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<h1 id="_1">计数<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h1>
<p><em>计数是计算概率的前提——在分配可能性之前,你必须先知道有多少种结果。本文涵盖乘法与加法规则、阶乘、排列、组合、二项式系数,以及支撑机器学习中采样、哈希和概率分析的基本组合工具。</em></p>
<ul>
<li>
<p>在计算概率之前,我们需要先数清结果的数量。如果你想知道在扑克中拿到一手赢牌的概率,你必须先知道一共有多少种可能的牌型,以及其中有多少种是赢牌。计数正是让概率精确化的基础工具。</p>
</li>
<li>
<p>最简单的计数原则是<strong>乘法规则</strong>。如果一个选择有 <span class="arithmatex">\(m\)</span> 种选项,另一个独立的选择有 <span class="arithmatex">\(n\)</span> 种选项,那么组合起来的总结果数为 <span class="arithmatex">\(m \times n\)</span></p>
</li>
<li>
<p>想象早上穿衣服的场景。你有 3 件衬衫和 4 条裤子。每件衬衫都能与每条裤子搭配,共有 <span class="arithmatex">\(3 \times 4 = 12\)</span> 种穿搭。</p>
</li>
</ul>
<p><img alt="树状图:3 件衬衫 × 4 条裤子 = 12 种穿搭" src="../../images/counting_outfits.svg" /></p>
<ul>
<li>
<p>乘法规则可以推广到任意数量的选择。如果你还有 2 双鞋,那么总穿搭数就变成 <span class="arithmatex">\(3 \times 4 \times 2 = 24\)</span>。每个新的独立选择都会乘到总计数中。</p>
</li>
<li>
<p><strong>加法规则</strong>处理"或"的场景。如果事件 A 有 <span class="arithmatex">\(m\)</span> 种发生方式,事件 B 有 <span class="arithmatex">\(n\)</span> 种发生方式,且它们不能同时发生(互斥),那么总的方式数为 <span class="arithmatex">\(m + n\)</span></p>
</li>
<li>
<p>假设你要从城市 X 前往城市 Y:开车有 3 条路线,坐火车有 2 条路线。你无法同时选择两者,因此总选项数为 <span class="arithmatex">\(3 + 2 = 5\)</span></p>
</li>
<li>
<p>当事件有重叠时,需要减去被重复计数的结果。如果 <span class="arithmatex">\(A\)</span><span class="arithmatex">\(B\)</span> 可以同时发生,计数为 <span class="arithmatex">\(|A \cup B| = |A| + |B| - |A \cap B|\)</span>。这就是容斥原理,它将在我们讨论概率加法规则时再次出现。</p>
</li>
<li>
<p>非负整数 <span class="arithmatex">\(n\)</span><strong>阶乘</strong>是从 1 到 <span class="arithmatex">\(n\)</span> 的所有正整数的乘积:</p>
</li>
</ul>
<div class="arithmatex">\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\]</div>
<ul>
<li>
<p>可以将阶乘理解为:将 <span class="arithmatex">\(n\)</span> 个不同的物体排成一列有多少种方式?三本书在书架上有 <span class="arithmatex">\(3! = 3 \times 2 \times 1 = 6\)</span> 种排列方式。按约定,<span class="arithmatex">\(0! = 1\)</span></p>
</li>
<li>
<p>阶乘的增长速度极快。<span class="arithmatex">\(10! = 3{,}628{,}800\)</span>,而 <span class="arithmatex">\(20!\)</span> 已经超过 <span class="arithmatex">\(2.4 \times 10^{18}\)</span>。这种爆炸式增长正是暴力搜索在组合问题中变得不切实际的原因。</p>
</li>
<li>
<p><strong>排列</strong>是对物体的有序安排。当你从 <span class="arithmatex">\(n\)</span> 个不同的物体中选取 <span class="arithmatex">\(r\)</span> 个且顺序重要时,排列数为:</p>
</li>
</ul>
<div class="arithmatex">\[P(n, r) = \frac{n!}{(n - r)!}\]</div>
<ul>
<li>
<p>想象从一个 10 人的俱乐部中选出会长、副会长和财务主管。第一个职位有 10 个候选人,第二个有 9 个,第三个有 8 个。因此 <span class="arithmatex">\(P(10, 3) = 10 \times 9 \times 8 = 720\)</span>。公式也印证了这一点:<span class="arithmatex">\(\frac{10!}{7!} = 720\)</span></p>
</li>
<li>
<p><strong>组合</strong>是无序的选择。当你从 <span class="arithmatex">\(n\)</span> 个中选取 <span class="arithmatex">\(r\)</span> 个且顺序无关紧要时,需要除去重复的排列顺序:</p>
</li>
</ul>
<div class="arithmatex">\[C(n, r) = \binom{n}{r} = \frac{n!}{r!(n - r)!}\]</div>
<ul>
<li>符号 <span class="arithmatex">\(\binom{n}{r}\)</span> 读作"n 选 r"。核心思想是:每个组合对应 <span class="arithmatex">\(r!\)</span> 种排列(选出的 <span class="arithmatex">\(r\)</span> 个物品可以有 <span class="arithmatex">\(r!\)</span> 种重新排列的方式),因此我们将排列数除以 <span class="arithmatex">\(r!\)</span></li>
</ul>
<p><img alt="并列对比:排列计数所有顺序,组合将相同集合合并" src="../../images/permutation_vs_combination.svg" /></p>
<ul>
<li>示例:从 10 人中组成一个 3 人委员会有多少种方式?顺序无关紧要(没有会长或副会长之分,只有成员),因此我们使用组合:</li>
</ul>
<div class="arithmatex">\[\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\]</div>
<ul>
<li>
<p>同样的 10 个人产生 720 种排列,但只有 120 种组合,因为每个 3 人组内部有 <span class="arithmatex">\(3! = 6\)</span> 种排序方式。</p>
</li>
<li>
<p>组合在概率中至关重要。二项式系数 <span class="arithmatex">\(\binom{n}{r}\)</span> 统计了在 <span class="arithmatex">\(n\)</span> 次试验中恰好获得 <span class="arithmatex">\(r\)</span> 次成功的方式数,这正是二项分布(见文件 03)的核心。</p>
</li>
<li>
<p>让我们通过一个经典的委员会问题来综合运用多种计数思路。</p>
</li>
<li>
<p><strong>问题</strong>:一个俱乐部有 8 名男性和 6 名女性。要组成一个 5 人委员会,其中恰好包含 3 名男性和 2 名女性,有多少种方式?</p>
