Five Pirates and 100 Coins

Five pirates, ranked $P_5$ (most senior) down to $P_1$ , must split 100 gold coins. The most senior living pirate proposes a split, then all living pirates vote. If at least half approve, the split stands; otherwise the proposer is thrown overboard and the next most senior takes over. Every pirate is perfectly rational and wants, in order, to survive, then to maximise his coins, then to see fewer pirates left aboard. How is the gold divided?

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#Solve the smallest game first

Work upward from the endgame, since each proposal is judged against what happens if it fails.

With one pirate, $P_1$ takes all 100. With two pirates, $P_2$ needs only his own vote, because one of two votes already clears the half mark, so $P_2$ keeps 100 and $P_1$ gets nothing.

#Bribe the pirates who would get nothing

A pirate approves only when the offer strictly beats his payoff in the next subgame, since equal coins plus a thinner crew is the outcome he prefers. So every proposer buys the cheapest votes he needs, the pirates whose continuation payoff is zero, with a single coin each.

Three pirates. If $P_3$ falls the game becomes the two-pirate game, where $P_1$ earns 0. So $P_3$ hands $P_1$ one coin and keeps 99, giving $(99, 0, 1)$ .
Four pirates. The continuation $(99, 0, 1)$ leaves $P_2$ as the one earning 0, so $P_4$ pays $P_2$ one coin and keeps 99, giving $(99, 0, 1, 0)$ .
Five pirates. $P_5$ needs three of five votes, his own plus two. The continuation $(99, 0, 1, 0)$ leaves $P_3$ and $P_1$ earning 0, so $P_5$ pays each one coin and keeps 98.

Each row solves the subgame for a crew of size n. The proposer sits on the left and keeps almost everything, paying a single coin to the pirates who would walk away with nothing if he were thrown overboard. Reading down to the five-pirate row gives the final split of 98, 0, 1, 0, 1 from most senior to most junior.

#The split

The most senior pirate keeps 98, the pirates two and four ranks below him take one coin each, and the rest get nothing. The final division is $(98, 0, 1, 0, 1)$ from $P_5$ down to $P_1$ , carried by the three votes of $P_5$ , $P_3$ , and $P_1$ . Seniority is worth almost the whole chest, precisely because everyone can see exactly how the bloodbath would otherwise unfold.