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Posers and Puzzles

Posers and Puzzles

  1. 16 Jun '04 17:01 / 2 edits
    There are 3 prisoners in a prison. Let’s call them A, B, and C. Tomorrow, 2 of them will be executed, but the prisoners don’t know which of them have been chosen. Prisoner A reasons that his chances of survival are 1/3 (a third). He then gets bored in his cell and goes to chat with a guard. “Tell me”, he says, “will I be executed tomorrow?” “I’m sorry,” the guard replies, “I am not allowed to tell you that.” “OK”, says Prisoner A, “you needn’t tell me anything about myself. But one of the other two, either B or C is sure to be executed. Perhaps both of them but, in any case, at least one of them.” “That’s right”, confirms the guard. “Very well”, says Prisoner A, “tell me the name of one person (B or C) who is to be executed.” “All right”, answers the guard, “B will be executed tomorrow.” On hearing this news Prisoner A feels much more cheerful. He reasons that tomorrow either he or Prisoner C will be executed. And that therefore his chances of survival have increased from 1/3 to 1/2. Is he right?

    .
  2. 16 Jun '04 18:04
    This seems to be a variation of the Monty Hall problem.

    -Ray.
  3. 16 Jun '04 18:28 / 1 edit
    Originally posted by rgoudie
    This seems to be a variation of the Monty Hall problem.

    -Ray.
    Correct, rgoudie.

    Funny thing is, while C has been sitting in his cell scratching his ass, his chances of surviving have doubled!

    .
  4. Standard member TheMaster37
    Kupikupopo!
    17 Jun '04 08:26
    Well actually...the prison director has alraedy decided that (for example B and C) will be executed. Therefor, the chance that A will be executed is 0.

    Chance depends greatly on the information you have. It measures the incertainty of events about to happen. Of course, when you know more, you'll be less uncertain.

    Ton
  5. 17 Jun '04 09:38
    Originally posted by TheMaster37
    Well actually...the prison director has alraedy decided that (for example B and C) will be executed. Therefor, the chance that A will be executed is 0.
    Well actually...the prison director has already decided that B OR C will be executed.

    .
  6. Standard member genius
    Wayward Soul
    17 Jun '04 09:46 / 2 edits
    Originally posted by TheMaster37
    Well actually...the prison director has alraedy decided that (for example B and C) will be executed. Therefor, the chance that A will be executed is 0.

    Chance depends greatly on the information you have. It measures the incertainty o ...[text shortened]... . Of course, when you know more, you'll be less uncertain.

    Ton
    i'm not great with words-so i'm probubly not the best to explain this, but i tried to get my brother to explain it, but he got confused when i tried to explain it to him, so i gave up...here goes!

    prisoner C's chances of survival have increased from 1/3 to 2/3 while prisoner A's are still 1/3.

    why? alrighty-one of thems going to survive, right? we can split this into two sections-A, and B&C. A has 1/3 chances of survival, while B&C has 2/3 chances of survival. but if B is definatly not going to survive, then C has still got 2/3 chances of survival. so, no-prisoner A's chances of surivival are still 1/3...

    i think i reeeeeeeeally can't explain stuff...i understand it, but i just can't get it accross. i did get a B for my higher english though?

    but whatever happens, prisoner B is screwed
  7. 17 Jun '04 18:23
    Originally posted by genius
    i'm not great with words-so i'm probubly not the best to explain this, but i tried to get my brother to explain it, but he got confused when i tried to explain it to him, so i gave up...here goes!

    prisoner C's chances of survival have increased from 1/3 to 2/3 while prisoner A's are still 1/3.

    why? alrighty-one of thems going to survive, right? w ...[text shortened]... s. i did get a B for my higher english though?

    but whatever happens, prisoner B is screwed
    I agree - this is the way I would put it.

    Before A asks the guard there are 3 possibilities, each with the probability stated below:

    A spared : 1/3
    B spared : 1/3
    C spared : 1/3

    After A asks the guard, but before the guard answers there are the following possibilities of who is spared and what the guard might say.

    A spared & Guard names B : 1/6
    A spared & Guard names C : 1/6
    B spared & Guard names C : 1/3
    C spared & Guard names B : 1/3

    After the guard names B we know it was either the first or last possibility that happened, but we have to normalise those two remaining probabilities so they add up to 1

    A spared & Guard names B: (1/6)/(1/6 + 1/3) = 1/3
    C spared & Guard names B: (1/3)/(1/6 + 1/3) = 2/3
  8. Standard member Otis
    Lucky Patzer
    18 Jun '04 08:15
    with b already chosen to be executed, that leaves either C or A being chosen to be executed with him. So A has 1/2 chance of surival.
  9. 18 Jun '04 08:23 / 1 edit
    Originally posted by Otis
    with b already chosen to be executed, that leaves either C or A being chosen to be executed with him. So A has 1/2 chance of surival.
    The above argument is equivalent to saying that it is better NOT to switch doors in the Monty Hall puzzle.

