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The Laws of Probability
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Bayes’s Theorem
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Probability Mass Functions
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Prediction
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Joint Probability
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The Price Is Right
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101 Bowls
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Two cookies
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Three cookies
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Probability Mass Functions
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The dice problem
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<div class="section" id="probability-mass-functions">
<h1>Probability Mass Functions<a class="headerlink" href="#probability-mass-functions" title="Permalink to this headline">¶</a></h1>
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<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">pandas</span> <span class="k">as</span> <span class="nn">pd</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="k">as</span> <span class="nn">np</span>
</pre></div>
</div>
</div>
</div>
<div class="section" id="review">
<h2>Review<a class="headerlink" href="#review" title="Permalink to this headline">¶</a></h2>
<p><a class="reference external" href="https://colab.research.google.com/github/AllenDowney/BiteSizeBayes/blob/master/05_test.ipynb">In the previous notebook</a> we used a Bayes table to interpret medical tests.</p>
<p>In this notebook we’ll solve an expanded version of the cookie problem with 101 Bowls. It might seem like a silly problem, but it’s not: the solution demonstrates a Bayesian way to estimate a proportion, and it applies to lots of real problems that don’t involve cookies.</p>
<p>Then I’ll introduce an alternative to the Bayes table, a probability mass function (PMF), which is a useful way to represent and do computations with distributions.</p>
<p>Here’s the function, from the previous notebook, we’ll use to make Bayes tables:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">pandas</span> <span class="k">as</span> <span class="nn">pd</span>
<span class="k">def</span> <span class="nf">make_bayes_table</span><span class="p">(</span><span class="n">hypos</span><span class="p">,</span> <span class="n">prior</span><span class="p">,</span> <span class="n">likelihood</span><span class="p">):</span>
<span class="sd">"""Make a Bayes table.</span>
<span class="sd"> </span>
<span class="sd"> hypos: sequence of hypotheses</span>
<span class="sd"> prior: prior probabilities</span>
<span class="sd"> likelihood: sequence of likelihoods</span>
<span class="sd"> </span>
<span class="sd"> returns: DataFrame</span>
<span class="sd"> """</span>
<span class="n">table</span> <span class="o">=</span> <span class="n">pd</span><span class="o">.</span><span class="n">DataFrame</span><span class="p">(</span><span class="n">index</span><span class="o">=</span><span class="n">hypos</span><span class="p">)</span>
<span class="n">table</span><span class="p">[</span><span class="s1">'prior'</span><span class="p">]</span> <span class="o">=</span> <span class="n">prior</span>
<span class="n">table</span><span class="p">[</span><span class="s1">'likelihood'</span><span class="p">]</span> <span class="o">=</span> <span class="n">likelihood</span>
<span class="n">table</span><span class="p">[</span><span class="s1">'unnorm'</span><span class="p">]</span> <span class="o">=</span> <span class="n">table</span><span class="p">[</span><span class="s1">'prior'</span><span class="p">]</span> <span class="o">*</span> <span class="n">table</span><span class="p">[</span><span class="s1">'likelihood'</span><span class="p">]</span>
<span class="n">prob_data</span> <span class="o">=</span> <span class="n">table</span><span class="p">[</span><span class="s1">'unnorm'</span><span class="p">]</span><span class="o">.</span><span class="n">sum</span><span class="p">()</span>
<span class="n">table</span><span class="p">[</span><span class="s1">'posterior'</span><span class="p">]</span> <span class="o">=</span> <span class="n">table</span><span class="p">[</span><span class="s1">'unnorm'</span><span class="p">]</span> <span class="o">/</span> <span class="n">prob_data</span>
<span class="k">return</span> <span class="n">table</span>
</pre></div>
</div>
</div>
</div>
</div>
<div class="section" id="bowls">
<h2>101 Bowls<a class="headerlink" href="#bowls" title="Permalink to this headline">¶</a></h2>
