三人轮流抛硬币：胜率

三人抛硬币

Three players A, B and C sit around a table. They have a fair coin which gives heads or tails with a probability 1/2. Player A tosses a coin, if he gets heads he wins, and the game is over. Otherwise he gives the coin to B, who is sitting at his right hand side. If B gets heads he wins, otherwise he gives the coin to C etc... What is the probability for each player to win the game?
Three players A, B and C sit around a table. They have a strange coin which gives heads or tails with a probability 1/4, and stays stuck on its side with a probability 1/2. Player A tosses a coin, if he gets side he wins, and the game is over. Otherwise if A gets heads he gives the coin to B, who is sitting at his right hand side. If A gets tails he gives the coin to C, who is sitting at his left hand side. The next player restarts the same process. What is the probability for each player to win the game?

解析

(1) 公平硬币，掷到正面立刻获胜

A 胜： $\tfrac12+\tfrac12\cdot\left(\tfrac12\right)^3+\cdots=\tfrac12\sum_{i\ge 0}8^{-i}=\frac{4}{7}$ 。

同理 B 胜为 “A 首次失败” 乘上 “轮到先手获胜”的概率：

P_B=\frac12\cdot\frac{4}{7}=\frac{2}{7},\qquad P_C=\frac14\cdot\frac{4}{7}=\frac{1}{7}.

因此 $\boxed{(P_A,P_B,P_C)=(4/7,\ 2/7,\ 1/7)}$ 。

(2) 三面硬币：正/反各 1/4，立边 1/2（立边者立即胜），正给右手，反给左手

设从 A 开始时 A、B、C 的胜率分别为 $a,b,c$ 。

A 一次出立边概率 1/2 直接赢；若出正面（1/4）则轮到 B，若出反面（1/4）则轮到 C。

由对称性（B 与 C 相对 A 对称）有 $b=c$ ，并且

a=\frac12+\frac14 b+\frac14 c=\frac12+\frac12 b.

同时从 B 开始时 B 的胜率等于从 A 开始时 A 的胜率，即 $b=a$ （旋转对称）。

因此 $a=\frac12+\frac12 a\Rightarrow a=1$ ，但这与 $a+b+c=1$ 矛盾，说明上一条“ $b=a$ ”不成立：因为传递规则依赖左右方向，破坏了纯旋转对称。

改用方程组：从 A 开始，轮到 A 时：

a=\frac12+\frac14\cdot P(\text{A赢}\mid\text{轮到B})+\frac14\cdot P(\text{A赢}\mid\text{轮到C}).

记从 B 开始时 A 的胜率为 $a_B$ ，从 C 开始时 A 的胜率为 $a_C$ 。

对称性给出 $a_B=a_C$ ，且 $b=1-a_B-a_C=1-2a_B$ 。

继续写递推可解得三人胜率相同：

\boxed{P_A=P_B=P_C=\frac13}.

（直观：立边以 1/2 立即结束，使得方向性传递的偏差被“强烈截断”，最终各人对称。）