</li>
<li>
<p><strong>第 1 步</strong>:从 8 人中选 3 名男性。</p>
</li>
</ul>
<div class="arithmatex">\[\binom{8}{3} = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\]</div>
<ul>
<li><strong>第 2 步</strong>:从 6 人中选 2 名女性。</li>
</ul>
<div class="arithmatex">\[\binom{6}{2} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15\]</div>
<ul>
<li><strong>第 3 步</strong>:应用乘法规则。每种男性选择可以与每种女性选择配对:</li>
</ul>
<div class="arithmatex">\[56 \times 15 = 840 \text{ 个委员会}\]</div>
<ul>
<li>
<p>这种将复杂计数问题分解为独立子选择再相乘的模式,是组合数学中的标准方法。</p>
</li>
<li>
<p>还有<strong>可重复的排列</strong>。当物品可以重复时,从 <span class="arithmatex">\(n\)</span> 种类型中选 <span class="arithmatex">\(r\)</span> 个会产生 <span class="arithmatex">\(n^r\)</span> 种结果。一个使用数字 0-9 的 4 位 PIN 码有 <span class="arithmatex">\(10^4 = 10{,}000\)</span> 种可能性。每一位都有 10 种选择,乘法规则即可解决。</p>
</li>
<li>
<p><strong>可重复的组合</strong>(也称"星条法")统计从 <span class="arithmatex">\(n\)</span> 种类型中选 <span class="arithmatex">\(r\)</span> 个、允许重复且顺序无关的方式数:</p>
</li>
</ul>
<div class="arithmatex">\[\binom{n + r - 1}{r} = \frac{(n + r - 1)!}{r!(n - 1)!}\]</div>
<ul>
<li>
<p>示例:从 4 种冰淇淋口味中选择 3 勺(允许重复)有 <span class="arithmatex">\(\binom{4 + 3 - 1}{3} = \binom{6}{3} = 20\)</span> 种选项。</p>
</li>
<li>
<p>总结计数工具箱:</p>
</li>
</ul>
<table>
<thead>
<tr>
<th>场景</th>
<th>公式</th>
</tr>
</thead>
<tbody>
<tr>
<td>有序,无重复(排列)</td>
<td><span class="arithmatex">\(P(n,r) = \frac{n!}{(n-r)!}\)</span></td>
</tr>
<tr>
<td>无序,无重复(组合)</td>
<td><span class="arithmatex">\(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)</span></td>
</tr>
<tr>
<td>有序,可重复</td>
<td><span class="arithmatex">\(n^r\)</span></td>
</tr>
<tr>
<td>无序,可重复</td>
<td><span class="arithmatex">\(\binom{n+r-1}{r}\)</span></td>
</tr>
</tbody>
</table>
<ul>
<li>每个涉及等可能结果(等概率结果)的概率计算都使用公式 <span class="arithmatex">\(P(\text{事件}) = \frac{\text{有利结果数}}{\text{总结果数}}\)</span>。计数为我们提供了这两个数字。有了这个基础,我们将在下一个文件中正式定义概率本身。</li>
</ul>
<h2 id="colab-notebook">编程练习(在 CoLab 或 notebook 中完成)<a class="headerlink" href="#colab-notebook" title="Permanent link">&para;</a></h2>
<ol>
<li>
<p>使用阶乘公式和直接计算两种方式计算 <span class="arithmatex">\(P(10, 3)\)</span><span class="arithmatex">\(\binom{10}{3}\)</span>。验证排列数总是组合数的 <span class="arithmatex">\(r!\)</span> 倍。
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">jax.numpy</span><span class="w"> </span><span class="k">as</span><span class="w"> </span><span class="nn">jnp</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a><span class="kn">from</span><span class="w"> </span><span class="nn">math</span><span class="w"> </span><span class="kn">import</span> <span class="n">factorial</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a><span class="n">n</span><span class="p">,</span> <span class="n">r</span> <span class="o">=</span> <span class="mi">10</span><span class="p">,</span> <span class="mi">3</span>
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a><span class="n">perm</span> <span class="o">=</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="o">//</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="n">r</span><span class="p">)</span>
<a id="__codelineno-0-7" name="__codelineno-0-7" href="#__codelineno-0-7"></a><span class="n">comb</span> <span class="o">=</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="o">//</span> <span class="p">(</span><span class="n">factorial</span><span class="p">(</span><span class="n">r</span><span class="p">)</span> <span class="o">*</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="n">r</span><span class="p">))</span>
<a id="__codelineno-0-8" name="__codelineno-0-8" href="#__codelineno-0-8"></a>
