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<h2 id="toc-title">Table of contents</h2>
<ul>
<li><a href="#introduction" id="toc-introduction" class="nav-link active" data-scroll-target="#introduction"><span class="header-section-number">11.1</span> Introduction</a></li>
<li><a href="#sec-one-pair" id="toc-sec-one-pair" class="nav-link" data-scroll-target="#sec-one-pair"><span class="header-section-number">11.2</span> Introducing a poker problem: one pair (two of a kind)</a></li>
<li><a href="#a-first-approach-to-the-one-pair-problem-with-code" id="toc-a-first-approach-to-the-one-pair-problem-with-code" class="nav-link" data-scroll-target="#a-first-approach-to-the-one-pair-problem-with-code"><span class="header-section-number">11.3</span> A first approach to the one-pair problem with code</a></li>
<li><a href="#sec-shuffling-deck" id="toc-sec-shuffling-deck" class="nav-link" data-scroll-target="#sec-shuffling-deck"><span class="header-section-number">11.4</span> Shuffling the deck with Python</a>
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<li><a href="#a-first-pass-computer-solution-to-the-one-pair-problem" id="toc-a-first-pass-computer-solution-to-the-one-pair-problem" class="nav-link" data-scroll-target="#a-first-pass-computer-solution-to-the-one-pair-problem"><span class="header-section-number">11.4.1</span> A first-pass computer solution to the one-pair problem</a></li>
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<li><a href="#finding-exactly-one-pair-using-code" id="toc-finding-exactly-one-pair-using-code" class="nav-link" data-scroll-target="#finding-exactly-one-pair-using-code"><span class="header-section-number">11.5</span> Finding exactly one pair using code</a></li>
<li><a href="#sec-bincount-tabulate" id="toc-sec-bincount-tabulate" class="nav-link" data-scroll-target="#sec-bincount-tabulate"><span class="header-section-number">11.6</span> Finding number of repeats using <code>np.bincount</code></a></li>
<li><a href="#looking-for-hands-with-exactly-one-pair" id="toc-looking-for-hands-with-exactly-one-pair" class="nav-link" data-scroll-target="#looking-for-hands-with-exactly-one-pair"><span class="header-section-number">11.7</span> Looking for hands with exactly one pair</a></li>
<li><a href="#two-more-tntroductory-poker-problems" id="toc-two-more-tntroductory-poker-problems" class="nav-link" data-scroll-target="#two-more-tntroductory-poker-problems"><span class="header-section-number">11.8</span> Two more tntroductory poker problems</a></li>
<li><a href="#the-concepts-of-replacement-and-non-replacement" id="toc-the-concepts-of-replacement-and-non-replacement" class="nav-link" data-scroll-target="#the-concepts-of-replacement-and-non-replacement"><span class="header-section-number">11.9</span> The concepts of replacement and non-replacement</a></li>
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<h1 class="title"><span id="sec-compound-probability" class="quarto-section-identifier"><span class="chapter-number">11</span> <span class="chapter-title">Probability Theory, Part 2: Compound Probability</span></span></h1>
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<section id="introduction" class="level2" data-number="11.1">
<h2 data-number="11.1" class="anchored" data-anchor-id="introduction"><span class="header-section-number">11.1</span> Introduction</h2>
<p>In this chapter we will deal with what are usually called “probability problems” rather than the “statistical inference problems” discussed in later chapters. The difference is that for probability problems we begin with a knowledge of the properties of the universe with which we are working. (See <a href="probability_theory_1a.html#sec-what-is-resampling" class="quarto-xref"><span>Section 8.9</span></a> on the definition of resampling.)</p>
