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Posers and Puzzles

Posers and Puzzles

  1. Standard member TheMaster37
    Kupikupopo!
    09 Mar '06 10:01
    Prisoners A, B and C talk to the guard, who tells them one of them will live, the other two will die.

    In a private talk with A, the guard tells him that B will die the next day.

    The chances of survival for A remain 1/3, while the chances for suvival for C goes up to 2/3. Though C does not know this. (This is TRUE, and I'm not here to discuss this part).

    My question is what happens if the guard tells C the exact same thing.

    The chances of suvival for C would be 1/3 and for A 2/3. Though A doesn't know this.

    So A knows: Me 1/3 ... Him 2/3
    And B knows: Me 1/3 ... Him 2/3

    Now A finds out that C got the same information from the guard as he did.

    Is his chance of survival now 1/2?
  2. Subscriber sonhouse
    Fast and Curious
    09 Mar '06 16:52
    Originally posted by TheMaster37
    Prisoners A, B and C talk to the guard, who tells them one of them will live, the other two will die.

    In a private talk with A, the guard tells him that B will die the next day.

    The chances of survival for A remain 1/3, while the chances for suvival for C goes up to 2/3. Though C does not know this. (This is TRUE, and I'm not here to discuss this p ...[text shortened]... hat C got the same information from the guard as he did.

    Is his chance of survival now 1/2?
    Sorry for diverting this but if prisoner A has been told who is
    to be executed, are you saying he still has one chance in three,
    or in other words no improvement or deprovement either way,
    based on the idea the guard may or may not be telling the truth?
    Just trying to undersand why A doesn't now have a 100 % chance of
    survival, assuming the guard is truthful. The truthful part seems the
    only link to a lower probability than 1 of survival.
  3. Standard member XanthosNZ
    Cancerous Bus Crash
    09 Mar '06 18:24 / 1 edit
    Originally posted by sonhouse
    Sorry for diverting this but if prisoner A has been told who is
    to be executed, are you saying he still has one chance in three,
    or in other words no improvement or deprovement either way,
    based on the idea the guard may or may not be telling the truth?
    Just trying to undersand why A doesn't now have a 100 % chance of
    survival, assuming the guard is truthful. The truthful part seems the
    only link to a lower probability than 1 of survival.
    Not only did you not read the question (two prisoners are killed, A knows that B will be killed, A could still be killed obviously) you also ignored TheMaster stating that the first situation is not under issue. Much like the Monty Hall problem this dilemma is counterintuitive and people with no math ability will argue till they are blue in the face about it.
  4. Standard member PBE6
    Bananarama
    09 Mar '06 19:48 / 1 edit
    Originally posted by TheMaster37
    Prisoners A, B and C talk to the guard, who tells them one of them will live, the other two will die.

    In a private talk with A, the guard tells him that B will die the next day.

    The chances of survival for A remain 1/3, while the chances for suvival for C goes up to 2/3. Though C does not know this. (This is TRUE, and I'm not here to discuss this p hat C got the same information from the guard as he did.

    Is his chance of survival now 1/2?
    It depends what question you're asking. If you're asking "does A's overall chance of survival increase from 1/3 to 1/2?", then the answer is no. If you're asking "does A's chance of survival increase from 1/3 to 1/2 at and for this stage?" then the answer is yes. But, overall A's chances of survival are still 1/3.

    For a thorough discussion of the first part of the question, check out the old "3 Prisoners" thread Thread 11542.
  5. Standard member PocketKings
    Banned from edits
    09 Mar '06 20:14
    Originally posted by TheMaster37
    Prisoners A, B and C talk to the guard, who tells them one of them will live, the other two will die.

    In a private talk with A, the guard tells him that B will die the next day.

    The chances of survival for A remain 1/3, while the chances for suvival for C goes up to 2/3. Though C does not know this. (This is TRUE, and I'm not here to discuss this p ...[text shortened]... hat C got the same information from the guard as he did.