<a id="__codelineno-0-9" name="__codelineno-0-9" href="#__codelineno-0-9"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;P(</span><span class="si">{</span><span class="n">n</span><span class="si">}</span><span class="s2">,</span><span class="si">{</span><span class="n">r</span><span class="si">}</span><span class="s2">) = </span><span class="si">{</span><span class="n">perm</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-0-10" name="__codelineno-0-10" href="#__codelineno-0-10"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;C(</span><span class="si">{</span><span class="n">n</span><span class="si">}</span><span class="s2">,</span><span class="si">{</span><span class="n">r</span><span class="si">}</span><span class="s2">) = </span><span class="si">{</span><span class="n">comb</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-0-11" name="__codelineno-0-11" href="#__codelineno-0-11"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;P / C = </span><span class="si">{</span><span class="n">perm</span><span class="w"> </span><span class="o">//</span><span class="w"> </span><span class="n">comb</span><span class="si">}</span><span class="s2"> (应等于 </span><span class="si">{</span><span class="n">r</span><span class="si">}</span><span class="s2">! = </span><span class="si">{</span><span class="n">factorial</span><span class="p">(</span><span class="n">r</span><span class="p">)</span><span class="si">}</span><span class="s2">)&quot;</span><span class="p">)</span>
</code></pre></div></p>
</li>
<li>
<p>通过程序解决委员会问题(8 人中选 3 名男性,6 人中选 2 名女性),并通过枚举所有有效委员会来验证。
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="kn">from</span><span class="w"> </span><span class="nn">itertools</span><span class="w"> </span><span class="kn">import</span> <span class="n">combinations</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a><span class="kn">from</span><span class="w"> </span><span class="nn">math</span><span class="w"> </span><span class="kn">import</span> <span class="n">factorial</span>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a><span class="k">def</span><span class="w"> </span><span class="nf">comb_count</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">r</span><span class="p">):</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a> <span class="k">return</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="o">//</span> <span class="p">(</span><span class="n">factorial</span><span class="p">(</span><span class="n">r</span><span class="p">)</span> <span class="o">*</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="n">r</span><span class="p">))</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a><span class="c1"># 公式法</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a><span class="n">men_ways</span> <span class="o">=</span> <span class="n">comb_count</span><span class="p">(</span><span class="mi">8</span><span class="p">,</span> <span class="mi">3</span><span class="p">)</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a><span class="n">women_ways</span> <span class="o">=</span> <span class="n">comb_count</span><span class="p">(</span><span class="mi">6</span><span class="p">,</span> <span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-1-10" name="__codelineno-1-10" href="#__codelineno-1-10"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;公式法: </span><span class="si">{</span><span class="n">men_ways</span><span class="si">}</span><span class="s2"> × </span><span class="si">{</span><span class="n">women_ways</span><span class="si">}</span><span class="s2"> = </span><span class="si">{</span><span class="n">men_ways</span><span class="w"> </span><span class="o">*</span><span class="w"> </span><span class="n">women_ways</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-1-11" name="__codelineno-1-11" href="#__codelineno-1-11"></a>
<a id="__codelineno-1-12" name="__codelineno-1-12" href="#__codelineno-1-12"></a><span class="c1"># 枚举法</span>