<p>In <a class="reference external" href="https://colab.research.google.com/github/AllenDowney/BiteSizeBayes/blob/master/05_dice.ipynb">Notebook 4</a>, we saw that the Bayes table works with more than two hypotheses. As an example, we solved a cookie problem with five bowls.</p>
<p>Now we’ll take it even farther and solve a cookie problem with 101 bowls:</p>
<ul class="simple">
<li><p>Bowl 0 contains no vanilla cookies,</p></li>
<li><p>Bowl 1 contains 1% vanilla cookies,</p></li>
<li><p>Bowl 2 contains 2% vanilla cookies,</p></li>
</ul>
<p>and so on, up to</p>
<ul class="simple">
<li><p>Bowl 99 contains 99% vanilla cookies, and</p></li>
<li><p>Bowl 100 contains all vanilla cookies.</p></li>
</ul>
<p>As in the previous problems, there are only two kinds of cookies, vanilla and chocolate. So Bowl 0 is all chocolate cookies, Bowl 1 is 99% chocolate, and so on.</p>
<p>Suppose we choose a bowl at random, choose a cookie at random, and it turns out to be vanilla. What is the probability that the cookie came from Bowl <span class="math notranslate nohighlight">\(x\)</span>, for each value of <span class="math notranslate nohighlight">\(x\)</span>?</p>
<p>To solve this problem, I’ll use <code class="docutils literal notranslate"><span class="pre">np.arange</span></code> to represent 101 hypotheses, numbered from 0 to 100.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">numpy</span> <span class="k">as</span> <span class="nn">np</span>
<span class="n">xs</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="mi">101</span><span class="p">)</span>
</pre></div>
</div>
</div>
</div>
<p>The prior probability for each bowl is <span class="math notranslate nohighlight">\(1/101\)</span>. I could create a sequence with 101 identical values, but if all of the priors are equal, we only have to probide one value:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">prior</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="mi">101</span>
</pre></div>
</div>
</div>
</div>
<p>Because of the way I numbered the bowls, the probability of a vanilla cookie from Bowl <span class="math notranslate nohighlight">\(x\)</span> is <span class="math notranslate nohighlight">\(x/100\)</span>. So we can compute the likelihoods like this:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">likelihood</span> <span class="o">=</span> <span class="n">xs</span><span class="o">/</span><span class="mi">100</span>
</pre></div>
</div>
</div>
</div>
<p>And that’s all we need; the Bayes table does the rest:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">table</span> <span class="o">=</span> <span class="n">make_bayes_table</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span> <span class="n">prior</span><span class="p">,</span> <span class="n">likelihood</span><span class="p">)</span>
</pre></div>
</div>
</div>
</div>
<p>Here’s a feature we have not seen before: we can give the index of the Bayes table a name, which will appear when we display the table.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">table</span><span class="o">.</span><span class="n">index</span><span class="o">.</span><span class="n">name</span> <span class="o">=</span> <span class="s1">'Bowl'</span>
</pre></div>
</div>
</div>
</div>
<p>Here are the first few rows:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">table</span><span class="o">.</span><span class="n">head</span><span class="p">()</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<div class="output text_html"><div>
<style scoped>
.dataframe tbody tr th:only-of-type {
vertical-align: middle;
}
.dataframe tbody tr th {
vertical-align: top;
}
.dataframe thead th {
text-align: right;
}
</style>
<table border="1" class="dataframe">
<thead>
<tr style="text-align: right;">
<th></th>
<th>prior</th>
<th>likelihood</th>
<th>unnorm</th>
<th>posterior</th>
</tr>
<tr>
<th>Bowl</th>
<th></th>
<th></th>
<th></th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<th>0</th>
<td>0.009901</td>
<td>0.00</td>
<td>0.000000</td>
<td>0.000000</td>
</tr>
<tr>
<th>1</th>
<td>0.009901</td>
<td>0.01</td>
<td>0.000099</td>
<td>0.000198</td>
</tr>
<tr>
<th>2</th>
<td>0.009901</td>
<td>0.02</td>
<td>0.000198</td>
<td>0.000396</td>
</tr>
<tr>
<th>3</th>
<td>0.009901</td>
<td>0.03</td>
<td>0.000297</td>
<td>0.000594</td>
</tr>
<tr>
<th>4</th>
<td>0.009901</td>
<td>0.04</td>
<td>0.000396</td>
<td>0.000792</td>
</tr>
</tbody>
</table>
</div></div></div>
</div>
<p>Because Bowl 0 contains no vanilla cookies, its likelihood is 0, so its posterior probability is 0. That is, the cookie cannot have come from Bowl 0.</p>
<p>Here are the last few rows of the table.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">table</span><span class="o">.</span><span class="n">tail</span><span class="p">()</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<div class="output text_html"><div>
