颜色切换期望

Expected number of color changes in 50 black & 50 white draws

专题: Probability / 概率
难度: L4
来源: QuantQuestion

个人笔记

题目详情

盒子里有 50 个黑球和 50 个白球，不放回依次抽出直至抽完。一旦相邻两次抽取的颜色发生变化，就记为一次“颜色切换”。

问：颜色切换次数的期望是多少？

英文原题

Same as the “runs” question with 50 black and 50 white marbles, drawn without replacement. A “color change” occurs when consecutive draws switch color. What is the expected number of changes?

解析

令 $I_i$ 表示第 $i$ 次与第 $i+1$ 次抽取颜色不同的指示变量（ $i=1,\dots,99$ ）。则切换次数为 $\sum_{i=1}^{99} I_i$ 。

在随机排列的模型下，任意相邻位置出现“黑白”或“白黑”的概率为

P(I_i=1)=\frac{2\cdot 50\cdot 50}{100\cdot 99}=\frac{50}{99}.

因此期望切换次数为

\mathbb{E}\left[\sum_{i=1}^{99} I_i\right]=99\cdot\frac{50}{99}=50.

英文解析

Let $I_i$ represent the indicator variable ( $i=1,\dots,99$ ) that is different in color from the $i$ and $i+1$ draws. The toggle count is $\sum_{i=1}^{99} I_i$ .

Under the randomly arranged model, the probability of "black and white" or "white and black" in any adjacent position is

P(I_i=1)=\frac{2\cdot 50\cdot 50}{100\cdot 99}=\frac{50}{99}.

Therefore, the expected number of switches is

\mathbb{E}\left[\sum_{i=1}^{99} I_i\right]=99\cdot\frac{50}{99}=50.