<p>We start with some basic problems in probability. To make sure we do know the properties of the universe we are working with, we start with poker, and a pack of cards. Working with some poker problems, we rediscover the fundamental distinction between sampling <em>with</em> and <em>without</em> replacement.</p>
</section>
<section id="sec-one-pair" class="level2" data-number="11.2">
<h2 data-number="11.2" class="anchored" data-anchor-id="sec-one-pair"><span class="header-section-number">11.2</span> Introducing a poker problem: one pair (two of a kind)</h2>
<p>What is the chance that the first five cards chosen from a deck of 52 (bridge/poker) cards will contain two (and only two) cards of the same denomination (two 3’s for example)? (Please forgive the rather sterile unrealistic problems in this and the other chapters on probability. They reflect the literature in the field for 300 years. We’ll get more realistic in the statistics chapters.)</p>
<p>We shall estimate the odds the way that gamblers have estimated gambling odds for thousands of years. First, check that the deck is a standard deck and is not missing any cards. (Overlooking such small but crucial matters often leads to errors in science.) Shuffle thoroughly until you are satisfied that the cards are randomly distributed. (It is surprisingly hard to shuffle well.) Then deal five cards, and mark down whether the hand does or does not contain a pair of the same denomination.</p>
<p>At this point, we must decide whether three of a kind, four of a kind or two pairs meet our criterion for a pair. Since our criterion is “two and only two,” we decide <em>not</em> to count them.</p>
<p>Then replace the five cards in the deck, shuffle, and deal again. Again mark down whether the hand contains one pair of the same denomination. Do this many times. Then count the number of hands with one pair, and figure the proportion (as a percentage) of all hands.</p>
<p><a href="#tbl-one-pair" class="quarto-xref">Table <span>11.1</span></a> has the results of 25 hands of this procedure.</p>
<div class="cell" data-layout-align="center">
<div id="tbl-one-pair" class="quarto-float quarto-figure quarto-figure-center anchored">
<figure class="quarto-float quarto-float-tbl figure">
<figcaption class="quarto-float-caption-top quarto-float-caption quarto-float-tbl" id="tbl-one-pair-caption-0ceaefa1-69ba-4598-a22c-09a6ac19f8ca">
Table 11.1: Results of 25 hands for the problem “one pair”
</figcaption>
<div aria-describedby="tbl-one-pair-caption-0ceaefa1-69ba-4598-a22c-09a6ac19f8ca">
<table class="caption-top table table-sm table-striped small">
<colgroup>
<col style="width: 13%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 16%">
</colgroup>
<thead>
<tr class="header">
<th>Hand</th>
<th>Card 1</th>
<th>Card 2</th>
<th>Card 3</th>
<th>Card 4</th>
<th>Card 5</th>
<th>One pair?</th>
</tr>
</thead>
<tbody>
<tr class="odd">
<td>1</td>
<td>King ♢</td>
<td>King ♠</td>
<td>Queen ♠</td>
<td>10 ♢</td>
<td>6 ♠</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>2</td>
<td>8 ♢</td>
<td>Ace ♢</td>
<td>4 ♠</td>
<td>10 ♢</td>
<td>3 ♣</td>
<td>No</td>
</tr>
<tr class="odd">
<td>3</td>
<td>4 ♢</td>
<td>5 ♣</td>
<td>Ace ♢</td>
<td>Queen ♡</td>
<td>10 ♠</td>
<td>No</td>
</tr>
<tr class="even">
<td>4</td>
<td>3 ♡</td>
<td>Ace ♡</td>
<td>5 ♣</td>
<td>3 ♢</td>
<td>Jack ♢</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>5</td>
<td>6 ♠</td>
<td>King ♣</td>
<td>6 ♢</td>
<td>3 ♣</td>
<td>3 ♡</td>
<td>No</td>
</tr>
<tr class="even">
<td>6</td>
<td>Queen ♣</td>
<td>7 ♢</td>
<td>Jack ♠</td>
<td>5 ♡</td>
<td>8 ♡</td>
<td>No</td>
</tr>
<tr class="odd">
<td>7</td>
<td>9 ♣</td>
<td>4 ♣</td>
<td>9 ♠</td>
<td>Jack ♣</td>
<td>5 ♠</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>8</td>
<td>3 ♠</td>
<td>3 ♣</td>
<td>3 ♡</td>
<td>5 ♠</td>
<td>5 ♢</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>9</td>
<td>Queen ♢</td>
<td>4 ♠</td>
<td>Queen ♣</td>
<td>6 ♡</td>
