, which is used especially for the bubble phase of tournaments.</p>
<p>-Another factor considered the most important to change those decisions is the <strong>knowledge of your rivals</strong>, which can serve to adapt and exploit their meta-game as much as possible, increasing or decreasing the bet size, the amount we have to pay to certain rivals, or simply to avoid or encourage certain clashes, which can give us an advantage.</p>
<p>It doesn&#8217;t seem easy to decide with so many things, so the experience also plays a fundamental role to be a good player, although <strong>are we really able to learn all the situations of Poker?</strong></p>
<p><a id="headsup" name="Ancla"></a></p>
<h2>6.Heads up Strategy</h2>
<p>Only in the Texas Holdem for 2 players there are <strong>10^161 different decisions</strong> possible, which <strong>is more than all the atoms that form the universe</strong>. This makes it impossible for a human to learn which is the most advantageous decision for all that immensity of situations. Although, as you have surely thought, there are programs where you can consult how to act in most of the possible cases to improve your strategy. These <strong>process millions of hands</strong> based on exploiting themselves, and offer us the <strong>most profitable actions at the statistical level</strong>, which you can use to adapt to your game and exploit your opponents, so currently to be a good player is also important to know how to interpret statistical data correctly.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/01/poker-snowie.jpg"><img decoding="async" class="aligncenter wp-image-741 size-full" src="https://www.math4all.es/wp-content/uploads/2020/01/poker-snowie.jpg" alt="" width="835" height="630" srcset="https://www.math4all.es/wp-content/uploads/2020/01/poker-snowie.jpg 835w, https://www.math4all.es/wp-content/uploads/2020/01/poker-snowie-300x226.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/01/poker-snowie-768x579.jpg 768w" sizes="(max-width: 835px) 100vw, 835px" /></a></p>
<p style="text-align: center;"><a href="http://www.pokersnowie.com/379-2-3-10.html" rel="nofollow">DOWNLOAD STATISTICAL CONSULTATION PROGRAM</a></p>
<p>One of the most important decisions that we must take in Poker is how to play our hand in the <strong>heads up</strong>, since that conditions us a lot during the following betting rounds, that&#8217;s why it&#8217;s important to have a <strong>previous well defined strategy</strong> that you can consult. That&#8217;s why I have included a series of <strong>tables</strong> based on the study of these programs made by a <strong>professional player</strong> (Zeros) and that could be helpful if you have them at hand during the game. In them you have the <strong>action plan</strong> based on the <strong>optimal return</strong> for any particular situation in the heads up, which can be very useful for online poker, so if you are interested, you have a link in the description, along with the link to these programs that I have commented in case you want to consult another situation. (And it&#8217;s not that I want you to play Poker but if you do, do it well).</p>
<p class="p1" style="text-align: center;"><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HEADS_UP_STRATEGY.jpg"><img decoding="async" class="aligncenter wp-image-1108 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HEADS_UP_STRATEGY-1024x581.jpg" alt="" width="1024" height="581" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HEADS_UP_STRATEGY.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_HEADS_UP_STRATEGY-300x170.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_HEADS_UP_STRATEGY-768x436.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><a href="http://math4all.es/poker/POKER_HEADS_UP_TABLE_STRATEGY(BEGINNERS_INTERMEDIATE).xlsx">DOWNLOAD HEADS UP TABLE</a></p>
<p>Now that you have seen that <strong>Poker is a very mathematical game</strong>, the most important question of all arises, <strong>is it possible to play Poker in a perfect way?</strong> Let&#8217;s see it:</p>
<p><a id="equilibrio" name="Ancla"></a></p>
<h2>7.Nash Equilibrium</h2>
<p>In 1951 a certain <strong>John Forbes</strong> Nash better known as John Nash elaborated a doctoral thesis demonstrating that any game with a finite number of mixed strategies has at least a <strong>Nash equilibrium</strong>. This balance is given when all the players have chosen in <strong>the best possible way</strong> and none of them can improve their strategy by much change it, as long as the others maintain theirs. This balance ensures that in a zero-sum game like poker, <strong>we will never lose in the long run</strong>.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-abstraccion.jpg"><img decoding="async" class="aligncenter wp-image-702 size-large" src="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-abstraccion-1024x578.jpg" alt="" width="1024" height="578" srcset="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-abstraccion-1024x578.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-abstraccion-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-abstraccion-768x433.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>How is this possible?</p>
<p>The search for this balance is not simple, because we must find an <strong>optimal combination</strong> that satisfies all players among many possible, which added to the amount of decisions that can be made makes it a <strong>computer challenge</strong>. Finding a balance in the heads up is less difficult but in the River it is much more complicated. (To understand this, try to imagine all the situations that can occur in this phase of the game)</p>
<p>As there are so many possible casuistry, <strong>it&#8217;s too expensive to compute a strategy for each hand or bet size</strong>, so scientists have chosen to abstract the full game to a more <strong>simplified version</strong> in terms of values and bet sizes, which allows the search for balance more efficiently, and that approximates in a very precise way to the full game through a set of <strong>algorithms</strong>. Using intelligently the calculation of this balance, they have built an artificial intelligence that plays poker for two players in a practically perfect way.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-algoritmo.jpg"><img decoding="async" class="aligncenter wp-image-703 size-large" src="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-algoritmo-1024x577.jpg" alt="" width="1024" height="577" srcset="https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-algoritmo-1024x577.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-algoritmo-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/01/Poker-Libratus-algoritmo-768x433.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In 2017 the <strong>Libratus</strong> software managed to beat the <strong>best world professional poker players</strong> for the first time with a significant statistical advantage, which is a historical fact for games of incomplete information. And all this thanks to mathematics.</p>
<p>Luckily or unfortunately <strong>Nash&#8217;s equilibrium</strong> is only possible for <strong>2 players</strong>, since it is possible to lose in long term expectation in Poker of 3 players or more, especially if one group of them gets together to eliminate another. Even if a balance is calculated for all three players, that combination does not have to be a Nash balance in itself. Consequently, if you play <strong>online poker with more than two people</strong>, you don&#8217;t have to worry for the moment, because their future is assured, although, who knows until when?</p>
<img src="https://www.math4all.es/wp-content/uploads/2020/11/poker-maths.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/the-mathematics-of-poker-range-position-and-ev/" data-pin-media="https://www.math4all.es/wp-content/uploads/2020/11/poker-maths.jpg" data-pin-description="The Mathematics of Poker &#8211; Range, Position and EV"/><p>The post <a href="https://www.math4all.es/the-mathematics-of-poker-range-position-and-ev/">The Mathematics of Poker – Range, Position and EV</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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		<title>The Mathematics of Poker &#8211; Odds and Outs</title>
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		<pubDate>Mon, 02 Nov 2020 12:28:38 +0000</pubDate>
				<category><![CDATA[The Mathematics of games]]></category>
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					<description><![CDATA[<p>Do you think math is important in poker? In this article I will analyze poker on a mathematical level so&#8230;</p>
<p>The post <a href="https://www.math4all.es/the-mathematics-of-poker-odds-and-outs/">The Mathematics of Poker – Odds and Outs</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="Las MATEMÁTICAS del PÓKER &#x2663;&#x2665;" width="1170" height="658" src="https://www.youtube.com/embed/tt7vdG6jP3g?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe></p>
<p class="p1">Do you think math is important in poker? In this article I will analyze poker on a mathematical level so that you know how to calculate your probabilities, analyze your actions, or how mathematics can make you a better player.</p>
<p><a href="#reglas">1.Texas Holdem Poker Rules</a><br />
<a href="#manos">2.Texas Holdem Poker Hands</a><br />
<a href="#acciones">3.Texas Holdem Poker Actions</a><br />
<a href="#probabilidades">4.Texas Holdem Poker Probabilities</a><br />
<a href="#informacion">5.Poker as a game of incomplete and imperfect information</a><br />
<a href="#oddsouts">6.Odds and Outs</a><br />
<a href="#calculadora">7.Texas Holdem Poker Calculator</a></p>
<p><a id="reglas" name="Ancla"></a></p>
<h2 class="p1">1.Texas Holdem Poker Rules</h2>
<p>Texas hold&#8217;em unlimited poker is a French card game with up to <strong>23 players simultaneously</strong>, where the objective is to eliminate all players from the table, to win all their chips, or simply to win as much money as possible in cash mode. For this, 2 non-visible cards are dealt to each player, and they will decide in turns and based on their cards, if they want to <strong>bet, check, fold, call or raise</strong>.</p>
<p>Once all players have made a decision, the <strong>dealer</strong> draws <strong>3 visible flop cards</strong>, which are used by each player to bind their cards in order to get <strong>the best possible hand</strong>, and proceeds to a new <strong>round of betting</strong>.</p>
<p>As the dealer uncovers a new card, the same procedure is repeated until finally <strong>5 cards are uncovered</strong>. Finally, the players still at the table show their cards and the winner of the pot is the player with the best hand. The players must use their knowledge and their <strong>strategy</strong> to make the best decisions that will allow them to win.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2019/07/mesa-de-poker.jpg"><img decoding="async" class="aligncenter wp-image-504 size-large" src="https://www.math4all.es/wp-content/uploads/2019/07/mesa-de-poker-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2019/07/mesa-de-poker.jpg 1024w, https://www.math4all.es/wp-content/uploads/2019/07/mesa-de-poker-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2019/07/mesa-de-poker-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><br />
<a id="manos" name="Ancla"></a></p>
<h2 class="p1">2.Texas Holdem Poker Hands</h2>
<ul>
<li><strong>Pair</strong>: First of all, we have the pair, which is the most basic hand in poker formed by <strong>2 cards</strong> of different suits that have the <strong>same value</strong>. If both players have a pair, the one with the highest value wins, and if that value is the same, the player with the highest card wins. The pair is the least valuable hand that can be linked.</li>
<li><strong>Double Pair</strong>: The double pair is composed of <strong>2 different pairs</strong> together with any other card. If both players have a double pair, the hand with the highest pair wins, and in case of a tie, the unmatched card with the highest value wins</li>
<li><strong>Three of a kind</strong>: As its name indicates, the trio is <strong>3 cards</strong> of different suits with the <strong>same value</strong>, together with 2 cards of any kind. In case both players have a trio, the one with the highest value wins. Note that in this modality of Poker there cannot be 2 trios of the same value, since there are only 4 possible suits and with one there are already 3.</li>
<li><strong>Straight</strong>: The Straight is made up of <strong>5 consecutive cards</strong> of the same suit or different suits, so all the values can be ordered one by one from lower to higher. Depending on the number of cards you start with, you will have a <strong>higher</strong> or <strong>lower</strong> straight, so in case of a tie, the one with higher values will always win.</li>
<li><b>Flush</b>: There are <strong>5 cards of the same suit</strong> regardless of their value. In case of a draw, the flush with the highest card wins.<br />
Full/Full house: Gather 3 cards of the same value and 2 cards of another, or what is the same, a trio and a pair simultaneously. When two players have a full house, the value of the highest trio wins.</li>
<li><b>Four of kind</b>: It&#8217;s composed of a quartet of <strong>4 cards</strong> of different suits that have the <strong>same value</strong>, together with any card. In case of a draw the value of that card decides the winner. Although this is the most popular hand, it is not the most valuable.</li>
<li><strong>Straight Flush</strong>: It&#8217;s a straight flush made up of all the lower cards of the same suit, which makes it a rather unlikely move. When two players have a straight flush, the straight with the highest values wins.</li>
<li><strong>Five of a kind (Royal Flush)</strong>: This is the best poker move and is composed of <strong>all the high cards of the same suit</strong> which are 10,J,Q,K and As. In case there are 2 simultaneous hands like this one, none of the players would win, although if you think about it, it is <strong>almost impossible</strong> to have 2 simultaneous royal flushes.</li>
</ul>
<p>&nbsp;</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS.jpg"><img decoding="async" class="aligncenter wp-image-1078 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS-1024x580.jpg" alt="" width="1024" height="580" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS-300x170.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS-768x435.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<ul>
