Three Prisoners problem

The explanation above may be summarised in the following table. As the warden is asked by A, he can only answer B or C to be executed (or "not pardoned").

Being pardoned	Warden: "not B"	Warden: "not C"	Sum
A	1/6	1/6	1/3
B	0	1/3	1/3
C	1/3	0	1/3

As the warden has answered that B will not be pardoned, the solution comes from the second column "not B". It appears that the odds for A vs. C to be pardoned are 1:2.

Call ${\displaystyle A}$, ${\displaystyle B}$ and ${\displaystyle C}$ the events that the corresponding prisoner will be pardoned, and ${\displaystyle b}$ the event that the warden tells A that prisoner B is to be executed, then, using Bayes' theorem, the posterior probability of A being pardoned, is:

${\displaystyle {\begin{aligned}P(A|b)&={\frac {P(b|A)P(A)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}\\&={\frac {{\tfrac {1}{2}}\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {1}{3}}.\end{aligned}}}$

The probability of C being pardoned, on the other hand, is:

${\displaystyle {\begin{aligned}P(C|b)&={\frac {P(b|C)P(C)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}\\&={\frac {1\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {2}{3}}.\end{aligned}}}$

The crucial difference making A and C unequal is that ${\displaystyle P(b|A)={\tfrac {1}{2}}}$ but ${\displaystyle P(b|C)=1}$. If A will be pardoned, the warden can tell A that either B or C is to be executed, and hence ${\displaystyle P(b|A)={\tfrac {1}{2}}}$; whereas if C will be pardoned, the warden can only tell A that B is executed, so ${\displaystyle P(b|C)=1}$.

The following scenarios may arise:

1. A is pardoned and the warden mentions B to be executed: 1/3 × 1/2 = 1/6 of the cases
2. A is pardoned and the warden mentions C to be executed: 1/3 × 1/2 = 1/6 of the cases
3. B is pardoned and the warden mentions C to be executed: 1/3 of the cases
4. C is pardoned and the warden mentions B to be executed: 1/3 of the cases

With the stipulation that the warden will choose randomly, in the 1/3 of the time that A is to be pardoned, there is a 1/2 chance he will say B and 1/2 chance he will say C. This means that taken overall, 1/6 of the time (1/3 [that A is pardoned] × 1/2 [that warden says B]), the warden will say B because A will be pardoned, and 1/6 of the time (1/3 [that A is pardoned] × 1/2 [that warden says C]) he will say C because A is being pardoned. This adds up to the total of 1/3 of the time (1/6 + 1/6) A is being pardoned, which is accurate.

It is now clear that if the warden answers B to A (1/2 of the time of case 1, and case 4), then 1/3 of the time C is pardoned and A will still be executed (case 4), and only 1/6 of the time A is pardoned (case 1). Hence C's chances are (1/3)/(1/2) = 2/3 and A's are (1/6)/(1/2) = 1/3.

The key to this problem is that the warden may not reveal the name of a prisoner who will be pardoned. If we eliminate this requirement, it can demonstrate the original problem in another way. The only change in this example is that prisoner A asks the warden to reveal the fate of one of the other prisoners (not specifying one that will be executed). In this case, the warden flips a coin chooses one of B and C to reveal the fate of. The cases are as follows:

1. A pardoned, warden says: B executed (1/6)
2. A pardoned, warden says: C executed (1/6)
3. B pardoned, warden says: B pardoned (1/6)
4. B pardoned, warden says: C executed (1/6)
5. C pardoned, warden says: B executed (1/6)
6. C pardoned, warden says: C pardoned (1/6)

Each scenario has a 1/6 probability. The original Three Prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1/6 probability of occurring. However, the warden in that case may not reveal the fate of a pardoned prisoner. Therefore, in the 1/6 of the time that case 3 occurs, since saying B is not an option, the warden says C instead (making it the same as case 4). Similarly, in case 6, the warden must say B instead of C (the same as case 5). That leaves cases 4 and 5 with 1/3 probability of occurring and leaves us with the same probability as above.