<td>4 ♢</td>
<td>No</td>
</tr>
<tr class="even">
<td>10</td>
<td>Queen ♠</td>
<td>3 ♣</td>
<td>7 ♠</td>
<td>7 ♡</td>
<td>8 ♢</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>11</td>
<td>8 ♡</td>
<td>9 ♠</td>
<td>7 ♢</td>
<td>8 ♠</td>
<td>Ace ♡</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>12</td>
<td>Ace ♠</td>
<td>9 ♡</td>
<td>4 ♣</td>
<td>2 ♠</td>
<td>Ace ♢</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>13</td>
<td>4 ♡</td>
<td>3 ♣</td>
<td>Ace ♢</td>
<td>9 ♡</td>
<td>5 ♡</td>
<td>No</td>
</tr>
<tr class="even">
<td>14</td>
<td>10 ♣</td>
<td>7 ♠</td>
<td>8 ♣</td>
<td>King ♣</td>
<td>4 ♢</td>
<td>No</td>
</tr>
<tr class="odd">
<td>15</td>
<td>Queen ♣</td>
<td>8 ♠</td>
<td>Queen ♠</td>
<td>8 ♣</td>
<td>5 ♣</td>
<td>No</td>
</tr>
<tr class="even">
<td>16</td>
<td>King ♡</td>
<td>10 ♣</td>
<td>Jack ♠</td>
<td>10 ♢</td>
<td>10 ♡</td>
<td>No</td>
</tr>
<tr class="odd">
<td>17</td>
<td>Queen ♠</td>
<td>Queen ♡</td>
<td>Ace ♡</td>
<td>King ♢</td>
<td>7 ♡</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>18</td>
<td>5 ♢</td>
<td>6 ♡</td>
<td>Ace ♡</td>
<td>4 ♡</td>
<td>6 ♢</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>19</td>
<td>3 ♠</td>
<td>5 ♡</td>
<td>2 ♢</td>
<td>King ♣</td>
<td>9 ♡</td>
<td>No</td>
</tr>
<tr class="even">
<td>20</td>
<td>8 ♠</td>
<td>Jack ♢</td>
<td>7 ♣</td>
<td>10 ♡</td>
<td>3 ♡</td>
<td>No</td>
</tr>
<tr class="odd">
<td>21</td>
<td>5 ♢</td>
<td>4 ♠</td>
<td>Jack ♡</td>
<td>2 ♠</td>
<td>King ♠</td>
<td>No</td>
</tr>
<tr class="even">
<td>22</td>
<td>5 ♢</td>
<td>4 ♢</td>
<td>Jack ♣</td>
<td>King ♢</td>
<td>2 ♠</td>
<td>No</td>
</tr>
<tr class="odd">
<td>23</td>
<td>King ♡</td>
<td>King ♠</td>
<td>6 ♡</td>
<td>2 ♠</td>
<td>5 ♣</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>24</td>
<td>8 ♠</td>
<td>9 ♠</td>
<td>6 ♣</td>
<td>Ace ♣</td>
<td>5 ♢</td>
<td>No</td>
</tr>
<tr class="odd">
<td>25</td>
<td>Ace ♢</td>
<td>7 ♠</td>
<td>4 ♡</td>
<td>9 ♢</td>
<td>9 ♠</td>
<td>Yes</td>
</tr>
<tr class="even">
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
</tr>
<tr class="odd">
<td><strong>% Yes</strong></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td>44%</td>
</tr>
</tbody>
</table>
</div>
</figure>
</div>
</div>
<p>In this series of 25 experiments, 44 percent of the hands contained one pair, and therefore 0.44 is our estimate (for the time being) of the probability that one pair will turn up in a poker hand. But we must notice that this estimate is based on only 25 hands, and therefore might well be fairly far off the mark (as we shall soon see).</p>
<p>This experimental “resampling” estimation does not require a deck of cards. For example, one might create a 52-sided die, one side for each card in the deck, and roll it five times to get a “hand.” But note one important part of the procedure: No single “card” is allowed to come up twice in the same set of five spins, just as no single card can turn up twice or more in the same hand. If the same “card” did turn up twice or more in a dice experiment, one could pretend that the roll had never taken place; this procedure is necessary to make the dice experiment analogous to the actual card-dealing situation under investigation. Otherwise, the results will be slightly in error. This type of sampling is “sampling without replacement,” because each card is <em>not replaced</em> in the deck prior to dealing the next card (that is, prior to the end of the hand).</p>
</section>
<section id="a-first-approach-to-the-one-pair-problem-with-code" class="level2" data-number="11.3">
<h2 data-number="11.3" class="anchored" data-anchor-id="a-first-approach-to-the-one-pair-problem-with-code"><span class="header-section-number">11.3</span> A first approach to the one-pair problem with code</h2>
<p>We could also approach this problem using random numbers from the computer to simulate the values.</p>
