In a prison, there are three men on death row. Let's call them A, B and C. As an act of mercy, the warden decides to release two of them. Without the participation of the prisoners, a lottery is held to determine which ones will be released, where all three prisoners are given equal chances. However, the warden wants to let the prisoners sweat for some time, so he decides not to tell them which one will be executed until next week.

However, when visiting prisoner A, the warden is persuaded to at least name one other prisoner that will be released. However, he doesn't remember the result of the lottery, so he walks back to his office in order to look it up.

On the way there, he decides that if prisoner A is to be executed and B and C are to be released, he will pick one of B and C with 1/2 probability each, and name that one to prisoner A. If prisoner A is to be released, he will simply name the other prisoner to also be released.

Coming back to prisoner A's cell, the warden says "Prisoner C will be released".

Question: What are the chances of each prisoner to be executed?

I've got the answer right by intuition pretty easily. Getting the formulas took me a bit longer, as I mixed things up. I still haven't found out how to solve this with set theory, so I would be very interested in seeing how that can be done.

But of course, please post your solutions, no matter how you arrive at them.

This is nothing more than the Monty Hall problem which has been discussed here countless times. I have actually asked this prisoner version of the problem, only with 7 prisoners.


Method 1:

A's chances of being executed from the beginning were 1/3, and this does not change upon learning the identity of one of the released prisoners. Therefore the probability that B is executed accounts for the remaining 2/3.


Method 2:

2 cases are possible:

a) B will be executed, in which case the guard had no choice but to tell prisoner C's fate (probability 1),

or

b) A will be executed, in which case the guard chose C randomly between B and C (probability 1/2).

The second case is half as likely as the first, therefore B is a 2:1 favorite to be executed (probability 2/3) while A will be executed with probability 1/3 as we said before.


Method 3:

This is really the same as method 2, but with the equations of Bayes' theorem worked out explicitly.
P(x | y) = the probability of event x occurring GIVEN the occurrence of event y.

Let A = event that A will be executed,
Let B = event that B will be executed
Let C = event that GUARD SAYS "C will be released"

By the definition of conditional proability we have:

P(A | C) = P(C and A) / P(C)

P(A | C) = P(C and A) / [ P(C and A) + P(C and B)]

By Bayes' therorem:

P(A | C) = P(C | A)*P(A) / [ P(C | A)*P(A) + P(C | B)*P(B) ]

P(A | C) = (1/2)*(1/3) / [ (1/2)*(1/3) + (1)*(1/3) ]

P(A | B) = (1/6) / (1/6 + 1/3)

P(A | B) = 1/3

So A is executed with probability 1/3, and B is executed with probability 2/3.

This is a pretty straight forward "If I flip a coin a thousand times and get tails everytime, what is the chance that my next flip will be heads.." type question.

The fact that Prisoner A cannot be named does not matter.

What you should consider is that Prisoner C will not be executed no matter what. Since there is an equal chance that it could be A or B, their odds of execution are 1/2 each.

Unfortunately, ay27aa, BruceZ is right, and you are wrong.

An alternative way to look at it is this:

A has a 1/3 a priori risk of being executed, and the a priority chance that either B or C is executed is 2/3. So, we group B and C together. When the warden says that C is to be released, we get no more information on the likelihood of either "group" to be executed, but we do get additional information on the intra-group probability of B and C. This is, B bears the whole risk of that group, since we now know that C bears none of that risk.

I'm still curious about the set theory solution. Anyone who knows how to do that?

I don't know if this is what you are looking for, but here is a simple way to think about the sample space for A:

1/3 of the time A is released after being told B will be released

1/3 of the time A is released after being told C will be released

1/6 of the time A is executed after being told B will be released

1/6 of the time A is executed after being told C will be released

You can see that the probabilities of these events sum to 1, so this covers all possibilities with no overlap. Now when A is told that "C will be released", A is either in the 2nd case or the 4th case, and his probability of being executed is (1/6) / [1/3 + 1/6] = 1/3.