<style scoped>
.dataframe tbody tr th:only-of-type {
vertical-align: middle;
}
.dataframe tbody tr th {
vertical-align: top;
}
.dataframe thead th {
text-align: right;
}
</style>
<table border="1" class="dataframe">
<thead>
<tr style="text-align: right;">
<th></th>
<th>prior</th>
<th>likelihood</th>
<th>unnorm</th>
<th>posterior</th>
</tr>
<tr>
<th>Bowl</th>
<th></th>
<th></th>
<th></th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<th>96</th>
<td>0.009901</td>
<td>0.96</td>
<td>0.009505</td>
<td>0.019010</td>
</tr>
<tr>
<th>97</th>
<td>0.009901</td>
<td>0.97</td>
<td>0.009604</td>
<td>0.019208</td>
</tr>
<tr>
<th>98</th>
<td>0.009901</td>
<td>0.98</td>
<td>0.009703</td>
<td>0.019406</td>
</tr>
<tr>
<th>99</th>
<td>0.009901</td>
<td>0.99</td>
<td>0.009802</td>
<td>0.019604</td>
</tr>
<tr>
<th>100</th>
<td>0.009901</td>
<td>1.00</td>
<td>0.009901</td>
<td>0.019802</td>
</tr>
</tbody>
</table>
</div></div></div>
</div>
<p>The posterior probabilities are substantially higher for the high-numbered bowls.</p>
<p>There is a pattern here that will be clearer if we plot the results.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="k">as</span> <span class="nn">plt</span>
<span class="k">def</span> <span class="nf">plot_table</span><span class="p">(</span><span class="n">table</span><span class="p">):</span>
<span class="sd">"""Plot results from the 101 Bowls problem.</span>
<span class="sd"> </span>
<span class="sd"> table: DataFrame representing a Bayes table</span>
<span class="sd"> """</span>
<span class="n">table</span><span class="p">[</span><span class="s1">'prior'</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">()</span>
<span class="n">table</span><span class="p">[</span><span class="s1">'posterior'</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">()</span>
<span class="n">plt</span><span class="o">.</span><span class="n">xlabel</span><span class="p">(</span><span class="s1">'Bowl #'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">ylabel</span><span class="p">(</span><span class="s1">'Probability'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">legend</span><span class="p">()</span>
</pre></div>
</div>
</div>
</div>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">plot_table</span><span class="p">(</span><span class="n">table</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">title</span><span class="p">(</span><span class="s1">'One cookie'</span><span class="p">);</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<img alt="_images/06_pmf_24_0.png" src="_images/06_pmf_24_0.png" />
</div>
</div>
<p>The prior probabilities are uniform; that is, they are the same for every bowl.</p>
<p>The posterior probabilities increase linearly; Bowl 0 is the least likely (actually impossible), and Bowl 100 is the most likely.</p>
</div>
<div class="section" id="two-cookies">
<h2>Two cookies<a class="headerlink" href="#two-cookies" title="Permalink to this headline">¶</a></h2>
<p>Suppose we put the first cookie back, stir the bowl thoroughly, and draw another cookie from the same bowl. and suppose it turns out to be another vanilla cookie.</p>
<p>Now what is the probability that we are drawing from Bowl <span class="math notranslate nohighlight">\(x\)</span>?</p>
<p>To answer this question, we can use the posterior probabilities from the previous problem as prior probabilities for a new Bayes table, and then update with the new data.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">prior2</span> <span class="o">=</span> <span class="n">table</span><span class="p">[</span><span class="s1">'posterior'</span><span class="p">]</span>
<span class="n">likelihood2</span> <span class="o">=</span> <span class="n">likelihood</span>
<span class="n">table2</span> <span class="o">=</span> <span class="n">make_bayes_table</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span> <span class="n">prior2</span><span class="p">,</span> <span class="n">likelihood2</span><span class="p">)</span>
<span class="n">plot_table</span><span class="p">(</span><span class="n">table2</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">title</span><span class="p">(</span><span class="s1">'Two cookies'</span><span class="p">);</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<img alt="_images/06_pmf_27_0.png" src="_images/06_pmf_27_0.png" />
</div>
</div>
<p>The blue line shows the posterior after one cookie, which is the prior before the second cookie.</p>
<p>The orange line shows the posterior after two cookies, which curves upward. Having see two vanilla cookies, the high-numbered bowls are more likely; the low-numbered bowls are less likely.</p>