<a id="__codelineno-1-13" name="__codelineno-1-13" href="#__codelineno-1-13"></a><span class="n">men</span> <span class="o">=</span> <span class="p">[</span><span class="sa">f</span><span class="s2">&quot;M</span><span class="si">{</span><span class="n">i</span><span class="si">}</span><span class="s2">&quot;</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">9</span><span class="p">)]</span>
<a id="__codelineno-1-14" name="__codelineno-1-14" href="#__codelineno-1-14"></a><span class="n">women</span> <span class="o">=</span> <span class="p">[</span><span class="sa">f</span><span class="s2">&quot;W</span><span class="si">{</span><span class="n">i</span><span class="si">}</span><span class="s2">&quot;</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">7</span><span class="p">)]</span>
<a id="__codelineno-1-15" name="__codelineno-1-15" href="#__codelineno-1-15"></a><span class="n">count</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">(</span><span class="mi">1</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="n">combinations</span><span class="p">(</span><span class="n">men</span><span class="p">,</span> <span class="mi">3</span><span class="p">)</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="n">combinations</span><span class="p">(</span><span class="n">women</span><span class="p">,</span> <span class="mi">2</span><span class="p">))</span>
<a id="__codelineno-1-16" name="__codelineno-1-16" href="#__codelineno-1-16"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;枚举法: </span><span class="si">{</span><span class="n">count</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
</code></pre></div></p>
</li>
<li>
<p>统计由 26 个小写字母组成的 4 位密码有多少种(允许重复)。然后统计没有重复字母的密码有多少种。
<div class="highlight"><pre><span></span><code><a id="__codelineno-2-1" name="__codelineno-2-1" href="#__codelineno-2-1"></a><span class="kn">from</span><span class="w"> </span><span class="nn">math</span><span class="w"> </span><span class="kn">import</span> <span class="n">factorial</span>
<a id="__codelineno-2-2" name="__codelineno-2-2" href="#__codelineno-2-2"></a>
<a id="__codelineno-2-3" name="__codelineno-2-3" href="#__codelineno-2-3"></a><span class="n">n</span> <span class="o">=</span> <span class="mi">26</span>
<a id="__codelineno-2-4" name="__codelineno-2-4" href="#__codelineno-2-4"></a><span class="n">r</span> <span class="o">=</span> <span class="mi">4</span>
<a id="__codelineno-2-5" name="__codelineno-2-5" href="#__codelineno-2-5"></a>
<a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a><span class="n">with_rep</span> <span class="o">=</span> <span class="n">n</span> <span class="o">**</span> <span class="n">r</span>
<a id="__codelineno-2-7" name="__codelineno-2-7" href="#__codelineno-2-7"></a><span class="n">without_rep</span> <span class="o">=</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="o">//</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="n">r</span><span class="p">)</span>
<a id="__codelineno-2-8" name="__codelineno-2-8" href="#__codelineno-2-8"></a>
<a id="__codelineno-2-9" name="__codelineno-2-9" href="#__codelineno-2-9"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;允许重复: </span><span class="si">{</span><span class="n">with_rep</span><span class="si">:</span><span class="s2">&gt;10,</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-2-10" name="__codelineno-2-10" href="#__codelineno-2-10"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;不允许重复: </span><span class="si">{</span><span class="n">without_rep</span><span class="si">:</span><span class="s2">&gt;10,</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-2-11" name="__codelineno-2-11" href="#__codelineno-2-11"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;含重复的比例: </span><span class="si">{</span><span class="mi">1</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">without_rep</span><span class="o">/</span><span class="n">with_rep</span><span class="si">:</span><span class="s2">.2%</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
