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The Prisoners’ Paradox

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Re: The Prisoners’ Paradox

#11

Post by gambling8nt »

randomtask wrote: post #8
jonsen wrote: post #6

Earlier quoted context omitted.

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/…

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.

Re: The Prisoners’ Paradox

#12

Post by huherto »

knowtheory wrote: post #9

Oh dear. So this is an implementation of the Monty Hall problem but now is crossing over into Metaphysics. There are several different ways to consider the ontological reality of probabilities (i guess wikipedia only has two listed, Frequentists and Bayesians http://en.wikipedia.org/wiki/Probability_interpretations ). The thought experiment in question probably should be answered with a Bayesian world view, and does…

But, did he discovered new information? What if the guard had said "Prisioner C is sure to die"?

Re: The Prisoners’ Paradox

#13

Post by gambling8nt »

huherto wrote: post #12
knowtheory wrote: post #9

Oh dear. So this is an implementation of the Monty Hall problem but now is crossing over into Metaphysics. There are several different ways to consider the ontological reality of probabilities (i guess wikipedia only has two listed, Frequentists and Bayesians http://en.wikipedia.org/wiki/Probability_interpretations ). The thought experiment in question probably should be answered with a Bayesian world view, and does…

But, did he discovered new information? What if the guard had said "Prisioner C is sure to die"?

He did discover new information. The probabilities of survival switched from: A: 1/3, B: 1/3, C:1/3 to A: 1/3, B:0, C: 2/3

He just didn't discover any information to change the probability of his own survival.

Re: The Prisoners’ Paradox

#15

Post by Lol_Lolovici »

mannicken wrote: post #3
mhb wrote: post #2

http://en.wikipedia.org/wiki/Monty_Hall_problem

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 incr…

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.

Re: The Prisoners’ Paradox

#16

Post by gort »

Before he asks the question, there are 4 possibilities:

  1. A lives, A gets told B dies
  2. A lives, A gets told C dies
  3. B lives, A gets told C dies
  4. C lives, A gets told B dies
Since in fact A got told that B dies, we are either in situation 1 or 4. However, these did not start off at equal probability. Lets define P(x) as the probability that x is the actual situation. It is obvious that, before the question is asked:

  P(1 or 2) == P(3) == P(4) == 1/3
Thus P(1) < P(4). This remains true once situations 2 and 3 have been ruled out.

Re: The Prisoners’ Paradox

#17

Post by huherto »

gambling8nt wrote: post #13
huherto wrote: post #12

Earlier quoted context omitted.

But, did he discovered new information? What if the guard had said "Prisioner C is sure to die"?

He did discover new information. The probabilities of survival switched from: A: 1/3, B: 1/3, C:1/3 to A: 1/3, B:0, C: 2/3 He just didn't discover any information to change the probability of his own survival.

nice. I think this sums it all.

Re: The Prisoners’ Paradox

#18

Post by jonsen »

gort wrote: post #16

Before he asks the question, there are 4 possibilities: 1. A lives, A gets told B dies 2. A lives, A gets told C dies 3. B lives, A gets told C dies 4. C lives, A gets told B dies Since in fact A got told that B dies, we are either in situation 1 or 4. However, these did not start off at equal probability. Lets define P(x) as the probability that x is the actual situation. It is obvious that, before the question is a…

Interesting setup. So

  P(1) == P(2) == 1/6
When A is told that B dies, it's down to 1. or 4.

  P(1 or 4) == 1/6 + 1/3 == 1/2
So chance A lives is now

  P(1) of P(1 or 4) == 1/6 of 1/2 == 1/3
And the chance C lives is now

  P(4) of P(1 or 4) == 1/3 of 1/2 == 2/3

Re: The Prisoners’ Paradox

#19

Post by moo »

gort wrote: post #16

Before he asks the question, there are 4 possibilities: 1. A lives, A gets told B dies 2. A lives, A gets told C dies 3. B lives, A gets told C dies 4. C lives, A gets told B dies Since in fact A got told that B dies, we are either in situation 1 or 4. However, these did not start off at equal probability. Lets define P(x) as the probability that x is the actual situation. It is obvious that, before the question is a…

A's survival chances don't change because the guard is filtering out the possible answers: B lives, A gets told A dies C lives, A gets told A dies

Re: The Prisoners’ Paradox

#20

Post by mhb »

The clearest explanation I've seen is this. Suppose there are 1,000 prisoners one of whom is to live. You ask the guard to tell you which 998 out of the 999 other prisoners will be killed. So there's one he didn't tell you about. Who do you think has a better chance of living - you or the prisoner in the other group whose fate the guard didn't reveal?
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