<p>I bet you can guess what’s coming next.</p>
</div>
<div class="section" id="three-cookies">
<h2>Three cookies<a class="headerlink" href="#three-cookies" title="Permalink to this headline">¶</a></h2>
<p>Suppose we put the cookie back, stir, draw another cookie from the same bowl, and get a chocolate cookie.</p>
<p>What do you think the posterior distribution looks like after these three cookies?</p>
<p>Hint: what’s the probability that the chocolate cookie came from Bowl 100?</p>
<p>We’ll use the posterior after two cookies as the prior for the third cookie:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">prior3</span> <span class="o">=</span> <span class="n">table2</span><span class="p">[</span><span class="s1">'posterior'</span><span class="p">]</span>
</pre></div>
</div>
</div>
</div>
<p>Now, what about the likelihoods? Remember that the probability of a vanilla cookie from Bowl <span class="math notranslate nohighlight">\(x\)</span> is <span class="math notranslate nohighlight">\(x/100\)</span>. So the probability of a chocolate cookie is <span class="math notranslate nohighlight">\((1 - x/100)\)</span>, which we can compute like this.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">likelihood3</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">-</span> <span class="n">xs</span><span class="o">/</span><span class="mi">100</span>
</pre></div>
</div>
</div>
</div>
<p>That’s it. Everything else is the same.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">table3</span> <span class="o">=</span> <span class="n">make_bayes_table</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span> <span class="n">prior3</span><span class="p">,</span> <span class="n">likelihood3</span><span class="p">)</span>
</pre></div>
</div>
</div>
</div>
<p>And here are the results</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">plot_table</span><span class="p">(</span><span class="n">table3</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">title</span><span class="p">(</span><span class="s1">'Three cookies'</span><span class="p">);</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<img alt="_images/06_pmf_36_0.png" src="_images/06_pmf_36_0.png" />
</div>
</div>
<p>The blue line is the posterior after two cookies; the orange line is the posterior after three cookies.</p>
<p>Because Bowl 100 contains no chocolate cookies, the posterior probability for Bowl 100 is 0.</p>
<p>The posterior distribution has a peak near 60%. We can use <code class="docutils literal notranslate"><span class="pre">idxmax</span></code> to find it:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">table3</span><span class="p">[</span><span class="s1">'posterior'</span><span class="p">]</span><span class="o">.</span><span class="n">idxmax</span><span class="p">()</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>67
</pre></div>
</div>
</div>
</div>
<p>The peak in the posterior distribution is at 67%.</p>
<p>This value has a name; it is the <strong>MAP</strong>, which stands for “Maximum Aposteori Probability” (“aposteori” is Latin for posterior).</p>
<p>In this example, the MAP is close to the proportion of vanilla cookies in the dataset: 2/3.</p>
<p><strong>Exercise:</strong> Let’s do a version of the dice problem where we roll the die more than once. Here’s the statement of the problem again:</p>
<blockquote>
<div><p>Suppose you have a 4-sided, 6-sided, 8-sided, and 12-sided die. You choose one at random, roll it and get a 1. What is the probability that the die you rolled is 4-sided? What are the posterior probabilities for the other dice?</p>
</div></blockquote>
<p>And here’s a solution using a Bayes table:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">hypos</span> <span class="o">=</span> <span class="p">[</span><span class="s1">'H4'</span><span class="p">,</span> <span class="s1">'H6'</span><span class="p">,</span> <span class="s1">'H8'</span><span class="p">,</span> <span class="s1">'H12'</span><span class="p">]</span>
<span class="n">prior</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="mi">4</span>
<span class="n">likelihood</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="mi">4</span><span class="p">,</span> <span class="mi">1</span><span class="o">/</span><span class="mi">6</span><span class="p">,</span> <span class="mi">1</span><span class="o">/</span><span class="mi">8</span><span class="p">,</span> <span class="mi">1</span><span class="o">/</span><span class="mi">12</span>
<span class="n">table</span> <span class="o">=</span> <span class="n">make_bayes_table</span><span class="p">(</span><span class="n">hypos</span><span class="p">,</span> <span class="n">prior</span><span class="p">,</span> <span class="n">likelihood</span><span class="p">)</span>