</code></pre></div></p>
</li>
<li>
<p>模拟生日问题:在 <span class="arithmatex">\(k\)</span> 人的群体中,至少两人共享生日的概率是多少?绘制 <span class="arithmatex">\(k = 1\)</span><span class="arithmatex">\(60\)</span> 的概率曲线,并找出概率超过 50% 的位置。
<div class="highlight"><pre><span></span><code><a id="__codelineno-3-1" name="__codelineno-3-1" href="#__codelineno-3-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">jax</span>
<a id="__codelineno-3-2" name="__codelineno-3-2" href="#__codelineno-3-2"></a><span class="kn">import</span><span class="w"> </span><span class="nn">jax.numpy</span><span class="w"> </span><span class="k">as</span><span class="w"> </span><span class="nn">jnp</span>
<a id="__codelineno-3-3" name="__codelineno-3-3" href="#__codelineno-3-3"></a><span class="kn">import</span><span class="w"> </span><span class="nn">matplotlib.pyplot</span><span class="w"> </span><span class="k">as</span><span class="w"> </span><span class="nn">plt</span>
<a id="__codelineno-3-4" name="__codelineno-3-4" href="#__codelineno-3-4"></a>
<a id="__codelineno-3-5" name="__codelineno-3-5" href="#__codelineno-3-5"></a><span class="k">def</span><span class="w"> </span><span class="nf">birthday_prob_exact</span><span class="p">(</span><span class="n">k</span><span class="p">):</span>
<a id="__codelineno-3-6" name="__codelineno-3-6" href="#__codelineno-3-6"></a> \<span class="s2">&quot;</span><span class="se">\&quot;\&quot;</span><span class="s2">k 人群体中至少有一对共享生日的概率。</span><span class="se">\&quot;\&quot;\&quot;</span>
<a id="__codelineno-3-7" name="__codelineno-3-7" href="#__codelineno-3-7"></a> <span class="n">p_no_match</span> <span class="o">=</span> <span class="mf">1.0</span>
<a id="__codelineno-3-8" name="__codelineno-3-8" href="#__codelineno-3-8"></a> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">k</span><span class="p">):</span>
<a id="__codelineno-3-9" name="__codelineno-3-9" href="#__codelineno-3-9"></a> <span class="n">p_no_match</span> <span class="o">*=</span> <span class="p">(</span><span class="mi">365</span> <span class="o">-</span> <span class="n">i</span><span class="p">)</span> <span class="o">/</span> <span class="mi">365</span>
<a id="__codelineno-3-10" name="__codelineno-3-10" href="#__codelineno-3-10"></a> <span class="k">return</span> <span class="mi">1</span> <span class="o">-</span> <span class="n">p_no_match</span>
<a id="__codelineno-3-11" name="__codelineno-3-11" href="#__codelineno-3-11"></a>
<a id="__codelineno-3-12" name="__codelineno-3-12" href="#__codelineno-3-12"></a><span class="n">ks</span> <span class="o">=</span> <span class="nb">list</span><span class="p">(</span><span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">61</span><span class="p">))</span>
<a id="__codelineno-3-13" name="__codelineno-3-13" href="#__codelineno-3-13"></a><span class="n">probs</span> <span class="o">=</span> <span class="p">[</span><span class="n">birthday_prob_exact</span><span class="p">(</span><span class="n">k</span><span class="p">)</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="n">ks</span><span class="p">]</span>
<a id="__codelineno-3-14" name="__codelineno-3-14" href="#__codelineno-3-14"></a>
<a id="__codelineno-3-15" name="__codelineno-3-15" href="#__codelineno-3-15"></a><span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">(</span><span class="n">figsize</span><span class="o">=</span><span class="p">(</span><span class="mi">8</span><span class="p">,</span> <span class="mi">4</span><span class="p">))</span>
