Ambiguity in pirates problem. I got a little side-tracked in the pirate puzzle, because I had never heard of a system where one could vote on the proposal he himself made!  Once I read teh beginning of the solution, I reworked it and got the proper result. So, here is a variation: what if a proposal for the gold distribution must be passed by at least 50% of the OTHER pirates.  So, Pirate 5 would need 2 out of 4 votes, pirate 4 would need 2 out of 3 votes, pirate 3 would need 1 out of 2 votes, pirate 2 would need 1 out of 1 votes. I'll send in my solution a little later on.  For more fun, what if all pirates agree that age dictates amounts? In other words, an older pirate must always get more gold than the pirate younger than him? This gem of a riddle has found a spot in my archives. Jim Sorenson Wednesday, July 24, 2002 Just to add clarification: a pirate wants to make sure he'll live, so his offer must be BETTER for his voters than if he were killed off.  He is not going to just match the amount and hope they are merciful. There is actually more than one solution. Jim Sorenson Wednesday, July 24, 2002 This is the same as requiring a strict majority of the pirates to agree on the issue, assuming that a pirate will always vote for his own scheme, which is the way I first interpreted this (though I see now that I got it wrong). Well, you'd solve this in exactly the same way as the others: Pirate 2 will be killed if it comes down to just the 2 of them, since even if he gives all the money to pirate 1, pirate 1 will kill him just for the hell of it. Pirate 3 therefore knows that he can count on pirate 2's vote no matter what, so he can take all the money for himself. { 1000 0 0 } Pirate 4 needs 2 votes. He obviously can't get pirate 3 to vote for him, so he must give 1 coin each to pirates 1 and 2 to give them a better deal than they'd get from pirate 3. { 998 0 1 1 } Pirate 5 also needs 2 votes. To maximise his take he should court the pirate who will be worst off if he is killed, i.e. pirate 3, so he should give 1 coin to pirate 3 and 2 to either one of the others. { 997 0 1 0 2 } | { 997 0 1 2 0 } As for the other problem, is "oldest pirate" the same as "highest-numbered pirate"? Jim Cameron Thursday, August 01, 2002 Having thought about these pirates a bit more (lovely problem, eh?) I'm not sure that we're being sufficiently piratical in our thinking. What happens if a pirate refuses to abide by the rules? Presumably the other pirates will kill him for trying to cheat them. But pirate 2 in the above scenario knows that this is going to happen anyway, so his best strategy from both a survival and a monetary point of view is to murder pirate 1 while pirate 1 is waiting for him to come up with a suggestion. Of course, pirate 1 knows this, so if it gets down to the two of them there's just going to be a fight. Now, what does this do to pirate 3's strategy ...? Jim Cameron Friday, August 02, 2002   Fog Creek Home