### Pirates!

Here’s a puzzle I just came across; I think I’ve seen it before but I don’t remember the answer:

You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it?

Presumably only the bottom 4 pirates vote, not all five, and presumably if the top pirate gets killed, the process begins anew with Pirate 4 proposing a division and Pirates 1-3 voting.

Also: suppose all the surviving pirates vote — including the one proposing the division. Does that change the answer?

## 10 thoughts on “Pirates!”

1. doctroid says:

My analysis:

Suppose there are only two pirates. Then Pirate 1 gets the only vote, so if he’s allocated anything less than all the gold, he’ll vote no. So Pirate 2 must propose 0 for himself, 100 for Pirate 1.

If there are three pirates, Pirate 3 must win the support of Pirate 1 or Pirate 2. So he can propose 100 for himself, 0 for Pirate 2, 0 for Pirate 1. Pirate 1 will vote no but what about Pirate 2? He gets 0 coins whether he objects or not; I assume he won’t object and kill Pirate 3 if there’s nothing to be gained. So he’ll say yes and the division is upheld.

With four pirates, Pirate 4 must gain the support of two others. So he can propose 100 for himself, 0 for Pirate 3, 0 for Pirate 2, and 0 for Pirate 1. Again Pirates 1 and 2 do no better but no worse than with three pirates, so they vote yes.

And likewise Pirate 5 proposes 100 for himself and 0 for the others. Pirate 4 is the only one who’s worse off this way and the others vote yes.

So with any number of pirates except 2, the top pirate gets everything.

If all pirates vote, including the one proposing, then it’s the same, except that in the 2-pirate case it’s Pirate 2 who gets 100 coins and Pirate 1 who gets none.

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2. I think you underestimate the bloodthirstiness of pirates, who would just as soon slit your throat as look at you. You have to throw at least a little gold their way to keep them from slitting your throat. So I make it:

Pirate 5: 97
Pirate 4: 0
Pirate 3: 1
Pirate 2: 0
Pirate 1: 2

(Pirate 3 would get nothing if Pirate 4 were in charge, and Pirate 1 would get one coin in that case. The other two would get more but you don’t need their support. Although I feel like the other pirates will eventually revolt or desert if they are only going to get one or two coins from any given plundering exercise, but I guess that’s beyond the purview of this puzzle.)

I am curious as to how the pirate rankings were established and enforced. Apparently Pirate 5 has no ability to keep the others from keelhauling him if they are in the mood.

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3. Actually there’s another way of looking at it, which is that Pirate 4 can offer to cut the other pirates a better deal if they don’t accept 5’s offer. So if the original proposal is what I said before, Pirate 4 can say, “Don’t listen to Pirate 5, me bonnie lads; I’ll give ye each five shiny doubloons when I’m captain.” Then Pirate 5 says, “Arrr! I’ll give ye six doubloons right here and now if ye listen not to this rogueish scoundrel.” (If 4 then later reneges the other pirates can rebel, although 4 can try bribing some of them not to …)

So really you’re bribing 4 to shut up, and bribing one of the others to go along with it. That makes this problem a lot trickier. Let me think about this some more.

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4. doctroid says:

Jake’s analysis seems correct, given the stated assumption. In the absence of information about this, I suppose (if I were Pirate 5) I’d sacrifice three gold pieces and do it Jake’s way to be safe. But it certainly seems to me a significant ambiguity in the puzzle.

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5. With two pirates the divvy is 0/100 (although really in that situation I see a daring swordfight breaking out, but that’s not part of the puzzle).

With three pirates pirate 3 proposes 99/1/0. Pirate 1 doesn’t like that so he proposes 0/2/98. Pirate 3 counterproposes 97/3/0 and so on. Eventually pirate 3, not wanting to be killed, proposes 0/100/0 and is accepted by pirate 2.

Pirate 4 proposes 98/1/0/1, but pirates 1 and 3 form a cabal and threaten to kill pirate 4 if they don’t get the goods. I think this has to turn out 0/50/0/50.

Pirate 5 proposes 97/1/0/1/0, but three of the pirates threaten to revolt. This gets tricky because there’s an odd doubloon left after you divide things three ways. I think pirate 2 probably has the advantage here, because if he can get things down to three pirates then he gets it all, and there’s no clear-cut advantage for any of the other pirates above that number, so I think it probably comes out 0/0/33/34/33.

What do you think?

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6. doctroid says:

But why would (in the 3 pirate case) Pirate 3 back down from 99/1/0? Pirate 2 will support that because if he doesn’t, Pirate 1 takes it all. (And kills Pirate 2, assuming bloodthirstiness.)

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7. My last scenario is faulty — pirates 1 and 3 will always favor revolt, because they do better under the four-pirate scheme. So in order to survive Pirate 5 has to offer 0/50/0/50/0. Although then pirate 3 (say) could propose 0/99/1/0/0. But pirate 3 doesn’t seem very trustworthy and might try to renegotiate after the coup, so I think pirate 4 would do well to consider the cap’n’s offer very seriously.

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8. Pirate 1 has to convince Pirate 2 that Pirate 1’s offer will be followed through on, and Pirate 2 has to convince Pirate 3 that he or she believes Pirate 1’s offer is legit. (Actually, the second part is more important than the first.) If that can’t be worked out then it all falls apart pretty quickly.

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9. doctroid says:

Ah, I see. But I think for this puzzle, we have to assume that in any situation each pirate will act (within the specified rules) to maximize his or her own take, and that they have to assume the same of the others. So Pirate 2 must assume that if Pirate 3 is out of the picture, Pirate 2 will be broke and dead, regardless of what Pirate 1 may have said to the contrary.

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10. My analysis is the same as Jake’s original. If not given an offer strictly better than the guaranteed minimum upon refusing, then you have a chance that the next proposer will offer you better than he has to. So, e.g., with three pirates, #2 should vote no on an offer of zero because he *might* get a better offer, and won’t get a worse one.

With your second question (the proposer gets to vote), I get (0,2,1,0,97) instead of (2,0,1,0,97).

As to coalitions and bargaining, that gets real interesting. I’ll have to think some more.

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