<li>All the hands you have seen are ordered by ranking from least to most, so you also know who would win when two different hands meet. In the event that no player manages to form either of these hands, the one with the highest card wins, and in the hypothetical case of having the same, the one with the next <strong>highest card</strong> wins.</li>
</ul>
<p><a id="acciones" name="Ancla"></a></p>
<h2 class="p1">3.Texas Holdem Poker Actions</h2>
<p>Now that you&#8217;ve seen all the hands and their priority, let&#8217;s review what <strong>actions you can take within the betting round</strong>, so don&#8217;t miss out!</p>
<ul>
<li><strong>Betting</strong>: Bets are placed in turns clockwise, and players who bet later must <strong>match the amount (call)</strong> bet at least to stay in the hand. But you can also <strong>raise</strong> if you want to force the other players to stay in the hand, and if you do so they must match the amount again until the raise is matched. (It&#8217;s like a tug-of-war).</li>
<li><strong>All-in betting</strong>: When a player bets <strong>all his chips</strong> in a hand, it is said that he is <strong>all-in</strong>, and (if another player bets a larger amount), a <strong>separate pot</strong> is created in which the player is not. If more than one player goes all in in the same hand, <strong>different pots</strong> will be accumulated. To go all in you need to have a good feeling because you risk losing all your chips.</li>
<li><strong>Betting Blinds</strong>: In the hold&#8217;em mode there are two players obliged to bet a <strong>minimum amount</strong> at the beginning of each round, which are the blinds. These are the blinds. The blinds change players in each hand, with the objective that all have equal losses/the same losses at the end. The <strong>Big Blind</strong> is equivalent to the minimum bet for that round and the <strong>Small Blind</strong> is usually half the Big Blind, but this can change. These blinds are posted consecutively to the left of the dealer, first the Small and then the Big, and are intended to force players who want to enter the hand with a minimum amount. In the <strong>tournaments</strong> the blinds are increased in quantity to accelerate the speed of the games.</li>
<li><strong>Check</strong>: When no player has made any bets in that round or the bet amount has already been matched, <strong>you can choose stay or &#8220;check&#8221;</strong> if you consider that it is not necessary to open or raise the bet with your cards. This allows you to remain in the hand, without making any extra bets.</li>
<li><strong>Checkout</strong>: The player always has the option to check out if he feels he has <strong>no options in that hand</strong>, and <strong>will lose his cards</strong> and the amount he had bet up to that point, with no possibility of re-entering.</li>
</ul>
<p><a href="https://www.math4all.es/wp-content/uploads/2019/07/acciones-de-poker.jpg"><img decoding="async" class="aligncenter wp-image-507 size-large" src="https://www.math4all.es/wp-content/uploads/2019/07/acciones-de-poker-1024x566.jpg" alt="" width="1024" height="566" srcset="https://www.math4all.es/wp-content/uploads/2019/07/acciones-de-poker-1024x566.jpg 1024w, https://www.math4all.es/wp-content/uploads/2019/07/acciones-de-poker-300x166.jpg 300w, https://www.math4all.es/wp-content/uploads/2019/07/acciones-de-poker-768x425.jpg 768w, https://www.math4all.es/wp-content/uploads/2019/07/acciones-de-poker.jpg 1080w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>You&#8217;ve seen what actions you can take, and what hands there are, but what chances do you think you have of linking any? Do you think it&#8217;s easy to get a straight or a poker?</p>
<p><a id="probabilidades" name="Ancla"></a></p>
<h2 class="p1">4.Texas Holdem Poker Probabilities</h2>
<p>Let&#8217;s calculate the <strong>probabilities of a Four of a Kind</strong>. To tie this hand you need <strong>4 cards of the same value</strong> out of 5 cards that are dealt. That is: The <strong>ways you have to get 4 equal cards</strong> (with 1 possible value of 13 and 4 possible suits of 4) by <strong>the ways to get another card that is not of that value</strong>, (with 1 possible value of the remaining 12 remaining, and logically 1 suit of 4). What it actually calculates for both, are the ways to select the values by the ways to select the suits.</p>
<p>This gives us exactly <strong>624 total possible ways to make Four of Kind</strong> from all the combinations you can make, but <strong>how many are there</strong>?</p>
<p>If we take into account that a French deck has <strong>52 cards</strong> and you must select <strong>5 cards</strong>, the total number of possibilities are all the <strong>possible combinations of 52 elements taken in 5</strong> since if you think about it carefully (and not like in this video) the order does not matter and they cannot be repeated. (but this is better thought out by you)</p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2020/11/FOUR_OF_A_KIND_ODDS.jpg"><img decoding="async" class="aligncenter wp-image-1080 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/FOUR_OF_A_KIND_ODDS-1024x583.jpg" alt="" width="1024" height="583" srcset="https://www.math4all.es/wp-content/uploads/2020/11/FOUR_OF_A_KIND_ODDS.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/FOUR_OF_A_KIND_ODDS-300x171.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/FOUR_OF_A_KIND_ODDS-768x437.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>This calculation indicates that the <strong>probability of linking a Four of a Kind</strong> before uncovering any cards is approximately <strong>0.02%</strong>, which means that you have very little chance a priori of linking it as it&#8217;s a <strong>very unlikely hand</strong>. But what do you think are the probabilities of the other hands?</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS_PROBABILITY.jpg"><img decoding="async" class="aligncenter wp-image-1081 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS_PROBABILITY-1024x582.jpg" alt="" width="1024" height="582" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS_PROBABILITY.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS_PROBABILITY-300x171.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_HANDS_PROBABILITY-768x437.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>As you see the important hands like <strong>Four of a Kind</strong> or <strong>Full House</strong> are the most difficult to get, and vice versa, <strong>the less important hands are much easier and more likely</strong>, which makes sense. Note also that the only hand you can often aspire to is a <strong>pair</strong>, as hands such as three of a kind, flush or even double pair are quite complicated to achieve.</p>
<p>The most important thing is that <strong>none of the hands reaches 50%</strong>, and not even the sum of all of them gets 50%, which means that <strong>most of the time you are not going to get anything</strong>, although this gives you some clues on how to approach the game. Was this what you imagined?</p>
<p>You have already seen that to get a good hand is not easy, then what you will have normally will be cards that can conform a hand with the cards that are left to uncover, or what is the same a <strong>hand draw</strong>, this will be able to be completed or not with certain probability. But how do you calculate?</p>
<p><a id="informacion" name="Ancla"></a></p>
<h2 class="p1">5.Poker as a game of incomplete and imperfect information</h2>
<p>If you have ever watched a <strong>poker tournament on TV</strong>, you will surely remember a <strong>percentage</strong> that is shown next to each player&#8217;s cards. This is nothing more than the <strong>probability</strong> each player has of winning and is based on both his cards and those of the other players, including the community cards. If we think about it, this calculation is only possible because <strong>the television knows the cards of each and every one</strong> of them, so only we can see what real probabilities they have and who is playing better.</p>
<p>If the players could know the hand of their rivals, they wouldn&#8217;t have any difficulty to play as they could calculate these probabilities in an exact way, but unfortunately this is not possible, because we don&#8217;t know what cards they have, that&#8217;s why poker is considered a game of <strong>incomplete and imperfect information</strong>. Consequently the players have to try to collect all the possible information, to calculate something that approaches as much as possible to those probabilities.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2019/07/torneo-de-poker.jpg"><img decoding="async" class="aligncenter wp-image-511 size-full" src="https://www.math4all.es/wp-content/uploads/2019/07/torneo-de-poker.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2019/07/torneo-de-poker.jpg 1024w, https://www.math4all.es/wp-content/uploads/2019/07/torneo-de-poker-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2019/07/torneo-de-poker-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><br />
<a id="oddsouts" name="Ancla"></a></p>
<h2 class="p1">6.Odds and Outs</h2>
<p>For this we must introduce two mathematical concepts that are used in the world of Poker that are the <strong>odds and outs</strong>, and that serve to calculate <strong>the probabilities of completing a draw</strong> or draws.</p>
<p>Imagine that you have a <strong>flush draw</strong> in your turn and you want to <strong>calculate the probabilities</strong> of being able to complete it in the last betting round.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS.jpg"><img decoding="async" class="aligncenter wp-image-1082 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p><strong>To complete</strong> you draw with the following card <strong>you need one that is of the same suit</strong>. Since a suit has <strong>13 cards</strong> and <strong>4</strong> of them have already come out, there are only <strong>9 cards</strong> left in the deck that <strong>would serve to complete your flush</strong>, and this gives a total of <strong>9 outs</strong>.</p>
<p>If we want to calculate the probabilities for those 9 outs, what we will do is calculate the <strong>odds</strong>, which is the probability used in poker and unlike the usual one, where favorable cases are calculated among the possible ones, here the unfavorable ones are calculated among the favorable ones. And why like that? Because normally we have more probabilities of <strong>not linking our draw</strong> than of linking it, but this you will see later.</p>
<p>The favorable cases are precisely the <strong>outs</strong> and the unfavorable are <strong>all possible cases except the favorable</strong> (ie all except the outs). Finally, the possible cases are the amount of cards that remain to be shown, which we calculate by subtracting from the <strong>52 total cards of the deck</strong>, those that have already been uncovered including also those of the rivals. If you do the calculation for your <strong>9 outs</strong>, you will get an <strong>odds of 4.11</strong>, or in other words, <strong>for every time you win you will lose 4.11 times</strong>, which is equivalent to a <strong>19.57% probability of linking the flush</strong> with the last card to be drawn. What do you think?</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS_2.jpg"><img decoding="async" class="aligncenter wp-image-1083 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS_2-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS_2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS_2-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_OUTS_2-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Calculating all this <strong>is not very practical</strong>, so <strong>I have created a chart</strong>, where you can check what odds you have with the outs of your draw or draws.</p>
<p>If you take a look you can see the table is basically composed of 3 columns, which are the probabilities that can be calculated in the <strong>stages of flop and turn</strong> ordered from left to right, where we are mainly interested in the first and last.</p>
<p>If we want to consult our flush draw with 9 outs in the turn we see the same 4.11 that we had calculated on our own.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART.jpg"><img decoding="async" class="aligncenter wp-image-1086 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART-1024x632.jpg" alt="" width="1024" height="632" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART-1024x632.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART-300x185.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART-768x474.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART.jpg 1775w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p style="text-align: center;"><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_ODDS_CHART.jpg">DOWNLOAD POKER ODDS CHART</a></p>
<p>With these tables you can see interesting things, like that a flush draw has more probabilities than any straight draw, or that a straight draw has more probabilities if it can be completed by two points instead of one.</p>
<p>Notice also that any draw that includes <strong>2 simultaneous hands</strong> (like a straight and flush draw at the same time) is always going to be <strong>more likely than either of those two separate draws</strong>. Since it can be completed in either way, it accepts a greater number of outs. Curious, isn&#8217;t it?</p>
<p>Another important detail is that you will almost never have more than a <strong>50% chance</strong>, which indicates as I said before that <strong>most of the time you will not link your draw</strong>. (Even if you don&#8217;t throw in the towel yet)</p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_RULE_OF_2_4.jpg"><img decoding="async" class="aligncenter wp-image-1088 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_RULE_OF_2_4-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_RULE_OF_2_4.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_RULE_OF_2_4-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_RULE_OF_2_4-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>On a practical level, you can calculate all this, without the need for tables, with the <strong>rule of 4</strong> and <strong>the rule of 2</strong> (multiplying by 4 in the flop a turn and river and by 2 in the turn). These rules provide <strong>a fairly close value</strong> and can be useful when we do not have anything at hand. If you use them for your draw, your <strong>9 outs in the turn</strong> are <strong>(9 x 2 = 18%)</strong> which is very close to the 19.57% shown in the table. This can be <strong>enough for many players</strong>, although there are other similar rules to approximate these values. There are also different <strong>odds variants</strong> if you are interested in the subject.</p>