<p>Let us first make some numbers from which to sample. We want to simulate a deck of playing cards analogous to the real cards we used previously. We don’t need to simulate all the features of a deck, but only the features that matter for the problem at hand. In our case, the feature that matters is the face value. We require a deck with four “1”s, four “2”s, etc., up to four “13”s, where 1 is an Ace, and 13 is a King. The suits don’t matter for our present purposes.</p>
<p>We first first make an array to represent the face values in one suit.</p>
<div class="cell" data-layout-align="center">
<div class="sourceCode cell-code" id="cb1"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb1-1"><a href="#cb1-1" aria-hidden="true" tabindex="-1"></a><span class="co"># Card values 1 through 13 (1 up to, not including 14).</span></span>
<span id="cb1-2"><a href="#cb1-2" aria-hidden="true" tabindex="-1"></a>one_suit <span class="op">=</span> np.arange(<span class="dv">1</span>, <span class="dv">14</span>)</span>
<span id="cb1-3"><a href="#cb1-3" aria-hidden="true" tabindex="-1"></a>one_suit</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-stdout">
<pre><code>array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13])</code></pre>
</div>
</div>
<p>We have the face values for one suit, but we need the face values for whole deck of cards — four suits. We do this by making a new array that consists of four repeats of <code>one_suit</code>:</p>
<div class="cell" data-layout-align="center">
<div class="sourceCode cell-code" id="cb3"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb3-1"><a href="#cb3-1" aria-hidden="true" tabindex="-1"></a><span class="co"># Repeat the one_suit array four times</span></span>
<span id="cb3-2"><a href="#cb3-2" aria-hidden="true" tabindex="-1"></a>deck <span class="op">=</span> np.repeat(one_suit, <span class="dv">4</span>)</span>
<span id="cb3-3"><a href="#cb3-3" aria-hidden="true" tabindex="-1"></a>deck</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-stdout">
<pre><code>array([ 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5,
5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9,
9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13,
13])</code></pre>
</div>
</div>
</section>
<section id="sec-shuffling-deck" class="level2" data-number="11.4">
<h2 data-number="11.4" class="anchored" data-anchor-id="sec-shuffling-deck"><span class="header-section-number">11.4</span> Shuffling the deck with Python</h2>
<p>At this point we have a complete deck in the variable <code>deck</code> . But that “deck” is <span class="python">ordered by value, first ones (Aces) then 2s and so on</span>. If we do not shuffle the deck, the results will be predictable. Therefore, we would like to select five of these “cards” (52 values) at random. There are two ways of doing this. The first is to use the ’rnd.choice`]{.python} tool in the familiar way, to choose 5 values at random from this strictly ordered deck. We want to draw these cards <em>without replacement</em> (of which more later). <em>Without replacement</em> means that once we have drawn a particular value, we cannot draw that value a second time — just as you cannot get the same card twice in a hand when the dealer deals you a hand of five cards.</p>
<div class="python">
<p>So far, each of our uses of <code>rnd.choice</code> has done sampling <em>with replacement</em>, where you <em>can</em> get the same item more than once in a particular sample. Here we need <em>without replacement</em>. <code>rnd.choice</code> has an argument you can send, called <code>replace</code>, to tell it whether to replace values when drawing the sample. We have not used that argument so far, because the default is <code>True</code> — sampling <em>with replacement</em>. Here we need to use the argument — <code>replace=False</code> — to get sampling <em>without replacement</em>.</p>
