(no subject)
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 :)
no subject
Date: 2007-08-09 04:42 pm (UTC)I remember reading about a game theory game where the optimum thing to do was found by iteration. After you do n steps of the fairly simple algorithm you find the optimum thing for you to do, however people tended not to do the optimum thing, they tended to do the thing that occurs if you do only a few iterations. The broad conclusion was that people are only able to do several iterations quickly in their head and go with that answer.
IIRC more intelligent people tended to do more iterations or something. It was all terribly interesting, and I wish I could remember it or link to it.