<p>After seeing all this you could ask yourself the following <strong>questions</strong>:</p>
<ul>
<li>If we normally have a <strong>small chance</strong>, what&#8217;s the point of betting, shouldn&#8217;t we do it only when we have a <strong>majority of possibilities</strong>?</li>
<li>Even if we are calculating the odds for our draw, can&#8217;t the rival beat us with a higher draw? Or in other words, <strong>how much difference is there between calculating our odds and the real odds</strong>?</li>
</ul>
<p><a id="calculadora" name="Ancla"></a></p>
<h2 class="p1">7.Texas Holdem Poker Calculator</h2>
<p>To solve your doubts I have created a <strong>Poker Calculator</strong> where we are going to experiment a little bit with all these concepts.</p>
<p>With it you can see that a simple <strong>pair of 77&#8217;s has more probabilities in front of an ace, k</strong>, or that a <strong>straight and flush draw has more probabilities</strong> than a brand new <strong>pair of kings</strong> on the flop.</p>
<p>You can also see that a <strong>pair with low values</strong> like 66 <strong>is not as good a hand either</strong>, as it can be easily outdone by any higher pair. You can consult any hand you need.</p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2019/07/calculadora-de-poker.jpg"><br />
</a><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR_2.jpg"><img decoding="async" class="aligncenter wp-image-1090 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR_2-1024x607.jpg" alt="" width="1024" height="607" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR_2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR_2-300x178.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR_2-768x455.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p style="text-align: center;"><a href="http://www.math4all.es/poker/poker-calculator.html">GO TO THE POKER CALCULATOR</a></p>
<p>Now let&#8217;s create with our Poker Calculator your <strong>flush draw</strong>. And let&#8217;s suppose that your opponent aspires to an inferior draw with a pair of J&#8217;s. We press the button to calculate the result and we see that your chances of winning against their cards are <strong>20%</strong> which is quite similar to what we had calculated.</p>
<p><strong>It may seem that the odds are enough to know for sure our probabilities</strong>. But imagine that the cards would allow your opponent to have a superior draw like Poker or Full House. If we recalculate your probabilities for this possible situation, we see that the <strong>calculation no longer fits the odds</strong>. This happens because before, your probability of winning depended only on whether or not you picked up your draw, but now you also depend on whether or not your opponent gets a <strong>higher draw</strong>, which decreases your options as you can be beaten with a better hand than the flush.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR.jpg"><img decoding="async" class="aligncenter wp-image-1091 size-large" src="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR-1024x603.jpg" alt="" width="1024" height="603" srcset="https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR-300x177.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/POKER_CALCULATOR-768x452.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>Things are getting interesting&#8230;</p>
<p>Note that the most important detail is that although the calculation is different, <strong>it&#8217;s not too big a difference</strong>, because it is still more difficult to link a Full or Four of a kind than a flush. This indicates that <strong>the stronger your hand draw is, the more reliable your odds will be on average</strong>, since it will be more difficult for another draw to modify them. Depending more on ourselves gives us <strong>more complete information</strong>, and we must take advantage of it.</p>
<p>But <strong>in practice it is easy that your opponents can have better cards than you</strong>, especially when there are many players at the table or big bets are made, that&#8217;s why you should also always have in mind the possible draws or hands that your opponents can have with the community cards, especially when a big amount of money is bet.</p>
<p>When your tied hand is superior to all those possible ones, your victory will be assured, but <strong>when you have only one draw and there can&#8217;t be higher hands</strong> with the cards that are left to reveal, <strong>your real probabilities will be at least your odds</strong> (calculated probability).</p>
<img src="https://www.math4all.es/wp-content/uploads/2020/11/poker-maths.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/the-mathematics-of-poker-odds-and-outs/" data-pin-media="https://www.math4all.es/wp-content/uploads/2020/11/poker-maths.jpg" data-pin-description="The Mathematics of Poker &#8211; Odds and Outs"/><p>The post <a href="https://www.math4all.es/the-mathematics-of-poker-odds-and-outs/">The Mathematics of Poker – Odds and Outs</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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		<title>The Monty Hall Problem</title>
		<link>https://www.math4all.es/the-monty-hall-problem/</link>
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		<dc:creator><![CDATA[math4all]]></dc:creator>
		<pubDate>Sun, 01 Nov 2020 08:09:09 +0000</pubDate>
				<category><![CDATA[The Mathematics of games]]></category>
		<category><![CDATA[game]]></category>
		<category><![CDATA[lets make a deal]]></category>
		<category><![CDATA[monty hall]]></category>
		<category><![CDATA[monty hall maths]]></category>
		<category><![CDATA[monty hall paradox]]></category>
		<category><![CDATA[monty hall probability]]></category>
		<category><![CDATA[monty hall problem]]></category>
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					<description><![CDATA[<p>Welcome to one of the most curious problems in mathematics, the Monty Hall problem. 1.Description of the problem 2.Intuitive analysis&#8230;</p>
<p>The post <a href="https://www.math4all.es/the-monty-hall-problem/">The Monty Hall Problem</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="&#x1f4e3; DESVELADA La Paradoja de MONTY HALL &#x1f410;&#x1f410; &#x1f697;" width="1170" height="658" src="https://www.youtube.com/embed/zzweTm8CwXQ?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe></p>
<p>Welcome to one of the most curious problems in mathematics, the Monty Hall problem.</p>
<p><a href="#descripcion">1.Description of the problem</a><br />
<a href="#analisisintuitivo">2.Intuitive analysis</a><br />
<a href="#simulador">3.Monty Hall Simulator</a><br />
<a href="#analisismatematico">4.Mathematical Analysis</a><br />
<a href="#analisiscondicional"> 4.1.Mathematical Analysis with Conditional Probability</a><br />
<a href="#solucion">5.Intuitive solution</a><br />
<a href="#explicacion">6.Explanation</a><br />
<a href="#conclusiones">7.Conclusions</a></p>
<p><a id="descripcion" name="Ancla"></a></p>
<h2>1.Description of the problem</h2>
<p>Imagine you are in a contest and I am the presenter</p>
<p>Behind one of these <strong>3 doors</strong> is a <strong>new car</strong>. And behind the other two there are goats (yes goats!!).</p>
<p>You have to <strong>choose a door</strong> and we will assume that for example you choose the door 1 (remember that you have to win the new car and not a goat &#8230;)</p>
<p><img decoding="async" class="aligncenter wp-image-405 size-large" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-descripcion-1024x576.png" alt="" width="1024" height="576" /></p>
<p>I who can see what is behind each door discard one of the doors that has a goat, opening the number 3 in this case, and now I ask you the following question, <strong>do you want to change the door 1 by 2?</strong>, or more importantly for the purpose of this video, <strong>do you think that matters?</strong></p>
<p><strong>The Monty Hall problem</strong> is one of the <strong>most famous problems of probability</strong> and comes from the television contest of the 70&#8217;s <strong>Let&#8217;s Make a Deal</strong>. The name refers to its famous presenter, and his well-known <strong>3-door game</strong>. Where a car was hidden between one of them, and the contestant had to choose a door. After discarding another of the doors, Monty Hall asked the contestant if he wanted to change the door or keep the one he had chosen. And this is where the real game begins.</p>
<p><a id="analisisintuitivo" name="Ancla"></a></p>
<h2 class="p1">2.Intuitive analysis</h2>
<p>I&#8217;m sure <strong>many of you think there&#8217;s no difference between keeping the door or changing it</strong>. If there are two doors and a car left, either option is the same as if you change the door you can win or lose the car and if you don&#8217;t change it you can also win or lose, which leaves each door with a 50% chance. But, is that true?</p>
<p>Your intuition may tell you that it is, but do you think your intuition has analyzed the problem correctly, <strong>do you think you can trust your instinct?</strong></p>
<p><a id="simulador" name="Ancla"></a></p>
<h2 class="p1">3.Monty Hall Simulator</h2>
<p>To get out of doubts I have created a <strong>simulator</strong> that recreates the rules of this game and we will see what results it offers us for both elections, so, don&#8217;t miss it:</p>
<p>It does exactly what we have described, so we start playing and <strong>ask us to choose a door</strong>, we choose it and the presenter opens another door where the car is not. Then he makes us choose between changing the door or keeping it.</p>
<p><strong>Always keeping the door:</strong></p>
<p>Let&#8217;s try first to stay the door we have chosen, so we play many games to the simulator, and we conclude that <strong>we win the car about 30% of the times</strong> we have played (which shows us the success rate).</p>
<p><strong>Always changing the door:</strong></p>
<p>If instead we choose to change our door for the other one, and we repeat this many times in the simulator, we come to the conclusion that this time <strong>we win the car about 60% of the times</strong>. Which makes both results very different.</p>
<p>&nbsp;</p>
<p style="text-align: center;"><strong>Don&#8217;t you believe it? Try it yourself</strong></p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_SIMULATOR.jpg"><img decoding="async" class="aligncenter size-large wp-image-1076" src="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_SIMULATOR-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_SIMULATOR.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_SIMULATOR-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_SIMULATOR-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p style="text-align: center;"><a href="http://www.math4all.es/monty-hall/monty-hall-simulator.html">PLAY MONTY HALL</a></p>
<p>We thought it wouldn&#8217;t matter, but we have seen that <strong>changing doors makes us winners in a higher percentage</strong> than keeping the door we had, and it seems to be quite superior. So what&#8217;s going on?</p>
<p><a id="analisismatematico" name="Ancla"></a></p>
<h2 class="p1">4.Mathematical Analysis</h2>
<p>If we analyze the problem at a mathematical level we will realize that this is a <strong>simple probability problem</strong>, where all the possible results of the experiment of choosing a door are 3:</p>
<p>choose car, choose a goat, or choose the other goat.</p>
<p>The <strong>events</strong> we need to solve the problem are:</p>
<p><strong>G: &#8220;Win the car without changing the door&#8221;</strong></p>
<p><strong>G&#8217;: &#8220;Win the car by changing the door&#8221;</strong><a href="https://www.math4all.es/wp-content/uploads/2020/11/monty_hall_door.jpg"><img decoding="async" class="aligncenter size-large wp-image-1066" src="https://www.math4all.es/wp-content/uploads/2020/11/monty_hall_door-1024x575.jpg" alt="" width="1024" height="575" srcset="https://www.math4all.es/wp-content/uploads/2020/11/monty_hall_door.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/monty_hall_door-300x168.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/monty_hall_door-768x431.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>We will study the <strong>probability of each event separately</strong>:</p>
<h3 class="p3">Without changing the door</h3>
<p>So let&#8217;s start by <strong>not changing the doors</strong>.</p>
<p><strong>If the contestant chooses the car</strong>, and doesn&#8217;t change the door, <strong>the car will win for sure</strong>, because it will hold its door until the end.</p>
<p>In the same way <strong>if the contestant chooses a goat</strong>, and doesn&#8217;t change the door, <strong>he will never be able to win the car</strong>.</p>
<p>As there is only <strong>1 case where he wins out of 3 possible choices</strong>, the probability of winning the car if he doesn&#8217;t change the door is 1/3. This is approximately a <strong>33.33% chance of winning</strong>. That&#8217;s 1 in 3 times, which leaves you at a disadvantage to win the car.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_WITHOUT_CHANGING_DOORS.jpg"><img decoding="async" class="aligncenter size-large wp-image-1068" src="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_WITHOUT_CHANGING_DOORS-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_WITHOUT_CHANGING_DOORS.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_WITHOUT_CHANGING_DOORS-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_WITHOUT_CHANGING_DOORS-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a>Let&#8217;s see now what the <strong>probability is if you decide to change doors</strong>:</p>
<h3 class="p3">Changing Door</h3>
<p><strong>If the contestant chooses the car</strong>, and changes the door, <strong>he will never be able to win</strong> it since he will always change it for a goat. It doesn&#8217;t seem like a good start.</p>
<p>Instead, notice that <strong>if he chooses a goat</strong>, the host always shows him the other goat. This automatically discards one of the goats, which means that the contestant <strong>will have to change the goat he has obligatorily for the car</strong>. As there are 2 goats as possible choices we will win the car 2 out of 3 times so the probability of winning the car by changing doors 2/3. This <strong>doubles the probabilities from 33.33 to 66.66%</strong>. This gives us a greater advantage to win the car.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_CHANGING_DOORS.jpg"><img decoding="async" class="aligncenter size-large wp-image-1069" src="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_CHANGING_DOORS-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_CHANGING_DOORS.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_CHANGING_DOORS-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_CHANGING_DOORS-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p><a id="analisiscondicional" name="Ancla"></a></p>