</div>
<div class="cell" data-layout-align="center">
<div class="sourceCode cell-code" id="cb5"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb5-1"><a href="#cb5-1" aria-hidden="true" tabindex="-1"></a><span class="co"># One hand, sampling from the deck without replacement.</span></span>
<span id="cb5-2"><a href="#cb5-2" aria-hidden="true" tabindex="-1"></a>hand <span class="op">=</span> rnd.choice(deck, size<span class="op">=</span><span class="dv">5</span>, replace<span class="op">=</span><span class="va">False</span>)</span>
<span id="cb5-3"><a href="#cb5-3" aria-hidden="true" tabindex="-1"></a>hand</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-stdout">
<pre><code>array([ 9, 4, 11, 9, 13])</code></pre>
</div>
</div>
<p>The above is one way to get a random hand of five cards from the deck. Another way is to use <span class="python">the <code>rnd.permuted</code> function</span> to shuffle the whole <code>deck</code> of 52 “cards” into a random order, just as a dealer would shuffle the deck before dealing. Then we could take — for example — the first five cards from the shuffled deck to give a random hand. See <a href="probability_theory_1a.html#sec-shuffling" class="quarto-xref"><span>Section 8.14</span></a> for more on <span class="python"><code>rnd.permuted</code></span>.</p>
<div class="cell" data-layout-align="center">
<div class="sourceCode cell-code" id="cb7"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb7-1"><a href="#cb7-1" aria-hidden="true" tabindex="-1"></a><span class="co"># Shuffle the whole 52 card deck.</span></span>
<span id="cb7-2"><a href="#cb7-2" aria-hidden="true" tabindex="-1"></a>shuffled <span class="op">=</span> rnd.permuted(deck)</span>
<span id="cb7-3"><a href="#cb7-3" aria-hidden="true" tabindex="-1"></a><span class="co"># The "cards" are now in random order.</span></span>
<span id="cb7-4"><a href="#cb7-4" aria-hidden="true" tabindex="-1"></a>shuffled</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-stdout">
<pre><code>array([12, 13, 2, 9, 6, 7, 7, 7, 11, 13, 2, 8, 6, 9, 4, 1, 5,
12, 11, 9, 1, 2, 4, 2, 3, 3, 11, 6, 4, 11, 8, 7, 13, 8,
12, 5, 4, 5, 9, 8, 5, 6, 3, 1, 1, 12, 3, 13, 10, 10, 10,
10])</code></pre>
</div>
</div>
<p>Now we can get our <code>hand</code> by taking the first five cards from the <code>deck</code>:</p>
<div class="cell" data-layout-align="center">
<div class="sourceCode cell-code" id="cb9"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb9-1"><a href="#cb9-1" aria-hidden="true" tabindex="-1"></a><span class="co"># Select the first five "cards" from the shuffled deck.</span></span>
<span id="cb9-2"><a href="#cb9-2" aria-hidden="true" tabindex="-1"></a>hand <span class="op">=</span> shuffled[:<span class="dv">5</span>]</span>
<span id="cb9-3"><a href="#cb9-3" aria-hidden="true" tabindex="-1"></a>hand</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-stdout">
<pre><code>array([12, 13, 2, 9, 6])</code></pre>
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</div>
<p>You have seen that we can use one of two procedures to a get random sample of five cards from <code>deck</code>, drawn without replacement:</p>
<ol type="1">
<li>Using <span class="python"><code>rnd.choice</code> with <code>size=5</code> and <code>replace=False</code></span> to take the random sample directly from <code>deck</code>, or</li>
<li>shuffling the entire <code>deck</code> and then taking the first five “cards” from the result of the shuffle.</li>
</ol>
<p>Either is a valid way of getting five cards at random from the <code>deck</code>. It’s up to us which to choose — we slightly prefer to shuffle and take the first five, because it is more like the physical procedure of shuffling the deck and dealing, but which you prefer, is up to you.</p>
<section id="a-first-pass-computer-solution-to-the-one-pair-problem" class="level3" data-number="11.4.1">
