Here’s my solution, adapted from my solution to “Greedy Remorseful Pirates”. I’m assuming no enforceable contracts. That makes this a noncooperative game theory problem. (I don’t yet know how to solve the cooperative case – what @byorgey and @zedmango were discussing – where you do have enforceable contracts but I agree it could be very interesting!)
Also, contra @zedmango’s and @daerdemandt’s solutions, I’m assuming no uncertainty. To achieve that I’m using one more assumption about how pirates break ties: they always favor pirates closest to their own height. I don’t think that assumption is strictly needed but it avoids a big mess of multiple Nash equilibria.
Also I’m ordering the pirates shortest to tallest. So pirate 1 is the shrimpy proposer.
In the 2-pirate case the short pirate (the proposer) is utterly doomed. There’s no way to appease the tall pirate. Even if you give them all the gold, they don’t lose any gold by killing you so they do.
In the 3-pirate case it’s the total 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 person will die if you die, they’ll vote yes and you have your majority. No need to give the tallest pirate anything.
In the 4-pirate case, you as shortest pirate have no way to win the 2nd pirate’s vote. They get everything if you die, by becoming the proposer in the 3-pirate case. So don’t give that pirate anything; they’re a lost cause. Instead, you need to win over the 2 tallest pirates (3 and 4) to get a majority. Giving them 1 coin each does the trick. That’s 1 coin better than they each get by killing you, so they’re on board.
In the 5-pirate case, the 2nd pirate is still a lost cause and you still need to appease 2 other pirates. Pirate 3 gets nothing if you die (they become pirate 2 in the 4-pirate case) so 1 coin suffices for them. But pirates 4 and 5 each get 1 coin if you die (by becoming pirates 3 and 4 in the 4-pirate case). So pick one of them – breaking the tie in favor of the shorter one, 4 – and give them 2 coins. That’s your 3rd of 5 votes so no need to waste any money on pirate 5.
Now we can see the pattern emerge. The shortest pirate allocates nothing to pirate 2, 1 coin to pirate 3, 2 coins to pirate 4, 0 coins to pirate 5, 1 coin to pirate 6, and continues alternating 0 and 1 coin for the remaining pirates.
Here’s a table showing the cases up to 9. Each column shows the allocation for a given number of pirates. A * means the remainder of the treasure, ! means certain death, and the tick marks are the yes votes. You can look up and to the left from each cell to see what that pirate gets if the proposer dies (if it’s worse they votes yes).
1 2 3 4 5 6 7 8 9
1 *' !' *' *' *' *' *' *' *'
2 * 0' 0 0 0 0 0 0
3 0 1' 1' 1' 1' 1' 1'
4 1' 2' 2' 2' 2' 2'
5 0 0 0 0 0
6 1' 1' 1' 1'
7 0 0 0
8 1' 1'
9 0
For a proper proof we just need to consider the allocation for the n-pirate case – {*, 0, 1, 2, 0, 1, 0, 1, 0, 1, …} – and show that the majority vote yes. We get 1 vote from the proposer (pirate 1), 1 vote from pirate 3, and 1 from pirate 4. That’s 3 out of 4 votes so far. For pirates 5 through n we get votes from all the 0’s in the allocation, which is half of them if n is even, so we’re set. If n is odd then the nth pirate is getting 0 and voting no. In that case we have 3 of 4 yeses, then half yeses for pirates 5 through n-1, then a no. Grouping the last no with the first 4, that’s still 3 out of 5 and then half of the rest, so still, barely, a majority. QED.