    Is his chance of survival now 1/2?
    in compound probability problems you would multiply the two probabilities together 1/3 * 1/2 = 1/6. the overall chance of survival for any of them would be 1/6. but if you take into account that two of them knowing that B is going to die that takes out the first probability of 1/3 and leaves the two remaining (A,C) with a probability of 1/2 for survival
  6. Subscriber sonhouse
    Fast and Curious
    09 Mar '06 20:21
    Originally posted by XanthosNZ
    Not only did you not read the question (two prisoners are killed, A knows that B will be killed, A could still be killed obviously) you also ignored TheMaster stating that the first situation is not under issue. Much like the Monty Hall problem this dilemma is counterintuitive and people with no math ability will argue till they are blue in the face about it.
    Yep I missed the two prisoners killed, thought it was one.
  7. Standard member TheMaster37
    Kupikupopo!
    09 Mar '06 21:27 / 1 edit
    Originally posted by PBE6
    If you're asking "does A's overall chance of survival increase from 1/3 to 1/2?", then the answer is no. If you're asking "does A's chance of survival increase from 1/3 to 1/2 at and for this stage?" then the answer is yes. But, overall A's chances of survival are still 1/3.
    Hmm, you are stating two situations, but I cannot see the difference. My thoughts were that after finding out what the guard told C the chances for A increase to 1/2 (and he knows that).

    Where's the flaw in A's reasoning by thinking: "Myself and C are in the same situation, with the same information. Therefore our chances of survival are equal."

    As with the original problem I'm easily thinking the wrong way. That's one of the reasons I posted this
  8. Subscriber AThousandYoung
    It's about respect
    10 Mar '06 09:36
    Originally posted by PBE6
    It depends what question you're asking. If you're asking "does A's overall chance of survival increase from 1/3 to 1/2?", then the answer is no. If you're asking "does A's chance of survival increase from 1/3 to 1/2 at and for this stage?" then the answer is yes. But, overall A's chances of survival are still 1/3.

    For a thorough discussion of the first pa ...[text shortened]... of the question, check out the old "3 Prisoners" thread Thread 11542.
    What do you mean by "this stage"?
  9. 10 Mar '06 17:00
    Originally posted by TheMaster37
    Prisoners A, B and C talk to the guard, who tells them one of them will live, the other two will die.

    In a private talk with A, the guard tells him that B will die the next day.

    The chances of survival for A remain 1/3, while the chances for suvival for C goes up to 2/3. Though C does not know this. (This is TRUE, and I'm not here to discuss this p ...[text shortened]... hat C got the same information from the guard as he did.

    Is his chance of survival now 1/2?
    Sorry, but I'm a bit confused here. I know you dont want to discuss the first part, ut it is integral to the understanding of the logic used in the question.

    Once the guard tells A that B will die, why does the chances for A surviving stay at 1/3, and C rise to 2/3?

    Surely, if B is gong to die, then still either one of A or C will die, therefor they both have a 50/50 chance of living.

    So, once the guard tells C that B will die, nothing changes. B will die, and either A or C will also die. They are still at 50/50, surely?
  10. Standard member PBE6
    Bananarama
    10 Mar '06 19:25
    Originally posted by TheMaster37
    Hmm, you are stating two situations, but I cannot see the difference. My thoughts were that after finding out what the guard told C the chances for A increase to 1/2 (and he knows that).

    Where's the flaw in A's reasoning by thinking: "Myself and C are in the same situation, with the same information. Therefore our chances of survival are equal."

    A ...[text shortened]... inal problem I'm easily thinking the wrong way. That's one of the reasons I posted this
    My conception of the problem is as follows:

    Each prisoner has an equal chance of survival initially, equal to 1/3. This is equivalent to having a dart board cut into three equal sections, with a single dart throw determining who survives. After prisoner A asks who's going to die, the guard throws the dart. The dart could have ended up anywhere on the board, but when the guard says "B is going to die", he reveals that the survival dart did not end up in B's section. So, prisoner A reasons that the dart had a 1/3 probability chance of ending up in his own section, and a 2/3 chance of ending up somewhere else. Since he knows that it didn't end up in B's section, the entire 2/3 chance gets shifted to C. Right now, prisoner is wishing he could play "let's make deal" with the guard, dropping the soap notwithstanding.

