Beeminder Forum

Greedy [Remorseful] Pirates

(I’m not making this one ungooglable but the standard version has enough differences that you’ll be wrong if you cheat! Also we’ll murder you. Don’t laugh, we’ve done it before!)

A bunch of pirates have stolen a treasure chest full of gold and they decide to divvy it up as follows:

The shortest pirate proposes an allocation and they all vote on it (including the proposer). If the proposed allocation passes the vote they go with it. If not – or even if there’s a tie – they heave the shortest pirate overboard and start over with the new shortest pirate.

Assume that when they kill the short person they feel a tinge of remorse, but they’ll gladly kill them if they stand to gain so much as a doubloon by it.

So if you’re the shortest pirate what allocation should you propose?

I interpret this to mean that the pirates’ preferences are a lexicographic ordering on the pair of (money gained, inverse of people killed). That is, a pirate will always prefer (and hence vote so as to ensure) a situation in which they end up with more money over one in which they end up with less, no matter how many other pirates will die; but between two scenarios in which they end up with the same amount of money, they will prefer the scenario in which fewer pirates die. Is that accurate?

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Assuming my interpretation above is correct, I think I know how to start, though I don’t have a complete solution yet. But this will probably help anyone who is stumped and wants a hint.

If there is only one pirate, obviously she will propose to give herself all the money, and vote for her own proposal.

If there are two pirates, the short one needs the tall one to vote for his proposal, or else he will walk the plank. However, simply voting against the short one’s proposal will guarantee that the tall one gets all the money (since she will be the only one left), so she clearly won’t accept any proposal in which she gets anything less. So the only thing for the short one to do is to propose giving all the money to the tall one, and none to himself. The tall pirate will vote for this; since she will get all the money either way she would prefer not to kill the short one.

After that it gets weird, and I haven’t worked out all the details yet.

Yup, that’s exactly what I intended. Thanks for making that nice and crisp! (PS: I tweaked the wording – “so much as a doubloon” – to possibly help nudge towards this interpretation but shall otherwise leave it to your cogent clarification.)

PS: I guess the normal-person way to say “lexicographic preference ordering” is that they only care about money but will break ties in favor of not killing their colleagues.

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If you established that the second pirate will get 0 but will live if it gets down to two, then the third pirate in a set of 3 will get a “yes” vote from the second by offering 1 coin. As in, when there’s 3 left the proposal is [N-1, 1, 0] which will get votes [yes, yes, no] and pass.

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That’s what I thought at first too, but that’s not what the third pirate should do.


Hmmm…well I guess if they want to avoid all loss of life not just their own, then the second pirate will always get 0 and so he would vote “yes” for [N, 0, 0] because it means 3 pirates live instead of just 2 or 1?

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Yes, this is what is meant by the “tinge of remorse” thing in the original problem statement. See the discussion that @dreev and I had above.

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Okay so if that’s true…the 3rd pirate can’t settle for any coinage less than all of it? So when there’s 4, if the 4th pirate offers anything less he’ll get a “no” vote from #3. Except that the 1st and 2nd pirates are getting 0 either way, so they would vote “yes” for [N, 0, 0, 0] to keep the 4th pirate alive.

So if there’s at least 3 pirates, the shortest pirate gets all the money?

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Oops, I seem to have made the problem slightly less interesting! Sorry about that! I made the same mistake, thinking the short pirate in the 3-pirate case would appease the 2nd one with a single coin.

I think we need to change the problem!

The pirates are Bloodthirsty. They still only care about money but will break ties in favor of killing their colleagues. I shall update the question now…

Ha, too funny: Greedy Sadistic Pirates

Ok, now undoing my edit. Back to the original Greedy [Remorseful] Pirates!

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I think that’s right. We can deal with the cases of 1, 2, or 3 pirates specially, as we did. Then for n >= 3 pirates, by induction the shortest pirate should always propose to give themselves all the money. The next shortest pirate will vote against it — they stand to get all the money otherwise — but all the other pirates will vote for it since they won’t get anything anyway, and they would prefer not to kill the shortest pirate.

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I don’t think it’s less interesting! It’s just a different problem than you thought.

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Looks like I’m late to the party, but here’s my attempt before I read the thread:

I take nothing if there are 2 pirates and take everything in any other case.

If there’s one pirate, then I’m golden.
If there are two pirates, then I’m screwed - the tallest pirate would only agree to get everything.
Now, important observation: there can be more than one proposal that would pass the vote!
The second tallest pirate knows that he gets nothing if he rejects my proposal, so if I give everything to the tallest pirate, we all live. However, if I take everything for myself, we all live too! So, with three pirates, I take everything once again.
With four pirates, I need three votes. I cannot get third-tallest guy on board, but the tallest and second-tallest gain nothing from screwing me over and will accept any deal. So, I take everything.
Now, here’s the pattern. The majority of pirates gets nothing out of voting against me, only second-shortest does. So they won’t vote against me. As long as there’s somebody else besides me and the second-shortest guy, I get the bank.

This does seem pretty counter-intuitive and not resembling real pirates though.


Here’s my writeup of the solution, nothing new from what @byorgey and @drtall figured out (nice work!):

(We’re making lots of the usual assumptions. Common knowledge of rationality, no enforceable contracts.)

In the degenerate case of 1 pirate, that pirate allocates all the gold to themself and vote yes. I guess that one went without saying.

(the rest in spoiler tags)

In the 2-pirate case the short pirate (the proposer) is super screwed. If they allocate a single coin for themself, the tall pirate will vote no and it will be a tie which means throwing the short pirate overboard. The only way to get the tall pirate to vote yes is to give them all the gold. So that’s what the short pirate will do.

In the 3-pirate case it’s the opposite: the shrimpy proposer can have everything. Why? Because if the proposer is killed, the 2nd pirate becomes the shortest pirate in the 2-pirate case. Not an enviable position, as we just saw. Since that pirate has nothing to gain by killing you, they’ll vote yes and you have your majority.

In the 4-pirate case, the proposer can still have everything. The 2nd pirate now wants to kill you, to become the lucky proposer in the 3-pirate case, but everyone else (pirates 3 and 4) gain nothing by killing you, so they vote yes.

And in fact that reasoning applies to every subsequent case. The shortest pirate gets everything and only the 2nd pirate votes nay.