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Comment by mannicken | Hacker News Reader

Yeah, except it's pretty obvious why there is a switch to 1/2 from 1/3 in Prisoner's Paradox and it's not obvious in MH.

What I understand happened in Prisoner's Paradox is that the very field of probabilistic distribution of pardons has shrinked from 3 to 2. It's like binary search. You cheated your way by knowing where _not to look_.

Another simple analogy: you can increase the value of rational number both by increasing the numerator and decreasing denominator. In this case you decreased denominator.

Monty Hall requires a slightly bigger leap of logic.

Actually, here's a more directly linked problem: http://en.wikipedia.org/wiki/Three_Prisoners_problem

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You mean it's pretty obvious why there is no switch to 1/2 from 1/3? As the Wikipedia article you linked points out: "The answer is he didn't gain information about his own fate, but he should switch with C if he can."

This isn't the same as Monty Hall. Monty Hall only increases the odds after switching doors. This actually seems like a trick meant to fool people who know about the Monty Hall problem but don't really understand it.

It's strange why my blatantly stupid post was upvoted.

I re-read all three problems after you replied and this is what I understood. Talking about Three Prisoners' problem since the article's wording doesn't include the very crucial ingredient for easier understanding: coin flip on warden's side.

For A, it doesn't really matter if B or C were executed. In fact, he knew that either B or C are going to be executed for sure. He knew that at least one person who is not him will be executed.

Warden just named the sure-to-be-executed-guy. Didn't change what he previously knew about his chances, since he still doesn't know for sure whether the-other-not-named-guy is going to be executed.

However, the distribution changed for C. Now A still knows that there 1/3 chance of him living but since we know for sure that the-guy-who-warden-named (B or C) will be killed the-guy-who-warden-didn't-name-and-who-isn't-A (C or B) will have 1 - 1/3 = 2/3 probability of living.

Correct.

That means that at least one of my cellmates is still sure to die. Give me his name

Prisoner A rejoices that his own chance of survival has improved from 1/3 to 1/2. But how is this possible? The guard has given him no new information. Has he?

No, Prisoner A still has a 1 in 3 chance of [edit: not] dieing. If he asked if Prisoner B had a pardon then you could argue his odds became 1/2. But, now he still only knows that at least one of his cellmates is going to die.

I think you've got it backwards. Only one is pardoned. His chance of survival is 1 in 3.

The initial chance of survival, after hearing that there is a chance, (for Prisoner A) is 1/3, since any one prisoner dies out of three. If he now knows, after consulting with the guard, that Prisoner B is sure to die that means that it's death for either A or C. Hence the chance of survival (and death since P(Death) = 1 - P(Survival)) is now 0.5. This results in a smaller chance of death (for him) 1/2 rather than 2/3. Hence the rejoicing.

Because of the way the information was gathered, the chances of survival for A and C are no longer equal (C is twice as likely to survive)--having only two possible outcomes does not guarantee that the probability is 50-50.

As I understand the problem it is identical to the Monty Hall problem if you map survival-prize and doors-prisoners. The opposite of survival is death and the opposite of prize is nothing.

So when the guard says B is sure to die he is actually opening door number 2 to show no prize is there.

Before this event the odds of survival for each prisoner were 1/3, 1/3, 1/3.

The odds that the guard would choose prisoner B as the non-surviving one (dead that is) in case prisoner A/B/C is the surviving one are: 1/2, 0, 1.

That is if A was pardoned then the guard can chose either of B or C so the odds of pointing out B are 1/2. The odds of pointing out B as dead when he is pardoned are 0 and the odds of pointing out B as dead when C is pardoned are 1 (because he cannot indicate A as dead because the guard is not allowed to tell A about his own fate). Hence the final odds for each prisoner must add up to 1 so through proportionality they are 1/3, 0, 2/3.

So prisoner A didn't find out anything new about his chance of being the pardoned one.

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