Answer: The Prisoner's Pardon: Three Boxes, One Clue

Key Idea / Intuition

This problem is a clever mirror image of the Monty Hall problem — but with the opposite conclusion. In Monty Hall, a door being opened shifts probability away from you. Here, because two out of three prisoners are pardoned, the guard always has someone other than Alice to name (whether or not Alice is pardoned), so the guard's answer carries no information about Alice's fate. Her probability stays exactly $2/3$.

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Formal Proof / Solution

Setup. There are $\binom{3}{2} = 3$ equally likely scenarios, each with probability $1/3$:

| Scenario | Who is pardoned | Alice pardoned? | |----------|----------------|-----------------| | S1 | Alice, Bob | Yes | | S2 | Alice, Carol | Yes | | S3 | Bob, Carol | No |

Alice is pardoned in 2 out of 3 scenarios, so $P(\text{Alice pardoned}) = 2/3$.

After the guard says "Bob." We apply Bayes' theorem. The guard must name someone other than Alice who is pardoned. Let $G = \text{"guard names Bob"}$.

• S1 (Alice, Bob pardoned): Guard can name Bob or Carol — wait, Carol is not pardoned. Guard must name Bob. So $P(G \mid S1) = 1$.
• S2 (Alice, Carol pardoned): Guard can only name Carol (Bob is not pardoned). So $P(G \mid S2) = 0$.
• S3 (Bob, Carol pardoned): Guard can name Bob or Carol. Assuming the guard picks uniformly at random, $P(G \mid S3) = 1/2$.

By the law of total probability: $P(G) = \frac{1}{3}\cdot 1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot\frac{1}{2} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}.$

Now update: $P(S1 \mid G) = \frac{P(G\mid S1)\,P(S1)}{P(G)} = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}.$

$P(S3 \mid G) = \frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3}.$

Alice is pardoned in scenario S1 (and not in S3), so: $P(\text{Alice pardoned} \mid G) = P(S1 \mid G) = \frac{2}{3}.$

Conclusion. Alice's probability of being pardoned is still $\boxed{2/3}$, unchanged. The guard's information tells her who else is safe, but nothing about herself — because there were always at least two pardoned prisoners and the guard could always find someone to name.

Contrast with Monty Hall. In Monty Hall, only one prize exists and the host must open an empty door other than yours. Here, two pardons exist and the guard must name a pardoned prisoner other than you. The symmetry works out oppositely: in Monty Hall you should switch (probability goes from $1/3$ to $2/3$); here, there is nothing to switch and your probability stays put.