The task of the three prisoners is the probability theory paradox, which has a common nature with the Monty Hall paradox . This paradox was first published by Martin Gardner in 1959.
Content
Formulation
Three prisoners, A, B and C, are imprisoned in solitary confinement and sentenced to death. The governor randomly chooses one of them and pardons him. The guard guarding the prisoners knows who is pardoned, but has no right to say this. Inmate A asks the guard to tell him the name of the (other) prisoner who will be executed for sure: “ If B is pardoned, tell me that C. will be executed. If pardoned by C, tell me that B. will be executed. If both of them are executed, I have mercy, toss a coin, and say the name B or C. "
The guard tells prisoner A that prisoner B will be executed. Prisoner A is glad to hear this, since he believes that now the probability of his survival is 1/2, not 1/3, as it was before. Prisoner A secretly tells Prisoner C that B will be executed. Inmate C is also happy to hear this, since he still believes that prisoner A's survival is 1/3, and his survival probability has increased to 2/3. How can this be?
Solution
The correct answer is that prisoner A did not receive information about his own fate. Prisoner A, before asking the guard, assesses his chances as 1/3, as well as B and C. When the guard says B will be executed, it’s like the likelihood that C is pardoned (1/3 probability) or A is pardoned (probability 1/3), and the coin that chooses between B and C chose B. (The probability is 1/2; overall, the probability that B is called is 1/6 because A is pardoned). Therefore, having learned that B will be executed, prisoner A estimates the chances of a pardon in this way: his chances are now 1/3, but now, knowing that B will be executed for sure, the chances of C for pardon are now 2/3.
Mathematical wording
Denote and as events indicating that the relevant prisoner will be pardoned, and an event signifying that the guard would call name B. Then, using Bayes theorem, the probability of a prisoner pardon A:
Intuitive Solution
Prisoner A has a chance of pardon 1/3. Knowing who from B and C will be executed does not change this chance. After prisoner A learns that B will be executed, he realizes that if he has not pardoned himself, then the chance that C will be pardoned is now 2/3.
Materials for understanding
Just as with the Monty Hall problem, it will be useful here to look at this problem from different points of view.
List of possible cases
The following cases may occur:
- A pardoned, and the guard announces that B will be executed: 1/3 × 1/2 = 1/6 of all cases
- A pardoned, and the guard announces that C will be executed: 1/3 × 1/2 = 1/6 of all cases
- B pardoned, and the guard declares that C will be executed: 1/3 of all cases
- C is pardoned, and the guard declares that B will be executed: 1/3 of all cases
With the proviso that in the situation when A is pardoned (the probability of such a situation is 1/3), the guard accidentally chooses the name of the executed person, he gets a chance 1/2, that he will say “B” and 1/2 that he will say “C”. This means that the probabilities are: 1/6 while (1/3 [A is really pardoned] * 1/2 [the guard calls B]), the guard calls B because A is pardoned, and (1/3 [A is really pardoned ] * 1/2 [the guard calls C]) the guard calls C, because A is pardoned. All this is 1/3 of all cases (1/6 + 1/6) when A is pardoned.
Now it is clear that the guard answers “Executed will be B” to the question of prisoner A (these are cases 1 and 4) in 1/2 of all cases; 1/3 is the probability that C is pardoned, but A will still be executed (case 4); and only 1/6 is the probability that A has been pardoned (case 1). Therefore, odds C: (1/3) / (1/2) = 2/3, odds A: (1/6) / (1/2) = 1/3.
The main snag here is that the guard cannot speak the name of the person who will be pardoned. If this condition is excluded, the original problem can be reformulated as follows: the prisoner asks the guard to tell him the fate of one of the two prisoners B and C, without specifying who will be executed. In this case, the guard throws up a coin to choose between B and C, and then tells the fate of one of them. With this wording, the following cases are possible.
- A pardoned, the guard says: B will be executed (1/6)
- A pardoned, the guard says: C will be executed (1/6)
- B pardoned, guard says: B pardoned (1/6)
- B pardoned, the guard says: C will be executed (1/6)
- C pardoned, the guard says: B will be executed (1/6)
- C is pardoned, the guard says: C is pardoned (1/6)
All outcomes are equally likely - 1/6. So: the guard in this situation still chooses from 6 cases, and he still cannot reveal the cards and say who is pardoned. In case 3, the guard cannot say that B is pardoned, so he will say that C will be executed (which will be true, because if B is pardoned, prisoners A and C will be executed). Also in case 6, when C is pardoned, but the guard who does not have the right to say this will call one of those who will be executed - he will call prisoner A the name of prisoner B. This brings the probability of cases 4 and 5 to 1/3, which leads us to the original results.
What is the paradox?
People think that the probability is 1/2, because they ignore the essence of the question that prisoner A asks the guard. If the guard could answer the question “ Will prisoner B be executed? ", Then in case of a positive answer, the probability of execution A would indeed decrease from 2/3 to 1/2.
The restriction that exists in the original problem of three prisoners makes the question of prisoner A useless, because with a probability of 100% two prisoners will be executed. That is, even if A is pardoned, they will call him any name; if A is sentenced to execution, it means that another prisoner will be executed with him, his name will be called prisoner A.
It turns out that prisoner A simply finds out with his question that one of prisoners B and C will be executed, which is clear from the conditions of the problem.
See also
- Prisoner's dilemma
- The problem of two envelopes
- The Parrondo Paradox
- Parado Monty Hall
- The problem of 100 prisoners
Links
- Mosteller, F. : Fifty Challenging Problems in Probability . Dover 1987 (reprint), ISBN 0-486-65355-2 , p. 28-29 ( restricted online version in Google Books )
- Richard Isaac: Pleasures of Probability . Springer 1995, ISBN 9780387944159 , p. 24-27 ( restricted online version in Google Books )