1. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
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    07 Dec '06 07:47
    100 prisoners are given the sort of crazy deal they always get in these problems. Each prisoner gets a number. There's a room with 100 identical boxes in a row, each containing a random number from 1-100, with no repeats. Each prisoner, in turn and isolated from all the others, will get a chance to open 50 boxes of his choosing, one at a time. If each prisoner manages to find his own number in one of the 50 boxes he opens, all prisoners go free. The prisoners can make a strategy together at the beginning, but there can be no communication afterwards. The room is completely unmarkable, and will be reset to the exact same starting state for each prisoner.
    Clearly if every prisoner picked randomly the probability that that prisoner would be successful would be 0.5 and therefore the probability they would all be successful would be 0.5^100 (read:practically zero, actually it's ~8*10^-31). What is the optimal strategy that the prisoners should use?
  2. SubscriberAThousandYoung
    All My Soldiers...
    tinyurl.com/y9ls7wbl
    Joined
    23 Aug '04
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    24791
    07 Dec '06 10:553 edits
    The prisoners don't get to watch other prisoners pick, right? Can they communicate via bonks on the wall or something?

    If the outside prisoners cannot get information from prisoners inside, then it's hard to see what they can do.
  3. Joined
    11 Nov '05
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    43938
    07 Dec '06 11:25
    A very interesting problem indeed.
    I have no idea about the solution so I eagerly wait for the correct answer and the explanation.
  4. Joined
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    07 Dec '06 11:53
    I think I see a way to make sure the probability they all succeed is 0.5. I won't post it yet - see what else comes up.
  5. Indiana
    Joined
    29 May '06
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    1694
    07 Dec '06 12:33
    I'm just throwing out this completely random response. Each prisoner takes their Sharpie Marker in with them and marks each number in the boxes they picked on the bottom of the box. the first prisoner marks half the boxes then if the second prisoner doesnt see his/her number he opens the other half of the boxes. 😀 I know this isn't the right answer.
  6. Joined
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    07 Dec '06 14:45
    I've seen this one before. I couldn't work it out and found the answer quite surprising - if I remember correctly they have roughly a 1 in 3 chance of surviving.
  7. Joined
    07 Sep '05
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    35068
    07 Dec '06 15:53
    Originally posted by mtthw
    I think I see a way to make sure the probability they all succeed is 0.5. I won't post it yet - see what else comes up.
    OK, I've re-read the question, and that was wrong. Glad I didn't post it now 🙂
  8. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
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    28719
    07 Dec '06 16:00
    Can the prisoners communicate in a roundabout way by picking the boxes in order, then waiting a specified length depending on the numbers they pick? I haven't checked yet, but there must be some function that makes the amount of time spent a unique sum given a unique set of numbers. Eventually, the whole set of boxes should be mapped out guaranteeing success for the remaining prisoners.
  9. Joined
    05 Oct '05
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    61209
    07 Dec '06 18:00
    All prisoners choose the same 50 boxes, say the left 50. You didn't say the prisoners received a unique number between 1-100. They might all have the same number.
  10. Standard memberPBE6
    Bananarama
    False berry
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    28719
    07 Dec '06 18:34
    Just did a quick Google, and found a detailed solution that I couldn't help but read. 😞 But a very interesting problem, with a very interesting and counter-intuitive answer.
  11. Joined
    25 Nov '06
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    1434
    07 Dec '06 19:441 edit
    A brilliant puzzle.
    🙂
    The answer is... well...(sweats)... um...
    OK I admit it, I have no clue!
  12. Joined
    27 Oct '05
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    72627
    07 Dec '06 20:21
    Someone post the answer?
  13. In your face
    Joined
    21 Aug '04
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    55993
    07 Dec '06 20:24
    Originally posted by dmnelson84
    Someone post the answer?
    Google it if you want to be enlightened.
  14. Joined
    27 Oct '05
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    72627
    07 Dec '06 21:01
    Fantastic riddle.
  15. Standard memberuzless
    The So Fist
    Voice of Reason
    Joined
    28 Mar '06
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    9908
    07 Dec '06 21:081 edit
    Originally posted by jimslyp69
    Google it if you want to be enlightened.
    someone post the google link? 😉
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