返回题库

红蓝相邻异色对的期望个数

Expected Red-Blue Pairs

专题
Probability / 概率
难度
L4

题目详情

将 aa 个红球与 bb 个蓝球（总数 n=a+bn=a+b）随机排成一行。求相邻两球颜色不同（红-蓝或蓝-红）的相邻对的期望数量。

英文原题

You randomly place aa red balls and bb blue balls in a row (total balls n=a+bn = a+b). What is the expected number of adjacent pairs of balls with different colors (red-blue or blue-red)?

解析

对每个相邻位置 i,i+1i,i+1（共 n−1n-1 对）设指示变量 IiI_i 表示该对异色，则期望为

E[∑i=1n−1Ii]=(n−1) P(相邻两位异色).\mathbb{E}\left[\sum_{i=1}^{n-1} I_i\right]=(n-1)\,\mathbb{P}(\text{相邻两位异色}).

在随机排列中

P(RB)=an⋅bn−1,P(BR)=bn⋅an−1,\mathbb{P}(RB)=\frac{a}{n}\cdot\frac{b}{n-1},\quad \mathbb{P}(BR)=\frac{b}{n}\cdot\frac{a}{n-1},

因此

P(异色)=2abn(n−1).\mathbb{P}(\text{异色})=\frac{2ab}{n(n-1)}.

代回得到

E[#异色相邻对]=(n−1)⋅2abn(n−1)=2aba+b.\mathbb{E}[\#\text{异色相邻对}]=(n-1)\cdot \frac{2ab}{n(n-1)}=\boxed{\frac{2ab}{a+b}}.

英文解析

For each adjacent pair of positions i,i+1i, i+1 (there are n−1n-1 such pairs), let the indicator variable IiI_i represent whether the pair has different colors. Then the expectation is

E[∑i=1n−1Ii]=(n−1) P(adjacent two are different colors).\mathbb{E}\left[\sum_{i=1}^{n-1} I_i\right]=(n-1)\,\mathbb{P}(\text{adjacent two are different colors}).

In a random permutation,

P(RB)=an⋅bn−1,P(BR)=bn⋅an−1,\mathbb{P}(RB)=\frac{a}{n}\cdot\frac{b}{n-1},\quad \mathbb{P}(BR)=\frac{b}{n}\cdot\frac{a}{n-1},

therefore

P(different colors)=2abn(n−1).\mathbb{P}(\text{different colors})=\frac{2ab}{n(n-1)}.

Substituting this back yields

E[#different color adjacent pairs]=(n−1)⋅2abn(n−1)=2aba+b.\mathbb{E}[\#\text{different color adjacent pairs}]=(n-1)\cdot \frac{2ab}{n(n-1)}=\boxed{\frac{2ab}{a+b}}.