I find this problem similar to the famous Monty Hall problem (or paradox if you will)
http://en.wikipedia.org/wiki/Monty_Hall_Paradox
In case you haven't heard about it I strongly suggest you study it a bit.
It is mathematically quite trivial, but does raise a few interesting questions about the way we think about probability and information and such things.
(When we were 2nd grade college students, our probability teacher asked this problem of the class and 90% of the people, who were all mathematician students, got it wrong. Which was interesting...as well as somewhat amusing.)

We must assume that all of their odds of execution are equal.

A = Prisoner A B = Prisoner B C = Prisoner C

E = odds of execution R = odds of release

A = 1/3E + 2/3R B = 1/3E + 2/3R C = 1/3E + 2/3R

B + C = 2/3E + 4/3R

Assuming that C is released:

C = 3/3R

B + C - C = 2/3E + 4/3R - 3/3R

B = 2/3E + 1/3R

A = 1/3E + 2/3R

Assuming that B is released:

B = 3/3R

B + C - B = 2/3E + 4/3R - 3/3R

C = 2/3E + 1/3R

A = 1/3E + 2/3R


I hope that helped clarify things for you Klyka.

If I read that properly (and I have no idea why the variable R and E have been added in, and what B+C actually -means-) this is a different answer to the one you gave a few posts ago.

Yes, I was wrong the first time. I was in a hurry but I wanted to submit the proof before I left so I didn't have a chance to write a little blurb.

I read the monty hall paradox on wikipedia, and the 100 doors if he opens the other 98 part really helped me to conceptualize it. After realizing my err, I then proceeded to write a formulaic approach to demonstrate the transfer of probability vs equalization to 1/2.

You were right Klyka, the part about the "intragroup" also really helped me to understand why i was wrong.

=)

I'll post my intuitive answer without reading the other posts.




Mathematical analysis:

Here is a solution I have not yet seen posted. Let

A = "Prisoner A will be executed."
C = "Warden tells A that C will be released."

It is clear from the problem statement that P(C | A) = 1/2 and P(C | not-A) = 1/2. Whenever two events satisfy P(C | A) = P(C | not-A), those two events must be independent. Hence, P(A | C) = P(A) = 1/3.

I think I've got a simpler way of looking at this.

Whatever happens, assuming the lottery is fair, Prisoner A has a 1/3 chance of being executed - on the basis that, going by the rules the warden has set, he won't find out either way, so it doesn't matter what the warden says.

We know Prisoner C has a 0 chance of being executed.

Which leaves Prisoner B with 2/3 chance.

EDIT: what darkmagus said.

C will be released
A has 2/3 change to be released, because his odds is only effected by lottery
B if C lost lottery so all 3 will be released. If C won lottery B has 1/2 change of being released. So 1/3 * 1 + 2/3 * 1/2 = 2/3

Answer A - 1/3, B - 1/3, C - 0.

There is no releasing all three. Exactly one prisoner will be executed, so A, B and C must add up to 1.

I see this sort of "solution" a lot. But really, it just amounts to stating the answer, without any justification. The heart of the problem, that so many people have difficulty with, is why do A's chances not change upon learning the identity of one of the released prisoners? This "solution" does nothing to address the question.

If we translate the above into mathematics, it says:

(i) P(A executed) = 1/3
(ii) P(A executed | learns C is released) = 1/3
(iii) Therefore, P(B executed | learns C is released) = 2/3

Step (ii) is presented without any justification at all.

On the other hand, DarkMagus presented an "intuitive" answer which, although superficially similar to the above, is actually a complete solution. I would like to illustrate how complete it is by translating each relevant sentence into mathematics. For this, let

A = "A will be executed"
B = "A learns that B will be released"
C = "A learns that C will be released"

The events B and C are complementary, i.e. B = not-C.

Since, by the symmetry of the problem, P(A | B) = P(A | C), i.e. P(A | not-C) = P(A | C)...

...A and C are independent.

Therefore, P(A | C) = P(A) = 1/3.

Nice!