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转盘随机游走：最后出现的扇区

A wheel of fortune

专题
Probability / 概率
难度
L4

题目详情

A wheel of fortune has n≥2n \geq 2 sectors labeled by 1,2,…,n1,2,\ldots ,n . In each step the wheel is rotated by one sector to the left or to the right with equal probability. The procedure is repeated until every number from the set {1,2,…,n}\{1,2,\ldots ,n\} appears at least once on top of the wheel. If the number 1 is on top at the beginning, determine the probability that the last number that appears on top is kk , where k∈{2,3,…,n}k \in \{2,3,\ldots ,n\} .

解析

把扇区重标为 0,1,…,n−10,1,\ldots,n-1，起始为 0。每步左右移动 1 等价于一维对称随机游走在环上探索新点。

对任意 k∈{1,2,…,n−1}k\in\{1,2,\ldots,n-1\}，最后一个“首次出现”的扇区在对称性下没有偏好，且总共有 n−1n-1 个候选（不可能是起始 0）。

因此

P(最后出现的是 k)=1n−1,k=2,3,…,n.\boxed{\mathbb{P}(\text{最后出现的是 }k)=\frac{1}{n-1}},\quad k=2,3,\ldots,n.