    .
  10. 18 Jun '04 16:31
    Can people do more complicated problems of this sort? lets see:

    Same scenario, A, B and C, two to be exectuted tomorrow.

    A says to the guard "Can you tell me someone who is going to be executed (b or c) please?"

    The guard says "No - I see the parallels of this situation with the monty hall problem and I will be helping either B or C to sleep easier tonight, which is not allowed".

    So A repies "Ok - but how about if once you have chosen someone to name as for the chop, you randomly select another of us (who might be me) and you tell me both names, not telling me which is the random one and which is the dead one"?

    "Ok" says the guard "That seems like it won't help much".

    What's the chance of any of the prisoners getting any useful information from the guards imminent answer to A's question?
  11. Donation richjohnson
    TANSTAAFL
    18 Jun '04 16:47
    Originally posted by iamatiger
    Can people do more complicated problems of this sort? lets see:

    Same scenario, A, B and C, two to be exectuted tomorrow.

    A says to the guard "Can you tell me someone who is going to be executed (b or c) please?"

    The guard says "No - I see the parallels of this situation with the monty hall problem and I will be helping either B or C to sleep ...[text shortened]... f the prisoners getting any useful information from the guards imminent answer to A's question?
    Zero. The prisoners can do nothing to alter their fate in this problem, so any information they get is not useful.
  12. 18 Jun '04 17:13 / 1 edit
    Originally posted by richjohnson
    Zero. The prisoners can do nothing to alter their fate in this problem, so any information they get is not useful.
    ok, ok . I meant useful as in helping them determine their fate, not helping them alter it!

    Although knowing they are more likely to be killed might make them want to escape.... which might help fate-wise.
  13. Donation Acolyte
    Now With Added BA
    18 Jun '04 21:58 / 1 edit
    Originally posted by iamatiger
    Can people do more complicated problems of this sort? lets see:

    Same scenario, A, B and C, two to be exectuted tomorrow.

    A says to the guard "Can you tell me someone who is going to be executed (b or c) please?"

    The guard say ...[text shortened]... eful information from the guards imminent answer to A's question?
    Assuming the guard uniformly picks one of the condemned (possibly including A), the guard's responses are symmetric, so we can consider what happens if the guard says 'B and C', and we'll call this event D. Let A, B and C be the events that the respective prisoner is to be spared. Then

    P(A|D) = P(AnD)/P(D) = P(A)*P(D|A)/P(D)

    P(A) = P(D) = 1/3
    P(D|A) = 1/2 (the guard chooses one of B and C, then chooses the other with prob. 1/2)
    => P(A|D) = 1/2

    By symmetry P(B|D) = 1/4, P(C|D) = 1/4

    So the guard is guaranteed to give information to all three prisoners about their fate, with the obvious result that the prisoner not named is more likely to be spared.
  14. Standard member Otis
    Lucky Patzer
    18 Jun '04 23:47
    trick questions no one is going to die
  15. Standard member Polynikes
    Spartiate
    19 Jun '04 01:51
    Originally posted by THUDandBLUNDER
    There are 3 prisoners in a prison. Let’s call them A, B, and C. Tomorrow, 2 of them will be executed, but the prisoners don’t know which of them have been chosen. Prisoner A reasons that his chances of survival are 1/3 (a third). He then gets bored in his cell and goes to chat with a guard. “Tell me”, he says, “will I be executed tomorrow?” “I’m sorry ...[text shortened]... ed. And that therefore his chances of survival have increased from 1/3 to 1/2. Is he right?

    .
    It seems to me that his chances do not increase from 1/3 to 1/2.

    Reason being, there is a 100% chance that B or C will be named. His question,

    “tell me the name of one person (B or C) who is to be executed.”

    tells him absolutely nothing at all. Of course, at least one of them is to die. The guard's answer only tells us what we already know.

    Lets rephrase the discussion.

    Prisoner: Will I die tomorrow?
    Guard: I cant tell you that.
    Prisoner: Ok, fair enough. Does one of them die tomorrow?
    Guard: Yes, of course fool. If two of you are to die, it's obvious that at least one of them will be chosen.
    Prisoner: What is the name of one prisoner who dies, not including me?
    Guard: No matter which name I say, you still dont know who dies with him, and no math or reasoning can possibly increase your odds of survival. Now lights out.