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Aug. 9th, 2007 02:24 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
jackPtc reminded us of the pirate loot-dividing puzzle.
(If you haven't heard it, five perfectly rational pirates want to divide some loot. The method is that the most senior pirate proposes a division, and the others vote on it. If a non-strict majority accept it, it stands, else he's thrown overboard and the others repeat the process. The pirates are assumed to care about, in order: surviving, gold, killing other pirates, and nothing else.
Read wikipedia for a filler description, and spoiler for the normal puzzle. If you haven't heard it, it's quite cute, so you probably want to read that rather than reading on here.)
Ian Stewart takes the calculation further, up past 200 pirates when the gold starts to run out, and finds amusing results: http://euclid.trentu.ca/math/bz/pirates_gold.pdf
But it occurred to me the standard solution possibly has another loophole.
As is traditional, game theory problems can improved with probabilities. In the standard solution, A offers C and E one gold piece each. But the logic works the same if he offers them a one-in-a-thousand chance of a gold piece. If they're trying to maximise their *certain* gold pieces, the original solution applies. But on the other hand, if you're trying to maximise gold pieces it's hard to say 1/1000 of a chance is better than none, which is what they'll certainly get if they try B's mercy :)
(If you haven't heard it, five perfectly rational pirates want to divide some loot. The method is that the most senior pirate proposes a division, and the others vote on it. If a non-strict majority accept it, it stands, else he's thrown overboard and the others repeat the process. The pirates are assumed to care about, in order: surviving, gold, killing other pirates, and nothing else.
Read wikipedia for a filler description, and spoiler for the normal puzzle. If you haven't heard it, it's quite cute, so you probably want to read that rather than reading on here.)
Ian Stewart takes the calculation further, up past 200 pirates when the gold starts to run out, and finds amusing results: http://euclid.trentu.ca/math/bz/pirates_gold.pdf
But it occurred to me the standard solution possibly has another loophole.
As is traditional, game theory problems can improved with probabilities. In the standard solution, A offers C and E one gold piece each. But the logic works the same if he offers them a one-in-a-thousand chance of a gold piece. If they're trying to maximise their *certain* gold pieces, the original solution applies. But on the other hand, if you're trying to maximise gold pieces it's hard to say 1/1000 of a chance is better than none, which is what they'll certainly get if they try B's mercy :)