>[!abstract] >The three prisoners problem is a [[Veridical paradox|veridical paradox]] and probability puzzle similar to the [[Monty Hall problem]], illustrating how new information changes conditional probabilities in counterintuitive ways. Three prisoners—A, B, and C—know that one will be pardoned and the other two executed. Prisoner A asks the warden to name one of the others who will be executed, and the warden says "B". A then reasons that his own chance of survival has risen from 1/3 to 1/2, since only A and C remain. However, this intuition is wrong: the probability that A is pardoned remains **1/3**, while C's probability rises to **2/3**, because the warden’s response contains information conditioned on A’s status. The problem highlights how selective disclosure reshapes probability distributions without changing the underlying odds. >[!quote] Problem >Three men—A, B and C—were in separate cells under sentence of death when the governor decided to pardon one of them. He wrote their names on three slips of paper, shook the slips in a hat, drew out one of them and telephoned the warden, requesting that the name of the lucky man be kept secret for several days. Rumor of this reached prisoner A. When the warden made his morning rounds, A tried to persuade the warden to tell him who had been pardoned. The warden refused. > >“Then tell me,” said A, “the name of one of the others who will be executed. If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, flip a coin to decide whether to name B or C.” > >“But if you see me flip the coin,” replied the wary warden, “you’ll know that you’re the one pardoned. And if you see that I don’t flip a coin, you’ll know it’s either you or the person I don’t name.” > >“Then don’t tell me now,” said A. “Tell me tomorrow morning.” >The warden, who knew nothing about probability theory, thought it over that night and decided that if he followed the procedure suggested by A, it would give A no help whatever in estimating his survival chances. So next morning he told A that B was going to be executed. > >After the warden left, A smiled to himself at the warden’s stupidity. There were now only two equally probable elements in what mathematicians like to call the “sample space” of the problem. Either C would be pardoned or himself, so by all the laws of conditional probability, his chances of survival had gone up from 1⁄3 to 1⁄2. > >The warden did not know that A could communicate with C, in an adjacent cell, by tapping in code on a water pipe. This A proceeded to do, explaining to C exactly what he had said to the warden and what the warden had said to him. C was equally overjoyed with the news because he figured, by the same reasoning used by A, that his own survival chances had also risen to 1/⁄2. > >Did the two men reason correctly? If not, how should each calculate his chances of being pardoned? ([[Gardner, 1959]]) >[!quote] Solution >The answer to the problem of the three prisoners is that A’s chances of being pardoned are 1⁄3 and that B’s chances are 2⁄3. Regardless of who is pardoned, the warden can give A the name of a man, other than A, who will die. The warden’s statement therefore has no influence on A’s survival chances; they continue to be 1⁄3. The situation is analogous to the following card game. Two black cards (representing death) and a red card (the pardon) are shuffled and dealt to three men: A, B, C (the prisoners). If a fourth person (the warden) peeks at all three cards, then turns over a black card belonging to either B or C, what is the probability that A’s card is red? There is a temptation to suppose it is 1⁄2 because only two cards remain face down, one of which is red. But since a black card can always be shown for B or C, turning it over provides no information of value in betting on the color of A’s card. This is easy to understand if we exaggerate the situation by letting death be represented by the ace of spades in a full deck. The deck is spread, and A draws a card. His chance of avoiding death is 51⁄52. Suppose now that someone turns face up 50 cards that do not include the ace of spades. Only two face-down cards are left, one of which must be the ace of spades, but this obviously does not lower A’s chances to 1⁄2. What about prisoner C? Since either A or C must die, their respective probabilities for survival must add up to 1. A’s chances to live are 1⁄3; therefore, C’s chances must be 2⁄3. This can be confirmed by considering the four possible elements in our sample space and their respective initial probabilities: > >- C is pardoned; warden names B. (Probability 1⁄3) >- B is pardoned; warden names C. (Probability 1⁄3) >- A is pardoned; warden names B. (Probability 1⁄6) >- A is pardoned; warden names C. (Probability 1⁄6) > >In cases 3 and 4, A lives, making his survival chances 1⁄3. Only cases 1 and 3 apply when it becomes known that B will die. The chances that it is case 1 are 1⁄3, or twice the chances (1⁄6) that it is case 3, so C’s survival chances are two to one, or 2⁄3. In the card-game model, this means that there is a probability of 2⁄3 that C’s card is red ([[Gardner, 1959]]). >[!related] >- **North** (upstream): [[Probability theory]], [[Veridical paradox]] >- **West** (similar): [[Monty Hall problem]] >- **East** (different): [[Intuition-based reasoning]] >- **South** (downstream): [[Bayesian updating]]