Thank you for visiting this site. This article covers “The Three Prisoners Problem.”
This paradox shares its essential structure with the Monty Hall problem and has long been known in Japan. It vividly illustrates the gap between intuition and calculation in conditional probability.
The Setup
Three prisoners — A, B, and C — have been sentenced to death. However, one of them, chosen at random, will be pardoned and released. The prisoners do not yet know who will be pardoned.
Prisoner A makes a request to the guard: “Of B and C, tell me the name of just one who will be executed. Since at least one of the three must be executed, surely you can share that much?”
The guard replies: “B will be executed.”
Prisoner A now reasons: “My odds of survival were 1/3 to begin with. But now that B is confirmed to be executed, only C or I can survive — so my odds must have risen to 1/2, right?”
Is A’s reasoning correct?
The Intuitive Guess and the Real Answer
A’s reasoning is wrong.
Even after hearing the guard’s information, A’s probability of survival remains 1/3. Meanwhile, C’s probability of survival rises to 2/3.
“Wait — why are A’s and C’s probabilities different?” That is precisely what makes this paradox fascinating.
Why A’s Probability Stays at 1/3
Let’s enumerate all cases. There are three scenarios depending on who receives the pardon.
Case 1: A is pardoned (probability 1/3) The guard may name either B or C. The probability that he says “B will be executed” is 1/2.
Case 2: B is pardoned (probability 1/3) The guard must say “C will be executed.” He cannot say “B will be executed” because B is pardoned.
Case 3: C is pardoned (probability 1/3) The guard says “B will be executed” (since B is being executed).
The cases where the guard says “B will be executed” are Case 1 (probability 1/3 × 1/2 = 1/6) and Case 3 (probability 1/3 × 1 = 1/3).
Given that the guard said “B will be executed,” the probability that A is pardoned is (1/6) ÷ (1/6 + 1/3) = (1/6) ÷ (1/2) = 1/3.
The probability that C is pardoned under the same condition is (1/3) ÷ (1/2) = 2/3.
The Connection to the Monty Hall Problem
You may have noticed that this is structurally identical to the Monty Hall problem.
- Three doors → Three prisoners
- The winning door → The pardoned prisoner
- Your initial choice → Prisoner A himself
- Monty opens a losing door → The guard names someone who will be executed
In the Monty Hall problem, the correct move is to switch doors. In the Three Prisoners Problem, the equivalent conclusion is: C has twice A’s probability of survival.
The guard only told A which one of B or C would be executed. This information does not change A’s probability, but it does shift C’s probability upward. The key insight is that information acts asymmetrically.
Why Prisoner A Is Wrong
The error A commits is the reasoning: “The candidates went from three to two, so the probability must be 1/2.”
Narrowing the candidates does sometimes change probabilities — but what matters here is that which name the guard says is not random. The guard can never name the pardoned prisoner, so his answer is systematically biased depending on who the pardon belongs to.
Ignoring this bias and simply thinking “two candidates, so 1/2” is exactly where the intuition goes wrong.
Summary
This article covered “The Three Prisoners Problem.”
Conditional probability consistently defies intuition. The lesson — that receiving information does not necessarily update probabilities symmetrically — is extremely important whenever you work with statistics.
To return to the full list of paradoxes, follow the link below.
Thank you for reading. We hope to see you in the next article.