<h2 class="p1">4.1.Mathematical Analysis with Conditional Probability</h2>
<p>The<strong> events</strong> we need to solve the problem are:</p>
<p><strong>A: The contestant chooses the door where the car is&#8221;</strong></p>
<p><strong>B: &#8220;The contestant chooses the door where there is a goat&#8221;</strong></p>
<p>And finally the event we are looking for that is:</p>
<p><strong>G: &#8220;The contestant win the car&#8221;</strong></p>
<p>The probability of choosing the door with the car as you see is 1 in 3, and that of choosing the door with a goat is 2/3 , but what is the probability of winning?</p>
<p class="p1"><strong>P(A)=1/3</strong></p>
<p class="p1"><strong>P(B)=2/3</strong></p>
<p class="p1"><strong>P(G)=?</strong></p>
<p>To win we can do it either way. <strong>Choose the car and win or choose a goat and win</strong>, therefore</p>
<p class="p1"><strong>P(G)= P((A and G) or (B and G))</strong></p>
<p>What translated into <strong>mathematical language</strong> is:</p>
<p class="p1"><strong>P(G)=P((A∩G) <span class="s1">∪</span> (B∩G))</strong><span class="Apple-converted-space">      (You simply have to change the symbols of and for intersection and the or for union)</span></p>
<p>Well, the <strong>probability of a union is the sum of probabilities</strong> so we can separate the probability into 2 simpler ones</p>
<p class="p1"><strong>P(G)=P(A∩G) + P(B∩G)</strong></p>
<p>If we use the <strong>conditional probability formula</strong> and clear the intersection we have left:</p>
<p class="p1"><strong>P(G)=P(G|A)·P(A) + P(G|B)·P(B)</strong></p>
<p>Now we only have to calculate the probability of winning for each strategy:</p>
<h3 class="p1">Probability without changing Door</h3>
<p>If we have decided to keep the door we had:</p>
<p><strong>In the case of having a car, we will have won for sure</strong>, since we are not going to change it, so the conditional probability of that event A, will be 100% or 1</p>
<p class="p1"><strong>P(G|A)=1</strong></p>
<p>But this will only happen when we have a car and <strong>we have 1/3 chance of choosing it</strong>, so the probability of choosing the car and winning is 1·1/3 which is 1/3</p>
<p><strong>P(A)=1/3</strong></p>
<p class="p1"><strong>P(A∩G)=P(G|A)·P(A)=1·1/3=1/3</strong></p>
<p><strong>In the case of having a goat</strong>, if we keep the door <strong>we will never win</strong> so the conditional probability of the event will be 0, and when we multiply it by the probabilities of having a goat it&#8217;s still 0, which is logical if we already see that it is impossible to choose the goat and win the car from the beginning.</p>
<p class="p1"><strong>P(G|B)=0</strong></p>
<p class="p1"><strong>P(B)=2/3</strong></p>
<p><strong>P(B∩G)=P(G|B)·P(B)=0·2/3=0</strong></p>
<p>If <strong>we add both amounts</strong> we have a 1/3 chance of winning if we keep the door.</p>
<p class="p1"><strong>P(G)=P(A∩G) + P(B∩G)=1/3 + 0=1/3</strong></p>
<p class="p1">This can be summarized as (Car,Goat,Goat)</p>
<h3 class="p1">Probability of Changing Door</h3>
<p>Let&#8217;s now choose to change the door always:</p>
<p><strong>In the case of having a car, we lose the car for sure</strong>, since we are going to change it for a goat, so the probability conditioned to that event A, will be 0</p>
<p class="p1"><strong>P(G|A)=0</strong></p>
<p class="p1">And the probability of taking the car and winning will be 0.</p>
<p><strong>P(A)=1/3</strong></p>
<p><strong>P(A∩G)=P(G|A)·P(A)=0·1/3=0</strong></p>
<p><strong>If instead we choose a goat</strong>, notice that the presenter always eliminates the other goat, so there will only be the car when we change the door, so <strong>we will always win the car</strong>.</p>
<p class="p1"><strong>P(G|B)=1</strong></p>
<p>This will only happen when we have a goat, but we have to consider that the probability of choosing a goat is 2/3, so the probability of having a goat and winning is 1·2/3 which is 2/3</p>
<p><strong>P(B)=2/3</strong></p>
<p class="p1"><strong>P(B∩G)=P(G|B)·P(B)=1·2/3=2/3</strong></p>
<p class="p1">If <strong>we add both amounts</strong> we have that the probability of winning if we change doors is 2/3</p>
<p class="p1"><strong>P(G)=P(A∩G) + P(B∩G)=0 + 2/3=2/3</strong></p>
<p class="p1">This can be summarized as (Goat,Car,Car)</p>
<h2 class="p1">5.Intuitive solution</h2>
<p><strong>If we don&#8217;t change the door</strong> we see clearly that as there are two goats and a car we have a <strong>1/3 chance of winning the car</strong>.</p>
<p><strong>If we change the door</strong>:</p>
<ul>
<li><strong>and we choose the car we always lose.</strong></li>
<li><strong>and we choose a goat we always win.</strong></li>
</ul>
<p>But as there is <strong>1 car and 2 goats, we will win twice out of 3 times</strong>, because 2 out of 3 times we will choose a goat.</p>
<p><a id="explicacion" name="Ancla"></a></p>
<h2 class="p1">6.Explanation</h2>
<p>For this we are going to imagine again <strong>all the possible cases that can occur for the 3 doors</strong>. (Note that we don&#8217;t differentiate the order of the goats, since that doubles the number of cases, but doesn&#8217;t change the probabilities)</p>
<p class="p5"><img decoding="async" class="aligncenter wp-image-394 size-large" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-posibilidades-1024x576.png" alt="" width="1024" height="576" />When we choose our door, <strong>we have a 1/3 chance of having the car</strong>. Even if the presenter opens a door, it is conditioned by the one we have chosen, since he can only open a door that is not ours. That implies that <strong>he always removes a goat from the other two doors, but never from ours</strong>.</p>
<p>Notice also that in those 2 doors there is always a goat to be eliminated, so the presenter only discards a goat that was safe and shows us where it is, but that does not increase the probabilities of our door, since we can still have the other goat the same times. This implies that <strong>the door we chose at the beginning will still have the car 1 out of 3 times and will not increase its probabilities, no matter how much we remove a door</strong>.</p>
<p><img decoding="async" class="aligncenter wp-image-417 size-large" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-nuestra-puerta-1024x576.png" alt="" width="1024" height="576" /></p>
<p>Surprising, that doesn&#8217;t change, but then what about the other door?</p>
<p><strong>If our door has the car 1 out of 3 times, the other 2 times will be in the other two doors</strong>, so that group of 2 is logically more likely to have the car than our door. We simply have to add up the probability of each one.</p>
<p class="p5"><img decoding="async" class="wp-image-383 size-large aligncenter" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-probabilidades-grupos-1024x576.png" alt="" width="1024" height="576" /></p>
<p>When the presenter opens a door, he knows what&#8217;s behind each one, so he always saves the car if he has one. <strong>If both doors have the car 2 out of 3 times and always discards a goat, the door that is saved will still have the car 2 out of 3 times</strong>, so its probability will be the same as both doors had but spread out in one. So you must change it for this one.</p>
<p>Note also that <strong>this door will not always be the same</strong> and therefore has more probability.</p>
<p><img decoding="async" class="aligncenter wp-image-384 size-large" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-descartes-1024x576.png" alt="" width="1024" height="576" /></p>
<p><img decoding="async" class="wp-image-385 size-large aligncenter" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-2-puertas-1024x576.png" alt="" width="1024" height="576" /></p>
<p class="p1">And that said, are you sure you don&#8217;t want to change doors?</p>
<h3 class="p5">Expansion of the idea</h3>
<p>Imagine now <strong>the same game but with 20 doors</strong> instead of 3. We keep choosing 1 door, and the presenter eliminates in the same way the doors that are goat until he gets one.</p>
<p><img decoding="async" class="aligncenter wp-image-421 size-large" src="http://www.math4all.es/wp-content/uploads/2018/11/monty-hall-20-puertas-1024x576.png" alt="" width="1024" height="576" /></p>
<p>&nbsp;</p>
<p>If we forget our door for a moment, and think only of the 19 that can be eliminated by the presenter. We will see very clearly that 19 doors are much more likely than one, so <strong>the car will be most of the time in one of the doors that the presenter has</strong> and only one in our door. And <strong>if the presenter eliminates all the doors that are goat of that group, the remaining door will have the car most of the time</strong>, in the same way.</p>
<p><a id="conclusiones" name="Ancla"></a></p>
<h2 class="p5">7.Conclusions</h2>
<p>In conclusion, the presenter only eliminates goats from his doors, and that can only give an advantage to some of his doors, but never to ours.</p>
<p>As we saw at the end it wasn&#8217;t that complicated either, but <strong>why haven&#8217;t we been able to see it before?</strong></p>
<p>At first the car can be equally on any of the doors, so we <strong>divide 1 prize between 3 doors</strong>, and so we calculated the probability for our door. So far so good.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_1.jpg"><img decoding="async" class="aligncenter size-large wp-image-1073" src="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_1-1024x577.jpg" alt="" width="1024" height="577" srcset="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_1.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_1-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_1-768x433.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>&nbsp;</p>
<p>The problem comes <strong>when they open a door for us</strong>.</p>
<p>When the presenter gives us the option of keeping our door or changing it, we must recalculate the probability of each door to know which one is more convenient for us. To do this, we <strong>divide 1 prize between 2 doors in the same way as before, as if the prize could be the same in each one of them</strong>, but this is not the case, since the discard that the presenter makes only increases the probability of the door he is going to offer you and not of both equally. This is why the calculation is incorrect. (66%&gt;33%)<a href="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_2.jpg"><img decoding="async" class="aligncenter size-large wp-image-1072" src="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_2-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_2-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/11/MONTY_HALL_PROBABILITY_2-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>&nbsp;</p>
<p>Note that <strong>the only way to have a 50% chance of winning, is to choose to change doors or not at random</strong>, because in this case although one door is more likely, we have the same options to choose it as the other, so the times we change doors, will compensate for the times we do not. In any case, this is still less than always changing the door, although more than always keeping the door. (66%&gt;50%&gt;33%)</p>
<p class="p5"><strong>Curious, isn&#8217;t it?</strong></p>
<p>We now know that <strong>Monty Hall</strong> was a very clever guy, using psychology and mathematics. <strong>If the contestant didn&#8217;t change doors, he had the chance to lose, and to gain an advantage he had to &#8220;take risks&#8221; and accept the change</strong>.</p>
<p>The reality is that most contestants lost their car, because very few people dared to change the door. Who would change a door when the question can be a trap? Only a madman, or a mathematician, and that&#8217;s what cheating is all about.</p>
<p>Who says math is useless?</p>
<p>If you are interested in the math of the game you can visit my youtube channel <a href="https://www.youtube.com/channel/UC9NzA04qApji9FO63msr_-w">math for all</a> on Youtube where I study mathematical games of chance like Poker or Blackjack.</p>
<img src="https://www.math4all.es/wp-content/uploads/2020/11/monty-hall-problem.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/the-monty-hall-problem/" data-pin-media="https://www.math4all.es/wp-content/uploads/2020/11/monty-hall-problem.jpg" data-pin-description="The Monty Hall Problem"/><p>The post <a href="https://www.math4all.es/the-monty-hall-problem/">The Monty Hall Problem</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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		<title>The Mathematics of Blackjack</title>
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		<pubDate>Sat, 31 Oct 2020 12:20:33 +0000</pubDate>
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					<description><![CDATA[<p>Do you think you can win at Blackjack? What is the best way to play? In this article we solve&#8230;</p>
<p>The post <a href="https://www.math4all.es/the-mathematics-of-blackjack/">The Mathematics of Blackjack</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="2&#x20e3;1&#x20e3; Las MATEMÁTICAS del BLACKJACK &#x2660;&#x2665;&#x2666;&#x2663;" width="1170" height="658" src="https://www.youtube.com/embed/YJZt-eAg_p8?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe></p>
<p>Do you think you can win at Blackjack? What is the best way to play? In this article we solve these and other questions with the help of mathematics.</p>
<p><a href="#reglas">1.Blackjack Rules</a><br />
<a href="#acciones">2.Blackjack Actions</a><br />
<a href="#premios">3.Blackjack Payouts</a><br />
<a href="#estudio">4.Mathematical Study of Blackjack</a><br />
<a href="#probabilidadespasarnos"> 4.1.Chances of exceeding 21</a><br />
<a href="#probabilidadescrupier"> 4.2.Chances of the Dealer exceeding 21</a><br />
<a href="#ventajabanca"> 4.3.Mathematical Advantage of the House</a><br />
<a href="#estrategiabasica">5.Basic Strategy</a><br />
<a href="#estrategiabasica1mazo"> 5.1.Blackjack Chart 1 Deck</a><br />
<a href="#estrategiabasicavariosmazos"> 5.2.Blackjack Chart </a><a href="#estrategiabasica1mazo">Table Several Decks</a><br />
<a href="#conteo">6.Counting cards</a><br />
<a href="#conclusiones">7.Conclusions</a></p>
<p><a id="reglas" name="Ancla"></a></p>
<h2 class="p1">1.Blackjack Rules</h2>
<p class="p1"><strong>Blackjack</strong> is a French deck game for up to <strong>7 players </strong>where the players face the house.</p>
