Three Prisoners Problem ("probability paradox")

Discussion in 'Archives' started by Cruel__Machine, Nov 2, 2007.

Three Prisoners Problem ("probability paradox")
  1. Unread #21 - Nov 4, 2007 at 2:49 AM
  2. eXact
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    Three Prisoners Problem ("probability paradox")

    Lol cruel!
     
  3. Unread #22 - Nov 4, 2007 at 2:58 AM
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    Three Prisoners Problem ("probability paradox")

    just kill em no1 would know
     
  5. Unread #23 - Nov 4, 2007 at 4:55 AM
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    Three Prisoners Problem ("probability paradox")

    Can't be bothered figuring it out lol, maybe if it had a joke for an answer then i would but it doesn't.
     
  7. Unread #24 - Nov 4, 2007 at 5:10 AM
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    Three Prisoners Problem ("probability paradox")

    oh man, its too late in the day to be working these things out.. try... to..morrow..
     
  9. Unread #25 - Nov 4, 2007 at 12:00 PM
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    Three Prisoners Problem ("probability paradox")

    Interesting.

    Second, its C.
     
  11. Unread #26 - Nov 4, 2007 at 3:35 PM
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    Three Prisoners Problem ("probability paradox")

    my theory: 50/50

    because: split the chances 1/3 up to 2/6 (its the same)
    If B is released, he HAD a chance of 2/6, which is now split equally to prisoner A and B, resulting each gaining 1/2 of the 2/6 which is 1/6

    They already had 2/6, so if they gain 1/6, they will eventually have 3/6, and 3/6 is equal to 1/2 resulting, A and B have the same chance.
     
  13. Unread #27 - Nov 4, 2007 at 4:11 PM
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    Three Prisoners Problem ("probability paradox")

    i still think its 50/50 because only 2 get released (1 beeing B)
    now your left with A and C now each of them have equal opportunity to get out because he cannot reveal the fate of A because he asked, there for they both have equal chance.


    or since he is able to speak to the warden then he is probably all ready beeing released he just wants to know with who.
     
  15. Unread #28 - Nov 4, 2007 at 7:59 PM
  16. Cruel__Machine
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    Three Prisoners Problem ("probability paradox")

    I'll try to make it more like the Monty Hall problem so it can be better understood...

    Ok... say you were to choose which prisoner was to ask the warden. You have a 2/3 chance of choosing a prisoner that is to be released, correct? And every time you choose a prisoner that is to be released, the warden will reveal the other prisoner that is to be released to them, and the person left over will therefore be the one sentenced to death.
    So if the situation is like that 2/3 of the time when you choose the prisoner to ask, then the probability of the third person to be sentenced to death would be the same, 2/3.

    But like Steph mentioned from the wiki... if A had asked "Will B be released?", the warden would've said yes, and A and C's probability would be the same, 1/2.
     
  17. Unread #29 - Nov 4, 2007 at 8:04 PM
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    Three Prisoners Problem ("probability paradox")

    Who cares what the answer is just say its 50% :D
     
  19. Unread #30 - Nov 4, 2007 at 8:05 PM
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    Three Prisoners Problem ("probability paradox")

    Um...

    All 3 don't know whose gonna die they all have a 66.6 chance of dieing.
    Then A knows B is gonna live. Doesn't that make it 100% that A and C are gonan die?
     
  21. Unread #31 - Nov 4, 2007 at 8:15 PM
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    Three Prisoners Problem ("probability paradox")

    ROFL... reread it.
    Only 1 of them will die.
     
  23. Unread #32 - Nov 4, 2007 at 8:16 PM
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    Three Prisoners Problem ("probability paradox")

    0_0 oh rofl! I was like.... wdf am I missing here?
     
  25. Unread #33 - Nov 4, 2007 at 9:58 PM
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    Three Prisoners Problem ("probability paradox")

    /win

    I love this problem. Nobody ever believes the correct answer :D
     
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