Answers to questions:
No there is no way for the prisoners to communicate once the first prisoner enters the room. The strategy must be completely decided before that happens.
The optimal strategy has around a 1/3 chance of suceeding.
And of course obviously if every prisoner picks the same 50 boxes the chance of suceeding is 0 (as 50 prisoners will not be in those 50 boxes).
Originally posted by XanthosNZ Answers to questions:
1. No there is no way for the prisoners to communicate once the first prisoner enters the room. The strategy must be completely decided before that happens.
2. The optimal strategy has around a 1/3 chance of suceeding.
3. And of course obviously if every prisoner picks the same 50 boxes the chance of suceeding is 0 (as 50 prisoners will not be in those 50 boxes).
Apologies for editing the quote (I hope I'm not spanked for this, as it seems to be the big no-no of the day). 🙄
Does point #1 imply that the prisoners cannot sort the boxes they check and leave them sorted?
Does point #3 imply that each prisoner has a different number? This seems to be what you're saying....
My best guess is that each prisoner chooses a different combination of boxes such that each box gets chosen the same number of times. 100 prisoners x 50 choices per means 5000 choices for 100 boxes. This means each one should get picked 50 times somehow...
Originally posted by XanthosNZ Answers to questions:
No there is no way for the prisoners to communicate once the first prisoner enters the room. The strategy must be completely decided before that happens.
The optimal strategy has around a 1/3 chance of suceeding.
And of course obviously if every prisoner picks the same 50 boxes the chance of suceeding is 0 (as 50 prisoners will not be in those 50 boxes).
From a practical standpoint, the prisoners could agree to bend the left side of the box if it was an odd number inside, and bend the right side of the box, if it was an even number.
Then if your number was even you would just open the boxes that had bends in the right side of the box until you found yours.
It would all depend on the first prisoner actually finding his number but the odds would be raised to 50% since 50 boxes would be even and 50 boxes would be odd. With 100 prisoners, the odds would be 100% if the first prisoner found his number.
Yes every prisoner has a different number and no the solution does not involve any method of communication. The room will be left exactly as it was when they entered. No moving the boxes.
The solution would work if there were in fact 100 identical rooms and the prisoners entered one each at the same time.
Uzeless, your reasoning is wrong. The first prisoner would only be able to determine the odd/evenness of half the boxes (and would have a 50% chance of finding his own number). So unless he managed to find all odds or all evens (very low odds) the second prisoner would need to find odds/evens as well and would have a 50% chance of finding his own number (it would be interesting if the second prisoner was even and the first had found 40 evens and 10 odds, does the second look at the 40 known evens first?) and you have around a 25% chance. So you broke the rules of the puzzle and your solution still isn't as good as the optimal solution.
Group huddle - everyone gets a visible number, 1 to 100
First person ( any number ) into the room then labels all the boxes 1 to 100.
That person then opens the box that matches her number. If that box does not contain her number then she opens the box of the number it does hold. She continues until she finds her number or until she has opened 49 boxes. In any event the next entrant goes to the numbered box of the last entrant and continues the process. That process should give better than a 1 in 3 chance of surviving.
No doubt Xanth will shoot it all to bits but I think the logic is there?
Originally posted by dmnelson84 Skeeter, your answer is so similar in verbage to the google yielded response. I think you cheated. Still not 100% right though.
Well knob jockey, if I googled it you'd think that I would get it 100% right, right?
Jesus wept.
Originally posted by skeeter Well knob jockey, if I googled it you'd think that I would get it 100% right, right?
Jesus wept.
skeeter
Well clearly if you worked it out yourself you should have no trouble explaining why it works.
And anyway how do the prisoners label the boxes? Surely you mean they agree to count them from the left so that the room contents doesn't need to change?
Prisoners and Boxes
08 Dec '06 03:56
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