### Pirate Game - A Variation

Posted:

**Sun Mar 26, 2017 10:58 pm UTC**You all know the classic Pirate Game - five pirates find a treasure of gold and have to divide it among themselves. The most senior/fiercest/whatever pirate proposes a division, and if it does not get 50% of the vote, that pirate is killed and the next pirate makes a proposal. It's an old one, it's a good one, but I would like to discuss a variation. I looked for it, and the earliest I could find it was by WarDaft on this thread, so I'll credit that as the source, but I'll restate the problem in full here.

As before, there are five pirates (in order, A, B, C, D, and E) dividing a treasure of 1000 gold. As before, the pirates will vote on pirate A's proposal first, followed if it fails by pirate B's, and so on, with the majority deciding, and the current proposer breaking any ties. However, the proposals will actually be announced in reverse order - i.e. first pirate E will announce his proposal (though it must be all 1000 to E), then pirate D will announce, then pirate C, then B, and finally A. Once all the proposals are announced, then the pirates vote on A's proposal, then B's if they need to, and so on. As in the original problem, all pirates first and foremost want to survive, and given that, want to get as much gold as possible (on average, if probabilities are involved), and given that, want to kill as many pirates as possible, and will never go against these priorities. If there are still two or more best options after those three criteria are considered, pirates will decide in an effectively random manner, beyond the realm of any possible deal-making or predictability on the part of the other pirates. Oh, and to close off a potential loophole, all of the facts in this paragraph are common knowledge between all pirates - and I mean the kind of infinitely-layered common knowledge that is not present in the Blue Eyes problem.

How will the gold get divided?

As a bonus question, does this generalize to N pirates, and if so how?

Another bonus question: if the pirates did have other criteria in the effectively-random-manner part that could be accounted for, how much could these payouts be affected?

To kick things off, I have analyses of the 1, 2, 3, and 4 pirate cases below.

As before, there are five pirates (in order, A, B, C, D, and E) dividing a treasure of 1000 gold. As before, the pirates will vote on pirate A's proposal first, followed if it fails by pirate B's, and so on, with the majority deciding, and the current proposer breaking any ties. However, the proposals will actually be announced in reverse order - i.e. first pirate E will announce his proposal (though it must be all 1000 to E), then pirate D will announce, then pirate C, then B, and finally A. Once all the proposals are announced, then the pirates vote on A's proposal, then B's if they need to, and so on. As in the original problem, all pirates first and foremost want to survive, and given that, want to get as much gold as possible (on average, if probabilities are involved), and given that, want to kill as many pirates as possible, and will never go against these priorities. If there are still two or more best options after those three criteria are considered, pirates will decide in an effectively random manner, beyond the realm of any possible deal-making or predictability on the part of the other pirates. Oh, and to close off a potential loophole, all of the facts in this paragraph are common knowledge between all pirates - and I mean the kind of infinitely-layered common knowledge that is not present in the Blue Eyes problem.

How will the gold get divided?

As a bonus question, does this generalize to N pirates, and if so how?

Another bonus question: if the pirates did have other criteria in the effectively-random-manner part that could be accounted for, how much could these payouts be affected?

To kick things off, I have analyses of the 1, 2, 3, and 4 pirate cases below.

**Spoiler:**