<span class="n">table</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<div class="output text_html"><div>
<style scoped>
.dataframe tbody tr th:only-of-type {
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}
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text-align: right;
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</style>
<table border="1" class="dataframe">
<thead>
<tr style="text-align: right;">
<th></th>
<th>prior</th>
<th>likelihood</th>
<th>unnorm</th>
<th>posterior</th>
</tr>
</thead>
<tbody>
<tr>
<th>H4</th>
<td>0.25</td>
<td>0.250000</td>
<td>0.062500</td>
<td>0.400000</td>
</tr>
<tr>
<th>H6</th>
<td>0.25</td>
<td>0.166667</td>
<td>0.041667</td>
<td>0.266667</td>
</tr>
<tr>
<th>H8</th>
<td>0.25</td>
<td>0.125000</td>
<td>0.031250</td>
<td>0.200000</td>
</tr>
<tr>
<th>H12</th>
<td>0.25</td>
<td>0.083333</td>
<td>0.020833</td>
<td>0.133333</td>
</tr>
</tbody>
</table>
</div></div></div>
</div>
<p>Now suppose you roll the same die again and get a 6. What are the posterior probabilities after the second roll?</p>
<p>Use <code class="docutils literal notranslate"><span class="pre">idxmax</span></code> to find the MAP.</p>
</div>
<div class="section" id="id1">
<h2>Probability Mass Functions<a class="headerlink" href="#id1" title="Permalink to this headline">¶</a></h2>
<p>When we do more than one update, we don’t always want to keep the whole Bayes table. In this section we’ll replace the Bayes table with a more compact representation, a probability mass function, or PMF.</p>
<p>A PMF is a set of possible outcomes and their corresponding probabilities. There are many ways to represent a PMF; in this notebook I’ll use a Pandas Series.</p>
<p>Here’s a function that takes a sequence of outcomes, <code class="docutils literal notranslate"><span class="pre">xs</span></code>, and a sequence of probabilities, <code class="docutils literal notranslate"><span class="pre">ps</span></code>, and returns a Pandas Series that represents a PMF.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">make_pmf</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span> <span class="n">ps</span><span class="p">,</span> <span class="o">**</span><span class="n">options</span><span class="p">):</span>
<span class="sd">"""Make a Series that represents a PMF.</span>
<span class="sd"> </span>
<span class="sd"> xs: sequence of values</span>
<span class="sd"> ps: sequence of probabilities</span>
<span class="sd"> options: keyword arguments passed to Series constructor</span>
<span class="sd"> </span>
<span class="sd"> returns: Pandas Series</span>
<span class="sd"> """</span>
<span class="n">pmf</span> <span class="o">=</span> <span class="n">pd</span><span class="o">.</span><span class="n">Series</span><span class="p">(</span><span class="n">ps</span><span class="p">,</span> <span class="n">index</span><span class="o">=</span><span class="n">xs</span><span class="p">,</span> <span class="o">**</span><span class="n">options</span><span class="p">)</span>
<span class="k">return</span> <span class="n">pmf</span>
</pre></div>
</div>
</div>
</div>
<p>And here’s a PMF that represents the prior from the 101 Bowls problem.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">xs</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="mi">101</span><span class="p">)</span>
<span class="n">prior</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="mi">101</span>
<span class="n">pmf</span> <span class="o">=</span> <span class="n">make_pmf</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span> <span class="n">prior</span><span class="p">)</span>
<span class="n">pmf</span><span class="o">.</span><span class="n">head</span><span class="p">()</span>
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<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>0 0.009901
1 0.009901
2 0.009901
3 0.009901
4 0.009901
dtype: float64
</pre></div>
</div>
</div>
</div>
<p>Now that we have a priod, we need to compute likelihoods.</p>
<p>Here are the likelihoods for a vanilla cookie:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">likelihood_vanilla</span> <span class="o">=</span> <span class="n">xs</span> <span class="o">/</span> <span class="mi">100</span>
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</div>
</div>
</div>
<p>And for a chocolate cookie.</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">likelihood_chocolate</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">-</span> <span class="n">xs</span> <span class="o">/</span> <span class="mi">100</span>
</pre></div>
