4 ms·

The Prisoners’ Paradox

10 points by TriinT 17y ago

mhb 17y ago
http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem
- mannicken 17y ago
  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 http://en.wikipedia.org/wiki/Three_Prisoners_problem
  - Retric 17y ago
    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.
    - jonsen 17y ago
      I think you've got it backwards. Only one is pardoned. His chance of survival is 1 in 3.
      - randomtask 17y ago
        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.
        gambling8nt 17y ago
        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.
  - mattmaroon 17y ago
    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.
    - mannicken 17y ago
      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.
      - mattmaroon 17y ago
        Correct.
  - Lol_Lolovici 17y ago
    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.
- furyg3 17y ago
  Came here to post this same link. That prisoner should immediately switch name-tags with the other prisoner who's fate is in question :)
pushingbits 17y ago
As non-intuitive puzzles go I think the hat guessing game (http://www.relisoft.com/science/hats.html http://www.relisoft.com/science/hats.html) takes the cake. I still remember a dailywtf thread where 19 pages worth of people claim that the correct solution is completely bogus (http://thedailywtf.com/Comments/Riddle-Me-An-Interview.aspx http://thedailywtf.com/Comments/Riddle-Me-An-Interview.aspx).
jonsen 17y ago
A's chances has not changed. But the chance of C being the pardoned has doubled to 2/3.
knowtheory 17y ago
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 http://en.wikipedia.org/wiki/Probability_interpretations ). The thought experiment in question probably should be answered with a Bayesian world view, and does not appear to be. Prisoner A's chances of being executed is not materially effected when he learns new information, because the decision as to who has been pardoned has already been made. Finding out new information, does change a probability, but not the probability that he'll be executed. It only changes the likelihood of whether or not the assertion that he will be executed is true. The nature of the universe has not actually changed, merely the state of belief of Prisoner A. The point being, he's got no cause to celebrate, upon merely discovering information, because learning more information does not actually change the state of the world, in this case.
- huherto 17y ago
  But, did he discovered new information? What if the guard had said "Prisioner C is sure to die"?
  - gambling8nt 17y ago
    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.
    - huherto 17y ago
      nice. I think this sums it all.
    - deleted 17y ago
      [deleted]
gort 17y ago
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.
- jonsen 17y ago
  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
- moo 17y ago
  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
mhb 17y ago
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?
devin 17y ago
Forgive me, but doesn't it seem odd that the problem defines Prisoner A from one perspective, and Prisoner B from another? Prisoner B could just as well be Prisoner A since A was defined by the narrator, and Prisoner B by the guard. I know we're descending into the theoretical stuff, probabilities and so on, but might there not even be enough information to go down any path at all given this dilemma?
- jonsen 17y ago
  Prisoner A is the one that asks. That sets the one an only perspective for the guards answer.
  - devin 17y ago
    Really? From whose perspective is Prisoner A described as asking the question? The narrator's. Please correct me if I'm wrong, but there is an implicit third person in my view.

10 points by TriinT 17y ago

mhb 17y ago

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

mannicken 17y ago
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 http://en.wikipedia.org/wiki/Three_Prisoners_problem
- Retric 17y ago
  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.
  - jonsen 17y ago
    I think you've got it backwards. Only one is pardoned. His chance of survival is 1 in 3.
    - randomtask 17y ago
      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.
      - gambling8nt 17y ago
        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.
- mattmaroon 17y ago
  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.
  - mannicken 17y ago
    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.
    - mattmaroon 17y ago
      Correct.
- Lol_Lolovici 17y ago
  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.
furyg3 17y ago
Came here to post this same link. That prisoner should immediately switch name-tags with the other prisoner who's fate is in question :)

pushingbits 17y ago

As non-intuitive puzzles go I think the hat guessing game (http://www.relisoft.com/science/hats.html http://www.relisoft.com/science/hats.html) takes the cake. I still remember a dailywtf thread where 19 pages worth of people claim that the correct solution is completely bogus (http://thedailywtf.com/Comments/Riddle-Me-An-Interview.aspx http://thedailywtf.com/Comments/Riddle-Me-An-Interview.aspx).

jonsen 17y ago

A's chances has not changed. But the chance of C being the pardoned has doubled to 2/3.

knowtheory 17y ago

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 http://en.wikipedia.org/wiki/Probability_interpretations ). The thought experiment in question probably should be answered with a Bayesian world view, and does not appear to be. Prisoner A's chances of being executed is not materially effected when he learns new information, because the decision as to who has been pardoned has already been made. Finding out new information, does change a probability, but not the probability that he'll be executed. It only changes the likelihood of whether or not the assertion that he will be executed is true. The nature of the universe has not actually changed, merely the state of belief of Prisoner A. The point being, he's got no cause to celebrate, upon merely discovering information, because learning more information does not actually change the state of the world, in this case.

huherto 17y ago
But, did he discovered new information? What if the guard had said "Prisioner C is sure to die"?
- gambling8nt 17y ago
  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.
  - huherto 17y ago
    nice. I think this sums it all.
  - deleted 17y ago
    [deleted]

gort 17y ago

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.

jonsen 17y ago
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
moo 17y ago
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

mhb 17y ago

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?

devin 17y ago

Forgive me, but doesn't it seem odd that the problem defines Prisoner A from one perspective, and Prisoner B from another? Prisoner B could just as well be Prisoner A since A was defined by the narrator, and Prisoner B by the guard. I know we're descending into the theoretical stuff, probabilities and so on, but might there not even be enough information to go down any path at all given this dilemma?

jonsen 17y ago
Prisoner A is the one that asks. That sets the one an only perspective for the guards answer.
- devin 17y ago
  Really? From whose perspective is Prisoner A described as asking the question? The narrator's. Please correct me if I'm wrong, but there is an implicit third person in my view.