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.

Spoiler:

1 pirate: Lucky A doesn't need to share, and gets all 1000 gold.

2 pirates: No matter what A proposes, A will break the tie in his favor, so A will simply propose and get all 1000 gold.

3 pirates: As shown in the 2 pirate case, B's proposal will win if it comes to that, and A needs either B's vote or C's vote. So whatever B's proposal, A will offer whoever's worse off from it 1 more gold piece than they would get from it, and take all the rest of the gold. Knowing that A will do this, B will propose B:499/C:501, so that A will propose A:500/B:500. (For the second bonus question, if B is confident enough that A favors B over C, B can propose B:500/C:500, and milk one more gold out of A in A's new proposal A:499/B:501.)

4 pirates: Again, A needs one other vote to win, and so will offer one of the other pirates 1 gold more than they would get should A's plan fail. B can always arrange to be the one who gets the least gold of the other three, so only A and B will get gold. Thus far, this has all just been the analysis of Gwydion back on the original thread, but here there is a subtlety that went unnoticed: if B's proposal would fail, B will vote for A's proposal regardless of how much gold he gets - better to be alive than dead, after all - so B's proposal must be a successful one. B's ideal proposal is B:332/C:334/D:334 so that B will get 333 gold in the end. C and D could also be offered 333 and 335 in either order without B's payout changing. However, if C's proposal gives both C and D at least 335 gold, this proposal will fail, so B's payout in a successful proposal will be reduced. Worst-case scenario, C proposes C:500/D:500, and B will need to propose B:249/C:501/D:250 (C and D can be swapped) to get a successful proposal, meaning B's payout here is only 250. In other words, B's payout depends on C's proposal, and as far as the problem is concerned, C is completely indifferent about how much gold A and B get. In the end, A's winning proposal will be somewhere between A:667/B:333 and A:750/B:250. So effectively, A gets 667 gold, B gets 250 gold, and the remaining 83 gold is divided between A and B entirely at C's discretion, whatever that winds up being. (For the second bonus question, in B's best-case scenario, if B is confident enough A favors B over one of the others, B can propose a 333/333/334 split and A will instead propose A:666/B:334, giving B one extra gold. Of course, this all depends on whether or not C allows B to do this - I leave it as an exercise for the reader to prove that C can force B to accept any value from 250 to 333, while preventing B from being able to take advantage of ties to get any extra.)

I solved it as far as you did, except that I didn't realize pirate 4-C would have literally no incentive to choose any distribution over another.

Intuitively, it feels like 4-C should still pick 499/501. I wonder if we can adjust the rules to make that happen. Let's see… What if every decision was tiebroken by what WOULD happen if the current first pirate was killed? With those rules, 4-C would see that all choices are tied initially, but then make their decision based on the 3-pirate case. So C would still be motivated to force B to allocate the most gold to C. And we can get rid of the arbitrary choices for splitting up the gold between the remaining pirates by, say, making each pirate infinitesimally biased in favor of the later pirates based on how late they are the ordering. Then, I think the proposals would always be:

Last pirate: 1000 second to last: 499/501 third to last: 249/500/251 fourth to last: 249/249/250/252 fifth to last: 166/250/250/166/168 sixth to last: 166/167/166/166/167/168 and so on,

Except for the very first pirate, who Just bribes half the pirates and gives all the extra gold to themselves instead of distributing it among the rest of the pirates.

The pattern continues in a pretty similar way up until... I think it's 669 pirates, where the second pirate can't safely give themselves even 1 gold. I'll think about that scenario more later.

Also known as Eli Dupree. Check out elidupree.com for my comics, games, and other work.

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