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</div>
</div>
<p>To compute posterior probabilities, I’ll use the following function, which takes a PMF and a sequence of likelihoods, and updates the PMF:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">bayes_update</span><span class="p">(</span><span class="n">pmf</span><span class="p">,</span> <span class="n">likelihood</span><span class="p">):</span>
<span class="sd">"""Do a Bayesian update.</span>
<span class="sd"> </span>
<span class="sd"> pmf: Series that represents the prior</span>
<span class="sd"> likelihood: sequence of likelihoods</span>
<span class="sd"> """</span>
<span class="n">pmf</span> <span class="o">*=</span> <span class="n">likelihood</span>
<span class="n">pmf</span> <span class="o">/=</span> <span class="n">pmf</span><span class="o">.</span><span class="n">sum</span><span class="p">()</span>
</pre></div>
</div>
</div>
</div>
<p>The steps here are the same as in the Bayes table:</p>
<ol class="simple">
<li><p>Multiply the prior by the likelihoods.</p></li>
<li><p>Add up the products to get the total probability of the data.</p></li>
<li><p>Divide through to normalize the posteriors.</p></li>
</ol>
<p>Now we can do the update for a vanilla cookie.</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">bayes_update</span><span class="p">(</span><span class="n">pmf</span><span class="p">,</span> <span class="n">likelihood_vanilla</span><span class="p">)</span>
</pre></div>
</div>
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</div>
<p>Here’s what the PMF looks like after the update.</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">pmf</span><span class="o">.</span><span class="n">plot</span><span class="p">()</span>
<span class="n">plt</span><span class="o">.</span><span class="n">xlabel</span><span class="p">(</span><span class="s1">'Bowl #'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">ylabel</span><span class="p">(</span><span class="s1">'Probability'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">title</span><span class="p">(</span><span class="s1">'One cookie'</span><span class="p">);</span>
</pre></div>
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<img alt="_images/06_pmf_59_0.png" src="_images/06_pmf_59_0.png" />
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<p>That’s consistent with what we got with the Bayes table.</p>
<p>The advantage of using a PMF is that it is easier to do multiple updates. The following cell starts again with the uniform prior and does updates with two vanilla cookies and one chocolate cookie:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">data</span> <span class="o">=</span> <span class="s1">'VVC'</span>
<span class="n">pmf</span> <span class="o">=</span> <span class="n">make_pmf</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span> <span class="n">prior</span><span class="p">)</span>
<span class="k">for</span> <span class="n">cookie</span> <span class="ow">in</span> <span class="n">data</span><span class="p">:</span>
<span class="k">if</span> <span class="n">cookie</span> <span class="o">==</span> <span class="s1">'V'</span><span class="p">:</span>
<span class="n">bayes_update</span><span class="p">(</span><span class="n">pmf</span><span class="p">,</span> <span class="n">likelihood_vanilla</span><span class="p">)</span>
<span class="k">else</span><span class="p">:</span>
<span class="n">bayes_update</span><span class="p">(</span><span class="n">pmf</span><span class="p">,</span> <span class="n">likelihood_chocolate</span><span class="p">)</span>
</pre></div>
</div>
</div>
</div>
<p>Here’s what the results look like:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">pmf</span><span class="o">.</span><span class="n">plot</span><span class="p">()</span>
<span class="n">plt</span><span class="o">.</span><span class="n">xlabel</span><span class="p">(</span><span class="s1">'Bowl #'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">ylabel</span><span class="p">(</span><span class="s1">'Probability'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">title</span><span class="p">(</span><span class="s1">'Three cookies'</span><span class="p">);</span>
</pre></div>
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<img alt="_images/06_pmf_63_0.png" src="_images/06_pmf_63_0.png" />
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</div>
<p>Again, that’s consistent with what we got with the Bayes table.</p>
<p>In the next section, I’ll use a PMF and <code class="docutils literal notranslate"><span class="pre">bayes_update</span></code> to solve a dice problem.</p>
</div>
<div class="section" id="the-dice-problem">
