转盘游戏

定义状态：

• $S_0$：出现过 0 次“5”，且尚未失败
• $S_1$：出现过 1 次“5”，且尚未失败
• $L$：失败状态（已经看到“4”或“6”）
• $W$：获胜状态（已经看到第 2 次“5”）

转移概率为： $$\begin{cases} S_0 \to S_1: 1/7, & S_0 \to L: 2/7, & S_0 \to S_0: 4/7 \\ S_1 \to W: 1/7, & S_1 \to L: 2/7, & S_1 \to S_1: 4/7 \end{cases}$$

设 $p_0$ 为从 $S_0$ 出发最终获胜的概率，$p_1$ 为从 $S_1$ 出发最终获胜的概率，则

$$p_0 = \frac{1}{7}p_1 + \frac{4}{7}p_0, \quad p_1 = \frac{1}{7} + \frac{4}{7}p_1$$

先解 $p_1$： $$p_1 - \frac{4}{7}p_1 = \frac{1}{7} \implies \frac{3}{7}p_1 = \frac{1}{7} \implies p_1 = \frac{1}{3}.$$

再解 $p_0$： $$p_0 - \frac{4}{7}p_0 = \frac{1}{7}p_1 \implies \frac{3}{7}p_0 = \frac{1}{7} \cdot \frac{1}{3} \implies p_0 = \frac{1}{9}.$$

$$\boxed{\frac{1}{9}}$$

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Original Explanation

Define states:

• $S_0$: 0 occurrences of "5", game not lost
• $S_1$: 1 occurrence of "5", game not lost
• $L$: Losing state (seen "4" or "6")
• $W$: Winning state (seen 2nd "5")

Transition probabilities: $$\begin{cases} S_0 \to S_1: 1/7, & S_0 \to L: 2/7, & S_0 \to S_0: 4/7 \\ S_1 \to W: 1/7, & S_1 \to L: 2/7, & S_1 \to S_1: 4/7 \end{cases}$$

Let $p_0$ = probability of winning from $S_0$, $p_1$ = probability of winning from $S_1$. Then

$$p_0 = \frac{1}{7}p_1 + \frac{4}{7}p_0, \quad p_1 = \frac{1}{7} + \frac{4}{7}p_1$$

Solve $p_1$: $$p_1 - \frac{4}{7}p_1 = \frac{1}{7} \implies \frac{3}{7}p_1 = \frac{1}{7} \implies p_1 = \frac{1}{3}.$$

Then $$p_0 - \frac{4}{7}p_0 = \frac{1}{7}p_1 \implies \frac{3}{7}p_0 = \frac{1}{7} \cdot \frac{1}{3} \implies p_0 = \frac{1}{9}.$$

$$\boxed{\frac{1}{9}}$$