<p class="p1">The goal is to <strong>get 21</strong> with our cards or a higher value than the dealer, <strong>without exceeding the number</strong>.</p>
<p class="p1"><strong>If we exceed 21</strong> or add a lower value than the dealer we&#8217;ll automatically <strong>lose our bet</strong>.</p>
<p class="p1">Once the bets are placed <strong>2 visible cards</strong> are dealt to each player and only <strong>one for the dealer</strong></p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/blackjack-table.jpg"><img decoding="async" class="aligncenter wp-image-1054 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/blackjack-table-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/10/blackjack-table-1024x576.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack-table-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack-table-768x432.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack-table.jpg 2000w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p1">Players must say based on their cards and the croupier&#8217;s if they want to <strong>hit, stand, double down, or split</strong> their hand</p>
<p class="p1">All cards are worth the numerical value they have, i.e. from 2 to 10, except for the <strong>figures</strong> that are worth <strong>10</strong>, and the <strong>Ace</strong> that can be worth <strong>1 or 11</strong> to our interest</p>
<p class="p1">Payments are made on same time, and we can get 21 with more than two cards but <strong>will only be considered blackjack when done with two.</strong></p>
<p><strong>The house</strong> can only hit or stand, and only <strong>hits when its hand is below 17</strong>, so if it reaches or exceeds that value will stand automatically.</p>
<p class="p1">In case the dealer goes over 21 the players who are still at the table will win their bets and in case of a <strong>tie</strong> the player will regain his bet.</p>
<p>You can play blackjack with one or more decks being <strong>6 decks</strong> the standard in the <strong>European Blackjack</strong>.</p>
<p>The <strong>American Blackjack</strong> has some different rules that allow you to modify if the dealer hits or how players can split their hands. In addition he deals himself a <strong>second card covered</strong> and you can consult it in case his first card give him the chance to do blackjack If effectively he gets 21, the players they&#8217;ll lose their bet before they even play. This speeds up the game and also decreases the house advantage.</p>
<p><a id="acciones" name="Ancla"></a></p>
<h2 class="p1">2.Blackjack Actions</h2>
<h3 class="p1">-Hit/Stand</h3>
<p class="p1">In the first place, we can hit, <strong>while our sum does not exceed 21</strong> depending on the card they give us we&#8217;ll be closer to 21, or instead we&#8217;ll lose the bet if we get over it.</p>
<h3 class="p1">-Double down</h3>
<p class="p1">Para doblar nuestra apuesta <strong>necesitamos una mano que sume obligatoriamente 9, 10 u 11</strong> y solo podrá hacerlo en el principio de turno. En caso de doblar tu apuesta solo podrás recibir una carta más.</p>
<p class="p1">To double down <strong>we need a hand which obligatorily sums 9, 10 or 11</strong> and you can only do it at the beginning of the round. If you double your bet, you can only receive one more card</p>
<p class="p1">Unlike in <strong>American Blackjack</strong> we can double up with either hand.</p>
<h3 class="p1">-Split</h3>
<p>If our cards have the <strong>same value</strong>, we can <strong>split each card into different hands</strong> to play them at the same time, but for that it is obligatory to add also a bet as the initial. You must also know that within a split hand, even if we sum 21 with the next card is not considered blackjack.</p>
<p>After splitting, most casinos allow to double the resulting new hands, and this gives us some advantage. (Not allowing it increases the house advantage by 0.12%).</p>
<h3 class="p1">-Insurance Betting</h3>
<p>If the card uncovered by the dealer is an <strong>Ace</strong>, we can bet to insurance, <strong>if we feel that the house will get blackjack</strong> with the next card. If we&#8217;ve bet and finally gets it, we&#8217;ll be rewarded.</p>
<h3 class="p1">-Leave</h3>
<p>And finally leave, which is only possible in the <strong>American Blackjack</strong> and if the player leaves  <strong>he will lose only half the bet</strong>.</p>
<p><a id="premios" name="Ancla"></a></p>
<h2 class="p1">3.Blackjack Payouts</h2>
<p class="p1">&#8211;<strong>1 to 1 </strong>(one chip for each chip bet) if we <strong>beat the house</strong>.</p>
<p class="p1">&#8211;<strong>3 to 2</strong> (three chips, for every two chips bet or one and haf for each chip bet) if we beat the dealer with <strong>Blackjack</strong> (as long as there is no tie). In the <strong>American Blackjack of Las Vegas</strong> you pay with <strong>6 to 5</strong>.</p>
<p class="p1">&#8211;<strong>2 to 1 </strong>(two chips for each chip bet) if we win an <strong>insurance bet</strong>.</p>
<p>In the case of <strong>American Blackjack</strong>, it is also usual that if the player has blackjack and the croupier has a visible Ace, he is offered the possibility of charging blackjack with a 2 to 1 (instead of the usual 3 to 2) as an alternative to the insurance bet.</p>
<p><a id="estudio" name="Ancla"></a></p>
<h2 class="p1">4.Mathematical Study of Blackjack</h2>
<p class="p1">To analyze Blackjack at a mathematical level we must know that a French Deck has <strong>52 cards</strong>.</p>
<p class="p1"><strong>&#8211; 4 of them are Ace, </strong></p>
<p class="p1"><strong>-16 of them are 10 or Face card.</strong></p>
<p class="p1"><strong>-The remaining 32 are numbers from 2 to 9</strong></p>
<p class="p1">So we can draw some conclusions:</p>
<p class="p1">-Approximately <strong>one third of the cards, have the value 10</strong> (which takes a fundamental weight).</p>
<p class="p1">-After that, there are only <strong>4 cards that are Ace</strong> so if we want to get a blackjack we&#8217;re gonna need one of them but&#8230; which is the probability?</p>
<p>The <strong>probability of getting Blackjack</strong> are the ways we have to get 21 with 2 cards of all possible combinations to pull 2 cards out of the deck.</p>
<p>Or put another way, <strong>favorable cases between possible cases</strong>, which is not the same than odds, even are related.</p>
<p>To get 21 with 2 cards we need an Ace and a 10. And in the deck there are <strong>4 Aces and 16 tens</strong> (between face cards and 10 value cards).</p>
<p>On the other hand, all combinations of 2 cards, <strong>combinations of 52 elements taken 2 at a time</strong> (where the order doesn&#8217;t matter, and they can&#8217;t be repeated),</p>
<p>so the final probability is:</p>
<p class="p1">4<span class="s1">.</span>16 / C52,2 which is equal to 64/1326 or a <strong>4,8% chance</strong> of getting Blackjack (which was logical since we know that to have blackjack we always need an ace and there are only 4 in the deck).</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack.jpg"><img decoding="async" class="aligncenter wp-image-1055 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack-1024x583.jpg" alt="" width="1024" height="583" srcset="https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack-300x171.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack-768x437.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In case of playing with <strong>several decks</strong> each figure or number maintains the same proportion, due to it&#8217;s the same 4 aces in 52 cards, than 8 in 104. But the probability of getting a particular result vary slightly, since it influences less to remove a card in a deck of 104 than in a deck of 52. So <strong>each game with different decks has its own probabilistic study</strong>.</p>
<p><a id="probabilidadespasarnos" name="Ancla"></a></p>
<h3 class="p1">4.1.Chances of exceeding 21</h3>
<p>To calculate the probability of going over 21, we must to calculate the probability of going over for any hand and for that, <strong>we need to analyse hand in hand</strong>.</p>
<p class="p1">Let us imagine that our hand value <strong>sums for example 12</strong> value:</p>
<p class="p1">When we hit, we are faced with all these cases:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21.jpg"><img decoding="async" class="aligncenter wp-image-1056 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21-1024x577.jpg" alt="" width="1024" height="577" srcset="https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21-768x433.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>As we see, initially there are <strong>4 cases of 13</strong> where we exceed 21. This is approximately a <strong>30% chance</strong>, which means that <strong>with a 12 value only 30% of the time we hit, we will exceed 21</strong>.</p>
<p class="p1">(The calculation of this probability is a <strong>simplification</strong> because we really would have to take into account that the cards that make up the 12 vary slightly the probability of the card we have asked, but the difference with this calculation is not very large and also is much more complex to study and understand).</p>
<p>If we calculate the same for <strong>all possible hands</strong> and accurately we will obtain the following <strong>table</strong>:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21_2.jpg"><img decoding="async" class="wp-image-1057 size-large aligncenter" src="https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21_2-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21_2.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21_2-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/blackjack_probability_exceeding_21_2-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p3">As you can see, <strong>from 13</strong>, both us and the house <strong>we&#8217;ll exceed the most of the time</strong>.</p>
<p>Also note that if our sum is 11 or less we can never surpass 21 with one more card That&#8217;s why probability is 0.</p>
<p><a id="probabilidadescrupier" name="Ancla"></a></p>
<h3 class="p1">4.2.Chances of the Dealer exceeding 21</h3>
<p>As we just saw, if we don&#8217;t want to lose most of the time we must hit as long as our hand sum less than 14. The question is&#8230; <strong>If we put a limit on 14, do we have a chance of winning?</strong></p>
<p>If we consider that the dealer always hits for a card until sum 17 or more, stand with a value less than 17 only makes you a winner when the dealer exceeds.</p>
<p>Or to put it another way, if the dealer doesn&#8217;t go over 21, his hand will always have a value equal to or greater than 17, so the players who stood with a hand less than 17 will lose their bet.</p>
<p>Well, according to the mathematical study carried out the <strong>chances that the dealer exceeds</strong> 21 depending on the card he has (and his strategy) are the following:</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack_dealer.jpg"><img decoding="async" class="aligncenter wp-image-1058 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack_dealer-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack_dealer.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack_dealer-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/probability_blackjack_dealer-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>We see that for a <strong>high card</strong> the probability of go over is quite low being the Ace the card that offers the most advantages to the house.</p>
<p>For a <strong>low card</strong> the probability is higher, but note that <strong>in no case does it exceed 50%</strong></p>
<p>If we average all of them we conclude that <strong>the dealer surpasses 21 in 28.35% of the times</strong> (368,67/13), which means that in the remaining 71.65% of the times you stand with less than 17 you&#8217;re gonna lose your bet.</p>
<h3 class="p3">House Limit</h3>
<p>As we have just seen it&#8217;s useless to set a limit below the dealer limit, since we will lose in most cases.</p>
<p>But thanks to this we&#8217;ve discovered that the dealer’s up card offers us valuable information to make decisions.</p>
<p>After this we have the following questions:</p>
<p class="p3"><strong>-Why is the house limit 17?</strong></p>
<p class="p3"><strong>-What if I copy the house strategy and I also stand from 17?</strong></p>
<p><a id="ventajabanca" name="Ancla"></a></p>
<h3 class="p3">4.3.Mathematical Advantage of the House</h3>
<p>If you realise the dealer always waits the player finishes his play before proceeding to play him. That means that <strong>if the player goes over 21, he loses his bet regardless of what the dealer does</strong>.<strong> </strong>So if the dealer also goes over 21 it is the player who loses anyway since he had been eliminated before. That makes the house wins in a case, which is really a tie and this happens <strong>7.9% of the time you play</strong>. That&#8217;s about an 8% advantage over the player which is finally reduced to <strong>5.6%</strong> if we consider that blackjacks are payed 3 to 2, instead of the equitable 2 to 1 .That&#8217;s <strong>the casino&#8217;s main advantage</strong>.</p>
<p>To fight this, the player has options that allow him to have a <strong>flexible strategy</strong>. He can double down when he&#8217;s interested, split hands, leave, etc. But most importantly, he knows a dealer&#8217;s card that will make him make those decisions (differents from hit or stand only) considering this card and this, is where math comes in once again.</p>
<p><a id="estrategiabasica" name="Ancla"></a></p>
<h2 class="p6">5.Basic Strategy</h2>
<p>In the 50s, a group of mathematicians led by <strong>Roger Baldwin</strong> developed the Basic Strategy. This is based on <strong>optimizing player decisions</strong> in such a way that the action he chooses it&#8217;s the optimal in terms of profit from among all the actions he can take. This bring us at first the <strong>maximum possible long-term gain</strong>.</p>
<p>And is achieved by developing a probabilistic model which calculates based on the player&#8217;s hand conditioned by the dealer&#8217;s card <strong>when it is more optimal to make the decision of stand, hit, double down, split or bet on insurance</strong>, either for hard hands that are inflexible as for soft hands (where there is an ace and can change the sum).</p>
