三名囚犯问题

关键在于狱卒在“可说两个人名”时如何选择。

常见假设：如果 A 被赦免（则 B、C 都会被处决），狱卒在说 B 或说 C 之间等概率随机；若 A 未被赦免，则狱卒只能说出那个必死者。

设 $H$ 为“狱卒说 B 会被处决”。

• $P(A\text{被赦免})=1/3$，且此时 $P(H\mid A\text{被赦免})=1/2$。
• $P(C\text{被赦免})=1/3$，此时 B 必死，$P(H\mid C\text{被赦免})=1$。
• $P(B\text{被赦免})=1/3$，此时狱卒必说 C，$P(H\mid B\text{被赦免})=0$。

因此

$$P(A\text{被赦免}\mid H)=\frac{(1/3)(1/2)}{(1/3)(1/2)+(1/3)(1)}=\frac{1}{3}.$$

所以答案为 $\boxed{1/3}$（在上述狱卒随机策略假设下）。

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英文解析

The key is how the jailer chooses when “two names can be said”.

Common assumptions: If A is pardoned (then B, C will be executed), the jailer is random in the probability of saying B or C; if A is not pardoned, the jailer can only say the mortal.

Set $H$ to “The jailer said B would be executed.”

-$P(A\text{is pardoned})=1/3$, and now$P(H\mid A\text{is pardoned})=1/2$.
-$P(C\text{is pardoned})=1/3$, at which point B must die,$P(H\mid C\text{is pardoned})=1$.
-$P(B\text{is pardoned})=1/3$, at which point the jailer must say C,$P(H\mid B\text{is pardoned})=0$.

Therefore,

$$P(A\text{is pardoned}\mid H)=\frac{(1/3)(1/2)}{(1/3)(1/2)+(1/3)(1)}=\frac{1}{3}.$$

So the answer is $\boxed{1/3}$ (under the above jailer random strategy assumption).