The Pirate Gold Splitting Puzzle, Solved Backwards

The pirate puzzle is solved backward from the smallest crew, and the senior pirate ends up keeping nearly the entire stash.

The pirate gold splitting puzzle asks how a group of ranked pirates should divide a stash of coins when each pirate proposes a split, all pirates vote, and a proposal only passes with enough support, otherwise the proposer is thrown overboard and the next pirate proposes instead. The puzzle is solved backward from the smallest possible crew upward, and interviewers use it to test the same backward induction skill that shows up in stopping problems and repeated games.

There are 5 pirates, ranked pirate 1 through pirate 5 by seniority, dividing 100 gold coins. The most senior pirate remaining always proposes a split of the coins among everyone still on the ship. Every pirate, including the proposer, then votes, and the proposal passes if at least half of the votes are in favor, which means the proposer's own vote is always enough to secure a tie. If the proposal fails, the proposer is thrown overboard and the next most senior pirate proposes instead. Each pirate is assumed to be perfectly rational, strictly prefers more gold to less, and prefers to see a rival thrown overboard whenever a proposal would leave them no better off than rejecting it.

The puzzle is solved backward, starting from the smallest possible crew. With only pirate 5 remaining alone, there is nobody left to outvote, so pirate 5 simply keeps all 100 coins. This trivial case is the fixed point every larger case gets compared against.

With pirate 4 and pirate 5 remaining, pirate 4 proposes and needs only half of the two votes, which their own vote already supplies. Pirate 4 keeps all 100 coins and offers pirate 5 nothing, and the proposal passes on pirate 4's vote alone, regardless of how pirate 5 votes.

With pirates 3, 4, and 5 remaining, pirate 3 needs at least 2 of the 3 votes. If pirate 3's proposal fails, the situation collapses to the two pirate case above, where pirate 4 would get 100 and pirate 5 would get 0. Buying pirate 4's vote would cost more than the entire stash, so pirate 3 ignores pirate 4 entirely and instead offers pirate 5 just 1 coin, since 1 coin beats the 0 coins pirate 5 would otherwise get. Pirate 3 keeps 99 coins, pirate 5 votes yes, and the proposal passes 2 votes to 1.