<p class="p1"> <a href="https://www.math4all.es/wp-content/uploads/2020/10/basic_strategy_formula.jpg"><img decoding="async" class="aligncenter wp-image-1060 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/basic_strategy_formula-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/10/basic_strategy_formula.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/basic_strategy_formula-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/basic_strategy_formula-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>In the mathematical language we can say that by means of a recursive function with some defined final values, and a table with all the probabilities for the sum of the dealer, <strong>the maximum between the expected gain if he stands and the expected gain is calculated taking into account all the possible cards that it can receive when it asks for a card</strong> (the average of all the possible expected gains if it asks for the next card). With this a table of results is elaborated that <strong>assures us mathematically which is the most convenient action</strong>, making our strategy optimal whatever the combination of cards between the player and the dealer. So if you want to play Blackjack well I recommend you to download this blackjack <b>chart</b>:</p>
<p><a id="estrategiabasica1mazo" name="Ancla"></a></p>
<p class="p1"><a href="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK-BASIC-STRATEGY-CHART-1-DECK.jpg"><img decoding="async" class="aligncenter wp-image-1061 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK-BASIC-STRATEGY-CHART-1-DECK-633x1024.jpg" alt="" width="633" height="1024" srcset="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK-BASIC-STRATEGY-CHART-1-DECK.jpg 633w, https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK-BASIC-STRATEGY-CHART-1-DECK-185x300.jpg 185w" sizes="(max-width: 633px) 100vw, 633px" /></a></p>
<p style="text-align: center;"><a href="http://www.math4all.es/blackjack/BLACKJACK_BASIC_STRATEGY_CHART_1_DECK.jpg">DOWNLOAD BLACKJACK CHART 1 DECK</a></p>
<p><a id="estrategiabasicavariosmazos" name="Ancla"></a></p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_BASIC_STRATEGY_CHART_6_DECK-1.jpg"><img decoding="async" class="aligncenter size-large wp-image-1175" src="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_BASIC_STRATEGY_CHART_6_DECK-1-633x1024.jpg" alt="" width="633" height="1024" srcset="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_BASIC_STRATEGY_CHART_6_DECK-1-633x1024.jpg 633w, https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_BASIC_STRATEGY_CHART_6_DECK-1-185x300.jpg 185w, https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_BASIC_STRATEGY_CHART_6_DECK-1-768x1242.jpg 768w" sizes="(max-width: 633px) 100vw, 633px" /></a></p>
<p style="text-align: center;"><a href="http://www.math4all.es/blackjack/BLACKJACK_BASIC_STRATEGY_CHART_6_DECK.jpg">DOWNLOAD BLACKJACK CHART SEVERAL DECKS</a></p>
<p>&nbsp;</p>
<p>As a detail, notice that in hard hands <strong>from 17</strong>, it gives us instructions to <strong>stand</strong>. That means that regardless of the card have the dealer, math tell us that the expected profit is always higher if we stand. That answers the question of why the house has 17 as a limit. if we understand that <strong>the house is also a player</strong>.</p>
<p>Look also, you should <strong>never bet on insurance</strong>. This is because although the dealer has an Ace mostly he won&#8217;t get blackjack.</p>
<p>Well, with this strategy we went from the previous 5.6% to a <strong>percentage of less than 1%</strong> for the bank that will vary depending on the rules of the table and the decks used (put table). We have to understand that even though it is <strong>an optimal strategy it doesn&#8217;t necessarily mean that it&#8217;s a winning strategy</strong>, since the balance of our advantage, despite being small, is still negative. And this is mainly due to the fact that the dealer&#8217;s high cards diminish our advantage more than the low cards, so even if we take the best decision, we will not necessarily win, although we will play in the best way possible.</p>
<p>In <strong>American Blackjack</strong> the percentages are slightly different but in no case exceed 1%, as long as you apply a basic strategy adapted to these rules, remember that <strong>each mode is different and needs a different basic strategy</strong>.</p>
<h3 class="p6">Basic Strategy Analysis</h3>
<p>If the optimal way to play only minimizes losses, <strong>is there a real way to win at blackjack?</strong></p>
<p class="p3">To answer this question we must understand what hypotheses have been made to solve this problem.</p>
<p>The <strong>basic strategy</strong> was developed taking into account that the odds of any card are <strong>invariable</strong> throughout the game, i.e. the probability that an ace, a number or a figure will appear are always the same. This is not entirely real since since we know that <strong>the odds are changing while the cards are being dealt</strong>.</p>
<p>But what would happen if I told you that we can have a system to count these cards, and anticipate more winning hands?</p>
<p><a id="conteo" name="Ancla"></a></p>
<h2 class="p3">6.Counting Cards (HI-LO system)</h2>
<p>This counting system is known as HI-Lo and was first written in 1962 by a <strong>former IBM employee mathematician</strong> in the book &#8220;Beat the dealer</p>
<p>The idea is to <strong>control the high cards in the deck</strong>, which are the only ones that can make up blackjack, so that if we know how many high cards have appeared we know if there are many or few left to appear, and so we can know what the chances are of getting blackjack in real time, with the additional advantage that this entails.</p>
<p>To know that advantage <strong>we need to have a control over the cards that are appearing</strong>, something that seems to be quite complicated if for this we need to memorize all the cards that are appearing. Fortunately, mathematics offers us a better solution.</p>
<p>The <strong>Hi-Lo</strong> system associates the cards to three different values<strong> -1, 0 and 1</strong>.</p>
<p class="p3"><a href="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_HILO_COUNTING_CARDS-1.jpg"><img decoding="async" class="aligncenter wp-image-1064 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_HILO_COUNTING_CARDS-1-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_HILO_COUNTING_CARDS-1.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_HILO_COUNTING_CARDS-1-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/BLACKJACK_HILO_COUNTING_CARDS-1-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p>1 negative to high cards, 1 positive to low cards and the remaining cards acquire a zero value or 0.</p>
<p>Our task is then to <strong>keep track of all the cards</strong> that have been dealt including the dealer&#8217;s.</p>
<p>Since in this system there is the same proportion of high and low cards, a positive or negative count tells us directly when there are no more high cards left in the deck.</p>
<p>So <strong>if our count is negative</strong> it is because we will have counted more high cards, and <strong>there will be fewer high cards left to appear</strong>, something that does not interest us.</p>
<p>On the other hand, <strong>if our count is positive</strong>, it is because we will have counted more low cards, and we will know for sure that <strong>there are more high cards to appear</strong>. It is here where we can take advantage.</p>
<p>Our objective with this system is to <strong>accumulate in that account a positive number as big as possible that offers us the biggest probabilities of obtaining blackjack</strong>. If we achieve this we can say that the table is hot and we can modify the basic strategy to our interest for example to double the bet or to bet for sure. As blackjack pays out 3 to 2 and cases of ties are unlikely we have for the first time a game system with an <strong>expected profit higher than the bank&#8217;s</strong>. And all thanks to mathematics.</p>
<p>Notice that the low cards balance out the high ones, which means that the total count once all the cards have been dealt has to be 0 obligatorily. Besides we also see that <strong>there are 3 cards that are not considered neither high nor low</strong>, this is because the system has to balance 5 high with 5 low, but in the deck there are more numbers than aces and tens so there are 3 leftover numbers that are grouped in a neutral value that does not interfere on the sum.</p>
<p>In the case of playing with <strong>several decks</strong> you have to take into account that it is not the same a sum with 4 pending decks than the same sum with 1. That&#8217;s why <strong>you have to go divide the current sum by the number of decks that we think are left to go out</strong>, and this will be the real account.</p>
<p>There are other counting methods a little more efficient but they assign more values and are a little more complicated to execute, but if you understand the HI-LO you will understand any other.</p>
<p>Although we can have an advantage with this system all this happens after <strong>thousands and thousands of hands</strong> and it gives us an advantage of tenths of percentage, so you are going to need a very big wallet and a lot of time.</p>
<h3 class="p6">Current measures against Card Counting</h3>
<p>Unfortunately, casinos have introduced measures to counteract card counters.</p>
<p>-The first one is to <strong>increase the number of decks</strong>, and that is that the more decks you use the more complicated game you will have to win by counting. Think that the fact that there are more decks pending makes it more difficult to accumulate only the best cards at the end than with a single deck, that makes the variance of the count you are carrying is smaller and consequently you have less chances.</p>
<p><strong>One: 0,17%</strong></p>
<p><strong>Two: 0.46%</strong></p>
<p><strong>Four: 0.60%</strong></p>
<p><strong>Six: 0.64%</strong></p>
<p><strong>Eight: 0.66%</strong></p>
<p>-The second measure that is currently applied are the <strong>mixers and automatic elements to shuffle</strong> the cards before the end of the game, which force the counter to restart the account that had been accumulated and therefore lose all the advantage.</p>
<p>-In addition some casinos like <strong>in Las Vegas pay 6 to 5 for blackjack</strong> which triples the house advantage over the player and even in some if they suspect that a player is a counter they can afford to modify the house bet.</p>
<p>You have to know that the counting is legal but <strong>if you are caught you will be expelled from the casino</strong>. The casinos know the profile of an accountant so it won&#8217;t be too difficult for them to discover you if you start to win money.</p>
<p><a id="conclusiones" name="Ancla"></a></p>
<h2 class="p6">7.Final Conclusions</h2>
<p>From the point of view of numbers if <strong>there is a mathematical chance of winning at blackjack</strong> that&#8217;s why many people have won money, but it&#8217;s not easy to execute, not fast, not even cheap so it&#8217;s <strong>probably not within your reach</strong>.</p>
<p>In any case <strong>if you are going to play</strong>:</p>
<ul>
<li>I recommend you to learn and adapt first the basic strategy to improve your chances.</li>
<li>Do not bet on tables with payouts lower than 3 to 2.</li>
<li>Play at tables with the smallest possible number of decks and check the table rules very well.</li>
<li>And if you are going to count cards, study how they are shuffled and test your strategy first before investing a large amount.</li>
</ul>
<p><strong>Remember that</strong>:</p>
<ul>
<li>The more decks, the more the house wins.</li>
<li>The worse the rules, the more the house wins.</li>
<li>And the less strategy or knowledge you have, the faster and safer the house wins.</li>
</ul>
<p>So the house always wins?</p>
<p>It does not always win, but it is clear that if it does not, it will change whatever is necessary to do so. Remember that this is a business.</p>
<img src="https://www.math4all.es/wp-content/uploads/2020/10/blackjack-maths.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/the-mathematics-of-blackjack/" data-pin-media="https://www.math4all.es/wp-content/uploads/2020/10/blackjack-maths.jpg" data-pin-description="The Mathematics of Blackjack"/><p>The post <a href="https://www.math4all.es/the-mathematics-of-blackjack/">The Mathematics of Blackjack</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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		<title>The Mathematics of Roulette</title>
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		<pubDate>Fri, 30 Oct 2020 17:11:41 +0000</pubDate>
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					<description><![CDATA[<p>Do you think you can win in Roulette? In this article I will analyze the game of roulette on a&#8230;</p>
<p>The post <a href="https://www.math4all.es/the-mathematics-of-roulette/">The Mathematics of Roulette</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></description>
										<content:encoded><![CDATA[<p><iframe title="&#x1f522; Las MATEMÁTICAS de la RULETA &#x1f534; &#x26ab;" width="1170" height="658" src="https://www.youtube.com/embed/euYyO3ZF_PE?feature=oembed" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe></p>
<p>Do you think you can win in Roulette? In this article I will analyze the game of roulette on a mathematical level so that you know whether or not it is possible to win in the long run. Are you ready?</p>
<p><a href="#reglas">1.Roulette Rules</a><br />
<a href="#apuestas"> 1.1.Roulette Bets</a><br />
<a href="#estudio">2.Mathematical Study of the Roulette</a><br />
<a href="#esperanzapleno"> 2.1.Expectation of Straight Up</a><br />
<a href="#gananciapleno"> 2.2.Profit Expectation of Straight Up</a><br />
<a href="#gananciaapuestas"> 2.3.Profit Expectation of All Bets</a><br />
<a href="#simulacion"> 2.4.Simulation with real Money</a><br />
<a href="#preguntas">3.Frequently Asked Questions</a><br />
<a href="#conclusiones">4.Conclusions</a></p>
<p><a id="reglas" name="Ancla"></a></p>
<h2 class="p3">1.Roulette Rules</h2>
<p class="p3">-The <strong>European Roulette</strong> is made up of <strong>37 spinning numbers</strong> and a ball is thrown which falls into one of them.</p>
<div id="attachment_1032" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2020/10/european-roulette.jpg"><img decoding="async" aria-describedby="caption-attachment-1032" class="wp-image-1032" src="https://www.math4all.es/wp-content/uploads/2020/10/european-roulette-300x200.jpg" alt="" width="1024" height="682" srcset="https://www.math4all.es/wp-content/uploads/2020/10/european-roulette-300x200.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/european-roulette.jpg 737w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1032" class="wp-caption-text">European Roulette</p></div>
<p class="p3">-Bets can be made on based on a number, or a group of numbers, and are rewarded differently.</p>
<p class="p3">&#8211;<strong>18 of the numbers are red</strong> and the other <strong>18 are black</strong> which added to 0 gives us 37.</p>