<h3 data-number="11.4.1" class="anchored" data-anchor-id="a-first-pass-computer-solution-to-the-one-pair-problem"><span class="header-section-number">11.4.1</span> A first-pass computer solution to the one-pair problem</h3>
<p>Choosing the shuffle deal way, the cell to generate one hand is:</p>
<div class="cell" data-layout-align="center">
<div class="sourceCode cell-code" id="cb11"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb11-1"><a href="#cb11-1" aria-hidden="true" tabindex="-1"></a>shuffled <span class="op">=</span> rnd.permuted(deck)</span>
<span id="cb11-2"><a href="#cb11-2" aria-hidden="true" tabindex="-1"></a>hand <span class="op">=</span> shuffled[:<span class="dv">5</span>]</span>
<span id="cb11-3"><a href="#cb11-3" aria-hidden="true" tabindex="-1"></a>hand</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-stdout">
<pre><code>array([ 7, 4, 12, 1, 2])</code></pre>
</div>
</div>
<p>Without doing anything further, we could run this cell many times, and each time, we could note down whether the particular <code>hand</code> had exactly one pair or not.</p>
<p><a href="#tbl-one-pair-numbers" class="quarto-xref">Table <span>11.2</span></a> has the result of running that procedure 25 times:</p>
<div class="cell" data-layout-align="center">
<div id="tbl-one-pair-numbers" class="quarto-float quarto-figure quarto-figure-center anchored">
<figure class="quarto-float quarto-float-tbl figure">
<figcaption class="quarto-float-caption-top quarto-float-caption quarto-float-tbl" id="tbl-one-pair-numbers-caption-0ceaefa1-69ba-4598-a22c-09a6ac19f8ca">
Table 11.2: Results of 25 hands using random numbers
</figcaption>
<div aria-describedby="tbl-one-pair-numbers-caption-0ceaefa1-69ba-4598-a22c-09a6ac19f8ca">
<table class="caption-top table table-sm table-striped small">
<colgroup>
<col style="width: 13%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 12%">
<col style="width: 16%">
</colgroup>
<thead>
<tr class="header">
<th>Hand</th>
<th>Card 1</th>
<th>Card 2</th>
<th>Card 3</th>
<th>Card 4</th>
<th>Card 5</th>
<th>One pair?</th>
</tr>
</thead>
<tbody>
<tr class="odd">
<td>1</td>
<td>10</td>
<td>5</td>
<td>7</td>
<td>12</td>
<td>12</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>2</td>
<td>6</td>
<td>9</td>
<td>2</td>
<td>6</td>
<td>8</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>3</td>
<td>11</td>
<td>8</td>
<td>9</td>
<td>6</td>
<td>1</td>
<td>No</td>
</tr>
<tr class="even">
<td>4</td>
<td>8</td>
<td>10</td>
<td>2</td>
<td>11</td>
<td>12</td>
<td>No</td>
</tr>
<tr class="odd">
<td>5</td>
<td>1</td>
<td>10</td>
<td>11</td>
<td>8</td>
<td>5</td>
<td>No</td>
</tr>
<tr class="even">
<td>6</td>
<td>8</td>
<td>10</td>
<td>3</td>
<td>9</td>
<td>5</td>
<td>No</td>
</tr>
<tr class="odd">
<td>7</td>
<td>10</td>
<td>9</td>
<td>13</td>
<td>1</td>
<td>9</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>8</td>
<td>13</td>
<td>4</td>
<td>3</td>
<td>11</td>
<td>5</td>
<td>No</td>
</tr>
<tr class="odd">
<td>9</td>
<td>7</td>
<td>1</td>
<td>4</td>
<td>13</td>
<td>6</td>
<td>No</td>
</tr>
<tr class="even">
<td>10</td>
<td>11</td>
<td>5</td>
<td>11</td>
<td>8</td>
<td>4</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>11</td>
<td>7</td>
<td>10</td>
<td>7</td>
<td>13</td>
<td>9</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>12</td>
<td>2</td>
<td>11</td>
<td>4</td>
<td>7</td>
<td>8</td>
<td>No</td>
</tr>
<tr class="odd">
<td>13</td>
<td>12</td>
<td>1</td>
<td>3</td>
<td>10</td>
<td>2</td>
<td>No</td>
</tr>
<tr class="even">
<td>14</td>
<td>10</td>
<td>2</td>
<td>11</td>
<td>8</td>
<td>1</td>
<td>No</td>
</tr>
<tr class="odd">
<td>15</td>
<td>1</td>
<td>6</td>
<td>12</td>
<td>12</td>
<td>5</td>
<td>Yes</td>
</tr>
<tr class="even">
<td>16</td>
<td>4</td>
<td>8</td>
<td>7</td>
<td>8</td>
<td>6</td>
<td>Yes</td>
</tr>
<tr class="odd">
<td>17</td>
<td>7</td>
<td>10</td>
<td>9</td>
<td>4</td>