    Now, prisoner C asks the same thing, and also finds out that B is going to die. He also correctly deduces that his own chances of survival remain 1/3, while A's go up to 2/3. While each analysis is correct on it's own, when taken together they appear to form a contradiction. However, there is no contradiction because neither A nor C have complete and consistent information about the situation. Therefore, the Law of Total Probability does not apply across both analyses as pointed out by Cribs in the 3 Prisoners thread Thread 11542.

    When A finds out that C knows B is going to die, A now has a complete picture and must re-evaluate his chances. I believe the main reason is that if A knows that C knows B is going to die, this eliminates some of the possibilities. Initially, the guard could have selected AB, BC or AC as the victims, and reported either B or C in the case of BC. This leaves A safe 1/3 of the time because he knew AC wasn't possible (CB out of AB, BC or CB). However, since he knows that C knows B is going to die, BC with C being reported is no longer a possibility, so A's chances are now 1/2 (CB out of AB, CB). C has no such restriction, 'cause he still don't know jack.

    So, at this stage in the game I would say A has a 1/2 chance of surviving. However, overall A's chances are still 1/3 because he had a 2/3 chance of being chosen in the first place (AB, AC out of AC, BC, AC), and a 1/2 chance of making it past the second part where he finds out B is going to die, so (2/3)*(1/2) = 1/3.

    I think that's right. Makes sense? Close enough, I hope.
  11. Donation richjohnson
    TANSTAAFL
    10 Mar '06 19:52
    Originally posted by tojo
    Sorry, but I'm a bit confused here. I know you dont want to discuss the first part, ut it is integral to the understanding of the logic used in the question.

    Once the guard tells A that B will die, why does the chances for A surviving stay at 1/3, and C rise to 2/3?

    Surely, if B is gong to die, then still either one of A or C will die, therefor they both ...[text shortened]... nothing changes. B will die, and either A or C will also die. They are still at 50/50, surely?
    A critical part of the problem, imho, is the fact that A is told by the guard that the guard can't tell A anything about A's own situation. This allows A to determine that C's chance of surviving is 2/3, since the chances of 1 of B&C surviving, and the guard has told him it will not be B. Without this piece of information, there is no justification for A to group B&C's chances of survival together, and the problem is not analogous to the Monty Hall problem.

    (Recall that in the Monty Hall problem, the host may only reveal one of the unpicked doors.)
  12. Subscriber AThousandYoung
    It's about respect
    12 Mar '06 06:17
    Originally posted by PBE6
    My conception of the problem is as follows:

    Each prisoner has an equal chance of survival initially, equal to 1/3. This is equivalent to having a dart board cut into three equal sections, with a single dart throw determining who survives. After prisoner A asks who's going to die, the guard throws the dart. The dart could have ended up anywhere on the boar ...[text shortened]... 2/3)*(1/2) = 1/3.

    I think that's right. Makes sense? Close enough, I hope.
    That doesn't work for me. The prisoner could equally reason that the dart has a 1/3 chance of landing in C, and therefore A must have a 2/3 chance of living.
  13. 12 Mar '06 13:35
    A has 1/3 chance of survival. C has 2/3 chance of survival. It is counterintuitive as Xanthos stated above, but it is absolutely true. I had difficulty with the logic as well initially, but it is more fun if you work it out yourself. As far as the questions at hand, I believe that the odds are even at 50% based on the new combined knowledge of A and C, since they cannot both have a 2/3 chance of survival or a 1/3 chance of survival.
  14. Standard member TheMaster37
    Kupikupopo!
    12 Mar '06 20:53
    First off, thanks to all that have replied so far. It's realy interesting to see your thoughts on this.

    About the original problem:

    You have to take into account that in the case A is the survivor, the guard chooses B or C with chance 1/2. If you then tally up the possibilities with their chances you'll see that A's chances remain 1/3.

    I'm very much inclined to agree with PBE6, tough I'm not sure if I agree with the last statement (the overall chance)
  15. Standard member Palynka
    Upward Spiral
    13 Mar '06 16:17
    PBE6 is right although I also don't understand what he means by overall probability.

    At each level of info for each individual (after A and C share the info):

    A knows he has a 50/50 chance shared with C and B is as good as dead.

    C knows exactly the same thing (he has the same info of A.

    B doesn't know about it and still puts the odds at 1/3 each for survival.

    Is there a catch here I'm not seeing? The original problem is much more counter-intuitive.