<p class="p3">-Half of these numbers are <strong>even</strong> and the other <strong>odd</strong>, except 0 or 00 which is considered neither even nor odd in this game.</p>
<p class="p3">-In <strong>American Roulette</strong> you have one more number which is <strong>double zero</strong> but in this video we will focus on European Roulette with one zero.</p>
<div id="attachment_1033" style="width: 1034px" class="wp-caption aligncenter"><a href="https://www.math4all.es/wp-content/uploads/2020/10/american-roulette.jpg"><img decoding="async" aria-describedby="caption-attachment-1033" class="wp-image-1033" src="https://www.math4all.es/wp-content/uploads/2020/10/american-roulette-300x200.jpg" alt="" width="1024" height="682" srcset="https://www.math4all.es/wp-content/uploads/2020/10/american-roulette-300x200.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/american-roulette.jpg 737w" sizes="(max-width: 1024px) 100vw, 1024px" /></a><p id="caption-attachment-1033" class="wp-caption-text">American Roulette</p></div>
<p class="p3">-The <strong>bets</strong> can be made with the roulette stopped or in movement since the croupier or the machine indicates &#8220;<strong>place your bets</strong>&#8221; and until indicates &#8220;<strong>no more bets</strong>&#8220;.</p>
<p class="p3">-Once the ball has landed on the number, those who have hit it are rewarded and the <strong>dealer</strong> keeps the remaining chips from the players who have lost their bets.</p>
<p><a id="apuestas" name="Ancla"></a></p>
<h3 class="p1">Roulette Bets</h3>
<p class="p1">Let&#8217;s see now all the <strong>bets or plays</strong> of the European roulette to which you can bet and what are their <b>probability</b> (not the odds as some pages show, odds and probability are related but are not the same):</p>
<h4 class="p3">Simple Bets</h4>
<ul>
<li class="p3"><strong>Red-Black:</strong> 1 to 1 (probability 18/37 =~ <strong>48.6%</strong>) (47.4% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Even-Odd:</strong> 1 to 1 (probability 18/37 =~ <strong>48.6%</strong>) (47.4% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Low-High (1-18/19-36):</strong><span class="Apple-converted-space">  </span>1 to 1 (probability 18/37 =~ <strong>48.6%</strong>) (47.4% American Roulette)</li>
</ul>
<h4 class="p3">Multiple Bets</h4>
<ul>
<li class="p3"><strong>Double Dozen:</strong> 1/2<span class="Apple-converted-space">  to</span> 1 (probability 24/37 =~ <strong>64.8%</strong>) (63,1% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Double Column:</strong> 1/2 to 1 (probability 24/37 =~ <strong>64.8%</strong>) (63,1% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Dozen:</strong> 2 to 1 (probability 12/37=~  <strong>32.4%</strong>) (31.6% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Column:</strong> 2 to 1 (probability 12/37 =~ <strong>32.4%</strong>) (31.6% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Line/6 number:</strong> 5 to 1 (probability 6/37 =~ <strong>16.2%</strong>) (15.8% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Corner:</strong> 8 to 1 (probability 4/37 =~ <strong>10.8%</strong>) (10.5% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Street:</strong> 11 to 1 (probability 6/37 =~ <strong>8.1%</strong>) (7.9% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Split:</strong> 17 to 1 (probability 2/37 =~ <strong>5.4%</strong>) (5.3% American Roulette)</li>
</ul>
<ul>
<li class="p3"><strong>Straight Up:</strong> 35 to 1 (probability 1/37 =~ <strong>2.7%</strong>) (2.60% American Roulette)</li>
</ul>
<p>As we have seen, both Simple and Multiple bets son realmente<strong> are the same at a mathematical level</strong>. We have also seen that if a bet is more difficult to hit it is rewarded better.</p>
<p class="p1">which raises a number of <strong>questions</strong>:</p>
<p class="p1"><strong>-The first one is: is the payout profitable?</strong></p>
<p class="p1"><strong>-If some bets are easier to hit, but the others have better payout ¿which one is best for us?</strong></p>
<p><a id="estudio" name="Ancla"></a></p>
<h2>2.Mathematical Study of Roulette</h2>
<p class="p1">If we study Roulette from the mathematical point of view we will realize that any bet is an <strong>experiment</strong> with 2 possible results <strong>success</strong> or <strong>failure</strong>, where the probability of hitting each number is the same regardless of the numbers we take, due to all the numbers have the same probability of appear.</p>
<p class="p1">Besides, <strong>each game does not depend on the previous one</strong> and it doesn&#8217;t matter the number that came out before.</p>
<p class="p1">In this ideal scenario, the number of hits for a game series, follow a <strong>Binomial n,p distribution</strong> where n is the game numbers and p the probability of win the bet.</p>
<p class="p1">The probability of hitting x times in n games is n combined with x, by p to the x, by (1-p) to the n-x, where <strong>p is the probability of winning</strong> and (1-p) the probability of loosing.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/binomial-formula.jpg"><img decoding="async" class="aligncenter wp-image-1048 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/binomial-formula-1024x575.jpg" alt="" width="1024" height="575" srcset="https://www.math4all.es/wp-content/uploads/2020/10/binomial-formula.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/binomial-formula-300x168.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/binomial-formula-768x431.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p1">With this we can calculate the probability of hitting 0,1,2 times or more&#8230;</p>
<p class="p1">But to know if a bet is profitable we have to know what is the <strong>average profit</strong> we expect from that bet, so we need to define first the concept of <strong>mathematical Expectation</strong>.</p>
<p class="p1">The expectation for a binomial variable is the number of average hits for n experiments and is calculated as p∙n, in our case it will be the average <strong>number of bets won by playing n times</strong>.</p>
<p><a id="esperanzapleno" name="Ancla"></a></p>
<h3 class="p3">Expectation of Straight Up</h3>
<p class="p1">Then the expectation to hit 1 number playing <strong>1 time/game</strong> is 1∙(1/37) which is approximately 0.027, that is to say that <strong>playing once we are expecting to hit 2.7% on average</strong>.</p>
<p class="p1">For <strong>2 games</strong> will be the same but by 2, which is the <strong>5.4%</strong> of hitting.</p>
<p class="p1">For <strong>3 games</strong>, for 3 games it is the same but by 3,… and thus we can calculate the Expectation for the games that we want.</p>
<p class="p1">Notice now that for <strong>37 games</strong> the expectation is 1, that is means that <strong>playing a Straight 37 times we expect to win once on average</strong>, which some of you may have sensed.</p>
<p><a id="gananciapleno" name="Ancla"></a></p>
<h3 class="p3">Profit Expectation of Straight Up</h3>
<p class="p1">Now that we know how to calculate the average number of hits for each bet, and the payout, calculating the <strong>expected profit</strong> for a bet is simple, as it is easy to calculate the average profit and loss.</p>
<p class="p1"><strong>Expected profit = Expected benefit  &#8211; Expected loss</strong></p>
<p>where:</p>
<p class="p1">Expected benefit = Expectation(of the bet) · payout · bet</p>
<p class="p1">Expected loss = Loss Expectation(of the bet) · payout · bet</p>
<p class="p1">So the <strong>expected profit for a Straight in 1 game</strong>:</p>
<p class="p1">is equal to the expectation for a 1 game · payout · bet less the loss expectation for a 1 game · bet:</p>
<p class="p1">This is 1/37 · 35 (the payout) · bet &#8211; 36/37 · bet, which is about <strong>&#8211;</strong>0.027, In other words, <strong>playing once you expect to lose on average the 2.7% of your bet</strong>.</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/expected-profit.jpg"><img decoding="async" class="aligncenter wp-image-1049 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/expected-profit-1024x576.jpg" alt="" width="1024" height="576" srcset="https://www.math4all.es/wp-content/uploads/2020/10/expected-profit.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/expected-profit-300x169.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/expected-profit-768x432.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p1">and logically the <strong>expected margin of the casino</strong> or house benefit will be <strong>2.7% of your bet, every time you play</strong>.</p>
<p>If you calculate it for <strong>2 games</strong> the expected profit is <strong>double negative</strong> so the casino margin is up to <strong>5.4%</strong>.</p>
<p class="p1">Then what if we play 37 times?, As we have seen we hope to win once but&#8230; what&#8217;s the benefit?</p>
<p>In <strong>37 games</strong> the winning expectation is 1 and the hope of losing is logically 36, so if you do the calculations you&#8217;ll see that the profit is -1 · bet. This means that in 37 games <strong>the casino expects to win on average the value of your bet</strong>.</p>
<p class="p1">This means that <strong>if you play a Straight 37 times betting, for example, 1 dollar each time you expect to lose an average of 1 dollar</strong>.</p>
<p><a id="gananciaapuestas" name="Ancla"></a></p>
<h3 class="p3">Profit Expectation for All Bets</h3>
<p>As you&#8217;ve seen, playing a Straight doesn&#8217;t seem like a very profitable bet, as you lose money in the short and long term. The more money you play, the more money you lose, and you need an average of 37 games to hit only once, but when you do, they pay you with 35 chips which is less money than you spend. Then it doesn&#8217;t seem to make much sense to play a single number.</p>
<p>Como vemos, jugar a pleno, además de ser difícil de acertar nos genera pérdidas a corto y a largo plazo. a más jugamos más dinero perdemos y necesitamos jugar en promedio 37 veces para acertar, pero cuando lo hacemos nos pagan 35 que es menos de lo que hemos perdido. Por lo que no parece una buena estrategia jugar a un solo número todo el rato.</p>
<p class="p1">The question is, is there a better bet than the Straight?</p>
<p>For this, you need to know what is the <strong>profit margin for all the bets</strong>, then I&#8217;m going to show you some results that I have calculated on my own, so let&#8217;s go with them!</p>
<h4 class="p3">Simple Bets</h4>
<ul>
<li class="p3"><strong>Red-Black</strong> = (18/37∙1)Ap-(18/37∙1)Ap-(1/37∙1/2)Ap = 18/37Ap-19/37Ap = -1/74Ap =~<span class="Apple-converted-space">  <strong>&#8211;</strong></span><strong>1.35% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Even-Odd</strong> = (18/37∙1)Ap-(18/37∙1)Ap-(1/37∙1/2)Ap = 18/37Ap-19/37Ap = -1/74Ap =~<span class="Apple-converted-space">  <strong>&#8211;</strong></span><strong>1.35% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Low-High</strong> = (18/37∙1)Ap-(18/37∙1)Ap-(1/37∙1/2)Ap = 18/37Ap-19/37Ap = -1/74Ap =~<span class="Apple-converted-space">  <strong>&#8211;</strong></span><strong>1.35% of bet</strong></li>
</ul>
<h4 class="p3">Multiple Bets</h4>
<ul>
<li class="p3"><strong>Double Column</strong> = (24/37∙0.5)Ap-(13/37∙1)Ap = 12/37Ap-13/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~<span class="Apple-converted-space">  <strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Double Dozen</strong> = (24/37∙0.5)Ap-(13/37∙1)Ap = 12/37Ap-13/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~<span class="Apple-converted-space"><span class="Apple-converted-space"> </span></span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Column</strong> = (12/37∙2)Ap-(25/37∙1)Ap = 24/37A-25/37Ap = -1/37Ap<span class="Apple-converted-space">    </span>=~<span class="Apple-converted-space">  </span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Dozen</strong> = (12/37∙2)Ap-(25/37∙1)Ap = 24/37A-25/37Ap = -1/37Ap<span class="Apple-converted-space">    </span>=~<span class="Apple-converted-space"><span class="Apple-converted-space"> </span></span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Line/6 number</strong> = (6/37∙5)Ap-(31/37∙1)Ap = 30/37Ap-31/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~<span class="Apple-converted-space"><span class="Apple-converted-space"> </span></span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Corner</strong> = (4/37∙8)Ap-(33/37∙1)Ap <span class="Apple-converted-space">  </span>= 32/37Ap-33/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~<span class="Apple-converted-space"><span class="Apple-converted-space"> </span></span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Street</strong> = (2/37∙17)Ap-(35/37∙1)Ap = 34/37Ap-35/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~<span class="Apple-converted-space"><span class="Apple-converted-space"> </span></span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><strong>Split</strong> = (3/37∙11)Ap-(34/37∙1)Ap = 33/37Ap-34/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~ <span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<ul>
<li class="p3"><span class="Apple-converted-space"><strong>Straight Up</strong> </span>= (1/37∙35)Ap-(36/37∙1)Ap = 35/37Ap-36/37Ap = -1/37Ap<span class="Apple-converted-space">  </span>=~<span class="Apple-converted-space"><span class="Apple-converted-space"> </span></span><span class="Apple-converted-space"><strong>&#8211;</strong></span><strong>2.70% of bet</strong></li>
</ul>
<p>To know what is the casino margin for all bets I have built a <strong>spreadsheet</strong> and applied the formula you&#8217;ve seen for all of them, taking the probability and the payout of each one, to see what is the profit playing once or more times. The most important result obtained is that any <strong>simple or multiple bet has the same expected profit</strong> in the European Roulette, and the same profit always in the American Roulette. This is because in the European Roulette in the simple bets, if it comes out 0 we lose only the half of the bet and that makes the winnings better. In this way, I have been able to calculate some results. I have caught my attention and that you are going to see next.</p>
<p>&nbsp;</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/spreadsheet-roulette.jpg"><img decoding="async" class="aligncenter wp-image-1038 size-full" src="https://www.math4all.es/wp-content/uploads/2020/10/spreadsheet-roulette.jpg" alt="" width="1024" height="596" srcset="https://www.math4all.es/wp-content/uploads/2020/10/spreadsheet-roulette.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/spreadsheet-roulette-300x175.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/spreadsheet-roulette-768x447.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p style="text-align: center;"><a href="http://www.math4all.es/ruleta/roulette_bets(math4all).xlsx">DOWNLOAD SPREADSHEET</a></p>
