转盘随机游走：最后出现的扇区

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

因此

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

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英文解析

Let us relabel the sectors to $0,1,\ldots ,n - 1$ , and let us assume that sector 0 is on top of the wheel at the beginning. The wheel rotations can be modeled by a symmetric random walk that starts at 0 and in each step changes its location by $+1$ or

• 1 with equal probabilities. In what follows, the negative locations $-1,
• 2,\ldots ,
• (n-
  1)$correspond to the sectors$n
• 1,n
• 2,\ldots ,1$, in that order. For$a\in \mathbb{Z}$, denote by$T_{a}$the hitting time of the set${a}$. Let$X(t)$denote the sector on top of the wheel after$t$steps. Let$\tau$denote the number of steps until for the first time every number from the set${0,1,\ldots ,n
• 1}$has appeared at least once on top of the wheel. Clearly,$\mathbb{P}(X(\tau) =0) = 0$. We will prove that$\mathbb{P}(X(\tau) = k) = 1 / (n - 1)$for every$k\in {1,2,\ldots ,n - 1}$

First, let us determine the probability of the event $\{X(\tau) = 1\}$ . The event $\{X(\tau) = 1\}$ is equal to the event that sector 2 appears on top of the wheel before 1 appears. This event, in turn, is equivalent to a symmetric random walk, starting at 0, visiting $- (n

• 2)$ before visiting 1. From (2.166), we know that

$$\mathbb{P}(X(\tau) = 1) = \mathbb{P}\left(T_{-(n - 2)}< T_{1}\right) = \frac{1}{n - 1}.$$

By symmetry, we also have that

$$\mathbb{P}(X(\tau) = n - 1) = \frac{1}{n - 1}.$$

Let $k\in \{2,3,\ldots ,n - 2\}$ . Denote by $A_{k}$ the event that a symmetric random walk starting at O visits $k - 1$ before visiting - $(n - k - 1)$ , and then, starting at $k - 1$ visits $- (n - k - 1)$ before visiting $k$ . Denote by $B_{k}$ the event that a symmetric random walk starting at O visits $- (n - k - 1)$ before visiting $k - 1$ , and $- (n - k)$ . Then $- (n - k - 1)$ visits $k - 1$ before $B_{k}$ are disjoint, and the event $X(\tau) =$ the events $A_{k}$ and $B_{k}$ are $A_{k}\cup B_{k}$ . Also, note that the event $k$ is equivalent to $A_{k}\cup B_{k}$ . of a symmetric random visiting $k$ is equivalent to the event of $- (n - k - 1)$ before visiting, starting at $(k - 1) - (k - 1) = 0$ , a symmetric random walk, $k - 1) = - (n - 2)$ before visiting $k - (k - 1) = 1$ . Similarly, the event of a symmetric random walk starting at $- (n - k - 1)$ , visiting $k - 1$ before visiting $- (n - k)$ is equivalent to the event of a symmetric random walk, starting at $- (n - k - 1) + (n - k - 1) = 0$ , visiting $k - 1 + (n - k - 1) = n - 2$ before visiting $- (n - k) + (n - k - 1) = - 1$

Then, by using (2.166) repeatedly, we obtain that

$$\begin{array}{r l} & {\mathbb{P}(X(\tau) = k) = \mathbb{P}(A_{k}) + \mathbb{P}(B_{k})}\\ & {= \mathbb{P}\left(T_{k - 1}< T_{-(n - k - 1)}\right)\cdot \mathbb{P}\left(T_{-(n - 2)}< T_{1}\right)}\\ & {\quad +\mathbb{P}\left(T_{-(n - k - 1)}< T_{k - 1}\right)\cdot \mathbb{P}\left(T_{n - 2}< T_{-1}\right)}\\ & {= \frac{n - k - 1}{n - 2}\cdot \frac{1}{n - 1} +\frac{k - 1}{n - 2}\cdot \frac{1}{n - 1}}\\ & {= \frac{1}{n - 1}\cdot \mathbb{P}} \end{array}$$