Three Prisoners Problem ("probability paradox")

There are three prisoners, A, B, and C. Two of them will be released and one will be executed. A asks the warden to tell him the name of one of the others in his cohort who will be released. As the question is not directly about A's fate, the warden obliges and says, "B will be released." Assuming the warden's truthfulness, what are A's and C's respective probabilities of dying now?


NOTE: A and C do not have equal probability. (that's what I was initially thinking)

I was reading some things on probability and found this (and other similar problems). I found it interesting and it gave me a better understanding of probability, so I thought I'd share.

Thinking about it a bit before looking for the answer.
Here a nice list of relating problems (includes this one as well):
http://en.wikipedia.org/wiki/Category:Probability_theory_paradoxes

 

It's a binomial I'm guessing?

 

^, I seriously don't get this stuff. I'm not really good with maths

 

If the warden is tell the truth he has 50% chance of being released if it is that simple, if not it is a still 2/3 or a chance for freedom... Also I looked at the Berskon's paradox on the page and that would take forever to figure out with only deductive reasoning... But it couldn't be to large of a number I assume, I don't want to do the problem so if you do it tell me.

 

What the fuck is this, that wiki doesn't make sense.

To sum the prison problem up, it says this is the solution.

A's estimated chances of dying are 1/3.

The warden tells him that B will be set free (which is 1/3 people)

Surely, if he knows B is going to be set free then that would leave two people who are up for execution. Which must give the chance 1/2 :S

Where as wiki says...
C's chances of dying are now 2/3, and A's are 1/3.

If i'm an idiot for this sorry, and if someone could explain i'll be grateful.

 

The key thing to note is this: The warden's answer has no effect on prisoner A's chance, meaning it will always remain at 1/3.

So, if prisoner B is released, prisoner C now has a 2/3 chance.

And don't forget this is a paradox: the answer seems to be something obvious (1/2), but it is not (1/3 or 2/3, depending on which prisoner you are referring to).

It might help you understand the solution a bit better if you write out all the possible outcomes, which Wiki handily does farther down the page. However, the explanation also adds in another possible case (the guard can say B or C is being freed), so that might confuse you more =S

 

I kind of understand it now, i was thinking more complecated than i should have.

The original predictive chances were already measured taking into account what the guard said to person A.

Thanks steph, and if you read this again read your PM's

 

The way your thinking never changes the probability of death, it was 2/3 of a chance before so if you know part of equation the outcome still comes out the same, if the evidence is true then it is 50% chance of freedom or death now, but that is if you factor in new evidence. Althoughhe chance are still the same as the start. Tell me if I am wrong.

 

I've read it all, and I can't understand how A's chance are any different from C's

I think math is just setting itself for epic fail

 

EDIT: Ok read the wiki article and come up with this:

It could very easily have been C that asked the question, since their circumstances do not differ except for the fact that A is the one that asked the question, hence p(A) = P(C) ?

Im confused....

 

I can understand why the probability is 1/3 for A, but, if the problem is viewed as a whole, not from any individuals perspective, then wouldn't the odds still be 50:50 for A and C?

 

Shredder, thats what i thought as well. Wiki the Monty Hall paradox, its kinda on the same lines and the diagrams make sense.

Basically, its got to do with the fact that we're assuming the warden had to give an answer. If A and B were to be set free, the warden could only say B (he cnat tip of A), if B and C were to be free he would say either B or C, while C and A cant be set free. Thus the fact he chose B , favours the fact A and not C will be released. Correct me if im wrong any

 

The problem arises when you exclude the third person and don't refer to the original probability.
I suppose I should've posted the Monty Hall problem, it's easier to understand:
http://en.wikipedia.org/wiki/Monty_Hall_problem

And if you still don't understand in that form, I could probably explain it.

 

But if he had said C is to be released wouldn't the same apply to B? If so then A is getting favored chances just by asking the question. (assuming his fate won't be revealed.

I think that after you look at all the crazy things you people are saying you'll still come back to 50:50.

 

Yeah...

Yes, asking the warden increases A's probability of being released.

 

hhhhmmmmmm. yes agreed

 

A's chances stay the same; its C's that decrease

 

I think math just looses. It's obviously 50:50

 

my brain hurts now lol.

 

noes :'(
math is winn0r.