<p>-We see that simple bets are better for us than a Straight or any multiple bet.</p>
<p>-We see that within the single or multiple ones <strong>it doesn&#8217;t matter the bet you take</strong>, if you take two dozens, you will win more times but when you lose you will also lose more, and if you bet a Straight, the payout is bigger, but it will cost you more money to bet it than the payout you will get, that&#8217;s why <strong>the house margin is the same for all the bets</strong>.</p>
<p><a id="simulacion" name="Ancla"></a></p>
<h3>Simulation with real Money</h3>
<p class="p1"><span class="s1">Here&#8217;s a simple simulation what would happen with real money:</span></p>
<ul>
<li class="p1"><span class="s1"><strong>1 game betting 1 dollar/euro</strong>:<strong> </strong>you will lose on average <strong>1 cent</strong> in the simple ones and almost <strong>3 cents</strong> in the multiple ones (this is because in the simple bets if the 0 comes out, you lose only half of the bet and you have to contemplate that possibility)</span></li>
</ul>
<ul>
<li><span class="s1"><strong>1 game betting 20 dollars/euros</strong>:<strong> </strong>you will lose <strong>27 cents</strong> in the simple ones and <strong>54 cents</strong> in the multiple ones, which means that the more money you bet, the more money you lose.</span></li>
</ul>
<ul>
<li class="p1"><span class="s1"><strong>20 games betting 1 dollar/euro per game</strong>: you lose again <strong>27 and 54 cents</strong>. (since it is the same to play 1 time betting 20 dollas that to play 20 times betting 1 dollar), although you have to keep in mind that playing 20 times, your real average will be more similar to those amounts than playing once which means <strong>the more times you play the more sure you&#8217;ll lose</strong>.</span></li>
</ul>
<ul>
<li class="p1"><span class="s1"><strong>37 games with any amount</strong>: you will lose on average that <strong>total amount</strong> in any of the multiple ones and <strong>half</strong> that amount in the simple ones. That is to say, that if you substitute the x by any number you will know how much you are going to lose in average, whatever the bet is.</span></li>
</ul>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/profits_roulette.jpg"><img decoding="async" class="aligncenter wp-image-1050 size-large" src="https://www.math4all.es/wp-content/uploads/2020/10/profits_roulette-1024x575.jpg" alt="" width="1024" height="575" srcset="https://www.math4all.es/wp-content/uploads/2020/10/profits_roulette.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/profits_roulette-300x168.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/profits_roulette-768x431.jpg 768w" sizes="(max-width: 1024px) 100vw, 1024px" /></a></p>
<p class="p1"><span class="s1">Keep in mind that these results are for the <strong>European Roulette</strong> but the American Roulette is even worse because the bets are rewarded the same but you are less likely to get it right because there is one more number. And if you get 0 or 00, you&#8217;ll lose everything in this case.</span></p>
<p><a id="preguntas" name="Ancla"></a></p>
<h2 class="p1">3.Frequently Asked Questions</h2>
<p class="p1">I have gathered the most interesting questions for a casino player, and thus solved some of your doubts:</p>
<p class="p1">QUESTION: <strong><em>If there have been many reds in a row, is it more likely that a black will come out on the next roll?</em></strong></p>
<p class="p1">ANSWER: NO, in fact it is a very common mistake of the players. You have to take into account that roulette is a physical mechanism, which does not remember the results that happend before. That&#8217;s why each game is independent from the previous one, and what has to come out will come out, but not the opposite of what has come out before just because we are waiting for it. This is why it is said that roulette has no memory. It is important that you learn to calculate the probabilities correctly. Before playing, many reds or many blacks in a row has a small probability, but once that has happened, what comes next has no more probability or less because it is the opposite, in fact still has the same probability.</p>
<p class="p1">QUESTION: <strong><em>What happens if I play several bets at once, such as several numbers at the same time?</em></strong></p>
<p class="p1">ANSWER: Combining bets means that you make several bets at the same time, but it doesn&#8217;t mean that they are related, in fact each bet is paid separately. So if you play several bets, the house always pays each bet separately. At a mathematical level if you cover more numbers your probability can increase, but not your profitability. Being independent bets you are making many bets that are not profitable at the same time, and in the end that is even less profitable. Remember that if you want to take a group, a single play also groups numbers and has less margin.</p>
<p class="p1">QUESTION: <strong><em>What if I play to double my bet every time I lose?</em></strong></p>
<p class="p1">ANSWER: Doubling the red or black when lost (or any other bet) is known as &#8220;martingale&#8221;, and has also been studied mathematically. The first thing you should take into account is how many chips you have, and how many times you can afford to double your bet. Keep in mind that doubling implies betting twice as much as you previously lost each time. If you lose a chip, you have to bet 2, then 4, then 8, 16, 32&#8230;. and this is an exponential loss growth, which can make you lose all your chips in a relatively short streak of bad luck. The more you maximize the number of times you can double the more you can hold out for the next double, but this can only be done in two ways: either by having a lot of money, or by betting very low value chips. In either case you could win most of the times you visit a casino, but just because you win most of the times doesn&#8217;t mean that your balance will be positive in time. Remember that as long as you win you will keep playing. The problem is that mathematically there is a combination of reds or blacks&#8230; in a row that exceeds the limit of times you can double with the capital you have (because nobody has infinite money). Surely you think it&#8217;s almost impossible to get 10 or 15 reds in a row, and you&#8217;re right, it&#8217;s quite difficult, but not impossible. The problem is that by playing longer you are increasing the chances of that combination manifesting itself. The more you play, the more likely it is that this will happen, and if it does, you will need to double with money that is no longer of benefit so you can lose all the benefit you had accumulated from playing this and much more. Keep in mind that before playing it is very difficult to get 10 reds in a row, but once you get 9 reds, getting another one is quite simple, because roulette has no memory. Believe me, from experience I know that it is not so complicated to get those strips of red and black in a row, so be very careful with the martingale.</p>
<p class="p1">QUESTION: <strong><em>You said earlier that the margin is the same for all multiple plays, but is there a difference?</em></strong></p>
<p class="p1">ANSWER: Actually yes, if you choose a bet difficult to hit like a Straight, you will lose more often and you will need to keep playing to reach the payout or to recover the money you have lost, with which you will play more times on average, and the more you play the more you lose.</p>
<p class="p1">QUESTION: <em><strong>And what if I have few chips to bet, what bet would you recommend?</strong></em></p>
<p class="p1">ANSWER: If you have few chips, never take the Straight or any low probability bet, because you won&#8217;t have enough chips to cover the average number of shots until you hit. Remember that for a full house you need 37 shots to get an average hit, if you don&#8217;t have enough chips most of the time you will leave before hitting one. If you have few chips, make single or probable bets.</p>
<p class="p1">QUESTION: <strong><em>Something that has surprised me is that playing a game with 20 dollars is the same as 20 games with 1 dollar, because I like to play, and I don&#8217;t understand why I was going to play only 1 game of 20 dollars being able to play 20 and spend more time playing with the same money?</em></strong></p>
<p class="p1">ANSWER: According to this formula, yes, you will earn the same on average, but there is one thing you have overlooked and that is that the average is a statistic, that means the more games you make the closer your real profit is to what we have calculated. Playing only once you will win or lose but that benefit will be very different from the calculated, instead by statistics playing many times you will get closer to that number. That&#8217;s why the casino already gives you a series of chips for those 20 dollars, so you can play several times with that amount. Because the more you play, the more you let mathematics do its work.</p>
<p class="p1">QUESTION: <strong><em>So by playing those 20 dollars from chip to chip I&#8217;m going to lose more money?</em></strong></p>
<p class="p1">ANSWER: Not more money but more surely, you will get closer to that number which is negative. But that also makes it harder to win or lose more than that amount. Playing in the long term corrects any positive or negative deviations you may have and brings you closer to the expected value, that means a more stable value, although I remind you that it is negative, so the more you play the harder it will be to win.</p>
<p class="p1">QUESTION: <strong><em>Are you telling me that the casino is interested in you playing more times even if it&#8217;s the same amount?</em></strong></p>
<p class="p1">ANSWER: That&#8217;s exactly what I&#8217;m telling you, playing more times will make it harder for luck to benefit you.</p>
<p class="p1">QUESTION: <em><strong>So if I want to win I have to put the 20 dollars on one bet?</strong></em></p>
<p class="p1">ANSWER: If you want to opt to win yes, but if you play the 20 dollars to a single bet you are going to win or lose everything, because you only have one bet, If you want to win (or lose) the same amount but playing the chips one by one, you are going to have to win all the times (or lose all the times), because for a same roll a prize of 20 dollars is the same as 4 prizes of 5. But it is not the same to hit (or miss) once, as to hit (or miss) 4 times in a row.</p>
<p class="p1">QUESTION: <em><strong>Good but I can also lose more money, because it is also easier to fail 1 time than to fail 4.</strong></em></p>
<p class="p1">ANSWER: If you take risks you can win more or lose more. And the other way you will lose more surely. (statistically speaking)</p>
<p><a id="conclusiones" name="Ancla"></a></p>
<h2 class="p3">8. Final Conclusions</h2>
<p class="p3">it makes <strong>no sense to play Roulette to win money.</strong></p>
<p class="p3">-The casino doesn&#8217;t need to cheat, just <strong>the numbers aren&#8217;t in your favor</strong></p>
<p class="p3">-It&#8217;s only possible to <strong>win at Roulette</strong> if you know some physical phenomenon, or some statistical data of that roulette that can change the probability of some numbers in your favor. . For that you need a study of thousands and thousands of numbers and an acceptable mathematical certainty. In addition, you must write down the appearances of each number in a file, so perhaps this <strong>template</strong> could help you with this:</p>
<p>&nbsp;</p>
<p><a href="https://www.math4all.es/wp-content/uploads/2020/10/roulette-template.jpg"><img decoding="async" class="wp-image-1040 aligncenter" src="https://www.math4all.es/wp-content/uploads/2020/10/roulette-template-300x298.jpg" alt="" width="720" height="716" srcset="https://www.math4all.es/wp-content/uploads/2020/10/roulette-template-300x298.jpg 300w, https://www.math4all.es/wp-content/uploads/2020/10/roulette-template-150x150.jpg 150w, https://www.math4all.es/wp-content/uploads/2020/10/roulette-template-768x763.jpg 768w, https://www.math4all.es/wp-content/uploads/2020/10/roulette-template-1024x1018.jpg 1024w, https://www.math4all.es/wp-content/uploads/2020/10/roulette-template-180x180.jpg 180w, https://www.math4all.es/wp-content/uploads/2020/10/roulette-template.jpg 1299w" sizes="(max-width: 720px) 100vw, 720px" /></a></p>
<p style="text-align: center;"><a href="http://www.math4all.es/ruleta/roulette_template(math4all).xlsx">DOWNLOAD TEMPLATE</a></p>
<p>&nbsp;</p>
<p>But beware, if you see that some number comes out more times, the most likely is that it is a product of luck and not that this number is actually more probable. so you will need thousands of prints, and an <strong>acceptable statistical certainty</strong>. Keep in mind that casinos always fight to randomize their mechanisms as much as possible. But as curiosity you can make a study of many games to see it.</p>
<p class="p3">In any case, if you are going to play <strong>I recommend you</strong>:</p>
<ul>
<li>Make only simple and probable bets.</li>
<li>Be fleeting and make few bets.</li>
<li>Do not combine bets.</li>
<li>Limit the money you will play before entering, and never exceed that amount.</li>
<li>Retire on time (in this last one believe me).</li>
</ul>
<p class="p3">Keep in mind that <strong>in the long run</strong>:</p>
<ul>
<li class="p3">The more money you play, the more the house wins.</li>
<li class="p3">The more you combine, the more the house wins.</li>
<li class="p3">The more you play, the more you are sure and the more the house wins.</li>
</ul>
<p>So the casino always wins?</p>
<p>It doesn&#8217;t always win, but if you go back it will end up winning.</p>
<p>Remember that the game never ends.</p>
<p>If you want to see more videos like <strong>Poker</strong> or <strong>Blackjack</strong> you can visit my <a href="https://www.youtube.com/channel/UC9NzA04qApji9FO63msr_-w">math for all</a> channel where I study them in detail.</p>
<img src="https://www.math4all.es/wp-content/uploads/2020/10/roulette-maths.jpg" class="ss-hidden-pin-image" alt="" data-pin-url="https://www.math4all.es/the-mathematics-of-roulette/" data-pin-media="https://www.math4all.es/wp-content/uploads/2020/10/roulette-maths.jpg" data-pin-description="The Mathematics of Roulette"/><p>The post <a href="https://www.math4all.es/the-mathematics-of-roulette/">The Mathematics of Roulette</a> first appeared on <a href="https://www.math4all.es">Math4all</a>.</p>]]></content:encoded>
					
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