<h2>The dice problem<a class="headerlink" href="#the-dice-problem" title="Permalink to this headline">¶</a></h2>
<p>As an exercise, let’s do one more version of the dice problem:</p>
<blockquote>
<div><p>Suppose you have a 4-sided, 6-sided, 8-sided, 12-sided, and a <strong>20-sided die</strong>. You choose one at random, roll it and <strong>get a 7</strong>. What is the probability that the die you rolled is 4-sided? What are the posterior probabilities for the other dice?</p>
</div></blockquote>
<p>Notice that in this version I’ve added a 20-sided die and the outcome is 7, not 1.</p>
<p>Here’s a PMF that represents the prior:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">sides</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">4</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">12</span><span class="p">,</span> <span class="mi">20</span><span class="p">])</span>
<span class="n">prior</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="mi">5</span>
<span class="n">pmf</span> <span class="o">=</span> <span class="n">make_pmf</span><span class="p">(</span><span class="n">sides</span><span class="p">,</span> <span class="n">prior</span><span class="p">)</span>
<span class="n">pmf</span>
</pre></div>
</div>
</div>
<div class="cell_output docutils container">
<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4 0.2
6 0.2
8 0.2
12 0.2
20 0.2
dtype: float64
</pre></div>
</div>
</div>
</div>
<p>In this version, the hypotheses are integers rather than strings, so we can compute the likelihoods like this:</p>
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<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">likelihood</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">/</span> <span class="n">sides</span>
</pre></div>
</div>
</div>
</div>
<p>But the outcome is 7, so any die with fewer than 7 sides has likelihood 0.</p>
<p>We can adjust <code class="docutils literal notranslate"><span class="pre">likelihood</span></code> by making a Boolean Series:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">too_low</span> <span class="o">=</span> <span class="p">(</span><span class="n">sides</span> <span class="o"><</span> <span class="mi">7</span><span class="p">)</span>
</pre></div>
</div>
</div>
</div>
<p>And using it to set the corresponding elements of <code class="docutils literal notranslate"><span class="pre">likelihood</span></code> to 0.</p>
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<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">likelihood</span><span class="p">[</span><span class="n">too_low</span><span class="p">]</span> <span class="o">=</span> <span class="mi">0</span>
<span class="n">likelihood</span>
</pre></div>
</div>
</div>
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<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>array([0. , 0. , 0.125 , 0.08333333, 0.05 ])
</pre></div>
</div>
</div>
</div>
<p>Now we can do the update and display the results.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">bayes_update</span><span class="p">(</span><span class="n">pmf</span><span class="p">,</span> <span class="n">likelihood</span><span class="p">)</span>
<span class="n">pmf</span>
</pre></div>
</div>
</div>
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<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4 0.000000
6 0.000000
8 0.483871
12 0.322581
20 0.193548
dtype: float64
</pre></div>
</div>
</div>
</div>
<p>The 4-sided and 6-sided dice have been eliminated. Of the remaining dice, the 8-sided die is the most likely.</p>
<p><strong>Exercise:</strong> Suppose you have the same set of 5 die. You choose a die, roll it six times, and get 6, 7, 2, 5, 1, and 2 again. Use <code class="docutils literal notranslate"><span class="pre">idxmax</span></code> to find the MAP. What is the posterior probability of the MAP?</p>
</div>
<div class="section" id="summary">
<h2>Summary<a class="headerlink" href="#summary" title="Permalink to this headline">¶</a></h2>
<p>In this notebook, we extended the cookie problem with more bowls and the dice problem with more dice.</p>
<p>I defined the MAP, which is the quantity in a posterior distribution with the highest probability.</p>
<p>Although the cookie problem is not particularly realistic or useful, the method we used to solve it applies to many problems in the real world where we want to estimate a proportion.</p>
<p><a class="reference external" href="https://colab.research.google.com/github/AllenDowney/BiteSizeBayes/blob/master/07_euro.ipynb">In the next notebook</a> we’ll use the same method on a related problem, and take another step toward doing Bayesian statistics.</p>
</div>
</div>
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