The Monty Hall Remix — Three Prisoners

Anand's pardon probability stays at 1/3. Chetan's probability jumps to 2/3. The warden's naming of Bina transfers her 1/3 probability entirely to Chetan (not split with Anand), because the warden was forced to name Bina when Chetan is pardoned but had a choice when Anand is pardoned. WOW Moment: - Before: Anand=1/3, Bina=1/3, Chetan=1/3. - Warden says Bina will be executed. - After: Anand=1/3, Chetan=2/3. - Anand was right — his own odds did not change. But he was wrong that nothing changed. Chetan's odds DOUBLED from 1/3 to 2/3. - If Anand and Chetan could swap fates right now, Chetan should refuse and Anand should beg to swap. - You can gain information about others without gaining information about yourself. Bayesian probability is not symmetric.

The Three Prisoners problem is a classic brain teaser in probability theory that is mathematically identical to the Monty Hall problem. The result is highly counterintuitive because our brains naturally want to divide the remaining probability equally between the two survivors (making it 1/2 each). Let's use Bayes' Theorem to see why the probability is asymmetric: 1. Define the hypotheses: - A: Anand is pardoned. P(A) = 1/3. - B: Bina is pardoned. P(B) = 1/3. - C: Chetan is pardoned. P(C) = 1/3. 2. Let the evidence E be: "The warden names Bina as one to be executed." 3. Calculate the conditional probabilities of the warden's action under each hypothesis: - If Anand is pardoned (A): The warden can name either Bina or Chetan. Since they are identical, he chooses Bina with probability 1/2. P(E | A) = 1/2. - If Bina is pardoned (B): The warden cannot name Bina (she is to be pardoned). P(E | B) = 0. - If Chetan is pardoned (C): The warden must name Bina (since he cannot name Anand who asked the question, and he cannot name Chetan who is pardoned). P(E | C) = 1. 4. Apply Bayes' Theorem to find Anand's posterior probability: P(A | E) = [P(E | A) × P(A)] / [P(E | A)P(A) + P(E | B)P(B) + P(E | C)P(C)] P(A | E) = [1/2 × 1/3] / [(1/2 × 1/3) + (0 × 1/3) + (1 × 1/3)] P(A | E) = [1/6] / [1/6 + 1/3] = (1/6) / (1/2) = 1/3. Anand's probability of being pardoned remains exactly 1/3! 5. Now apply Bayes' Theorem for Chetan: P(C | E) = [P(E | C) × P(C)] / [P(E | A)P(A) + P(E | B)P(B) + P(E | C)P(C)] P(C | E) = [1 × 1/3] / [1/2] = 2/3. Chetan's probability of being pardoned has jumped to 2/3! To explain the WOW part: Why did Chetan benefit while Anand did not? The warden's statement carried no information about Anand because the warden would have named *someone* (either Bina or Chetan) regardless of whether Anand was pardoned or not. However, the warden's statement was highly informative about Chetan. If Chetan were the one pardoned, the warden was *forced* to name Bina. If Bina were pardoned, the warden would have named Chetan. By naming Bina, the warden completely eliminated Bina's chance of being pardoned, and because the warden's choice was constrained, all of Bina's prior probability (1/3) was transferred to Chetan, not Anand.