<a id="__codelineno-3-16" name="__codelineno-3-16" href="#__codelineno-3-16"></a><span class="n">plt</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">ks</span><span class="p">,</span> <span class="n">probs</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s2">&quot;#3498db&quot;</span><span class="p">,</span> <span class="n">linewidth</span><span class="o">=</span><span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-3-17" name="__codelineno-3-17" href="#__codelineno-3-17"></a><span class="n">plt</span><span class="o">.</span><span class="n">axhline</span><span class="p">(</span><span class="n">y</span><span class="o">=</span><span class="mf">0.5</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s2">&quot;#e74c3c&quot;</span><span class="p">,</span> <span class="n">linestyle</span><span class="o">=</span><span class="s2">&quot;--&quot;</span><span class="p">,</span> <span class="n">alpha</span><span class="o">=</span><span class="mf">0.7</span><span class="p">,</span> <span class="n">label</span><span class="o">=</span><span class="s2">&quot;50%&quot;</span><span class="p">)</span>
<a id="__codelineno-3-18" name="__codelineno-3-18" href="#__codelineno-3-18"></a><span class="n">cross</span> <span class="o">=</span> <span class="nb">next</span><span class="p">(</span><span class="n">k</span> <span class="k">for</span> <span class="n">k</span><span class="p">,</span> <span class="n">p</span> <span class="ow">in</span> <span class="nb">zip</span><span class="p">(</span><span class="n">ks</span><span class="p">,</span> <span class="n">probs</span><span class="p">)</span> <span class="k">if</span> <span class="n">p</span> <span class="o">&gt;=</span> <span class="mf">0.5</span><span class="p">)</span>
<a id="__codelineno-3-19" name="__codelineno-3-19" href="#__codelineno-3-19"></a><span class="n">plt</span><span class="o">.</span><span class="n">axvline</span><span class="p">(</span><span class="n">x</span><span class="o">=</span><span class="n">cross</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s2">&quot;#e74c3c&quot;</span><span class="p">,</span> <span class="n">linestyle</span><span class="o">=</span><span class="s2">&quot;--&quot;</span><span class="p">,</span> <span class="n">alpha</span><span class="o">=</span><span class="mf">0.7</span><span class="p">)</span>
<a id="__codelineno-3-20" name="__codelineno-3-20" href="#__codelineno-3-20"></a><span class="n">plt</span><span class="o">.</span><span class="n">xlabel</span><span class="p">(</span><span class="s2">&quot;群体大小 (k)&quot;</span><span class="p">)</span>
<a id="__codelineno-3-21" name="__codelineno-3-21" href="#__codelineno-3-21"></a><span class="n">plt</span><span class="o">.</span><span class="n">ylabel</span><span class="p">(</span><span class="s2">&quot;P(至少两人共享生日)&quot;</span><span class="p">)</span>
<a id="__codelineno-3-22" name="__codelineno-3-22" href="#__codelineno-3-22"></a><span class="n">plt</span><span class="o">.</span><span class="n">title</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;生日问题(在 k=</span><span class="si">{</span><span class="n">cross</span><span class="si">}</span><span class="s2"> 时超过 50%&quot;</span><span class="p">)</span>
<a id="__codelineno-3-23" name="__codelineno-3-23" href="#__codelineno-3-23"></a><span class="n">plt</span><span class="o">.</span><span class="n">legend</span><span class="p">()</span>
<a id="__codelineno-3-24" name="__codelineno-3-24" href="#__codelineno-3-24"></a><span class="n">plt</span><span class="o">.</span><span class="n">grid</span><span class="p">(</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">)</span>
<a id="__codelineno-3-25" name="__codelineno-3-25" href="#__codelineno-3-25"></a><span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</code></pre></div></p>
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