Five pirates, 100 gold coins.
The five pirates puzzle: the senior pirate proposes a split of 100 coins, the crew votes, and a losing proposer is thrown overboard. What does the first pirate propose? The answer, 98 coins for himself, looks absurd until you solve the game from the end.
Five pirates, strictly ranked from most senior (A) to most junior (E), must divide 100 gold coins. The most senior pirate proposes a division. Every pirate, including the proposer, votes to accept or reject. If at least half the votes are to accept, the coins are divided as proposed and the game ends. Otherwise the proposer is thrown overboard and the next most senior pirate makes a proposal, under the same rules, with one fewer pirate.
The pirates are perfectly rational and they all know it. Each one's priorities, in strict order: survive; then get as many coins as possible; then, all else being equal, prefer to see another pirate thrown overboard. Nobody can make binding promises or side deals.
What does pirate A propose, and does it pass?
Instinct says A should offer a generous, fair-looking split to avoid the plank. Instinct is about to be wrong by a wide margin. The game has a definite end state, and the way to solve it is not to imagine what A wants but to work out what would happen with two pirates left, then three, and so on back to five. Each step reuses the previous answer.
Stuck? Open hints one at a time
Each hint gives away one more layer. Stop the moment something clicks.
Hint 1 a nudge
Start at the end. With two pirates left, D proposes. D's own vote is half the votes, so D keeps everything. E gets nothing and cannot prevent it.
Hint 2 the key move
With three pirates, C needs one more vote. E knows that rejecting leads to the two-pirate game where E gets zero, so E accepts anything positive. C offers E a single coin and keeps 99.
Hint 3 nearly the answer
At each level, the proposer buys the cheapest votes: the pirates who would get nothing in the next round, one coin each. Track who gets zero at each stage and the five-pirate proposal follows.
The solution
Reveal the full solution
A proposes: A 98, B 0, C 1, D 0, E 1. It passes with the votes of A, C, and E, three of five.
Solve from the end, one pirate at a time:
| Pirates left | Proposal (senior first) | Why it passes |
|---|---|---|
| 2 (D, E) | 100, 0 | D's own vote is half; E is powerless |
| 3 (C, D, E) | 99, 0, 1 | E gets 0 in the next round, so 1 coin buys E |
| 4 (B, C, D, E) | 99, 0, 1, 0 | D gets 0 in the next round, so 1 coin buys D; B plus D is half |
| 5 (A to E) | 98, 0, 1, 0, 1 | C and E get 0 in the next round; 1 coin each buys both |
The logic at every level is identical. The proposer needs enough votes to reach half, counting their own. The cheapest votes belong to the pirates who would receive nothing if the proposer were thrown overboard, because a rational pirate prefers one coin to zero. Since the pirates who get nothing alternate between rounds (E, then D, then C and E), the proposer always knows exactly whom to pay and pays them the minimum. Everyone else gets zero and votes no, which does not matter.
The third priority, preferring to see someone thrown overboard when coins are equal, is what makes one coin necessary rather than zero: a pirate offered nothing will vote to reject, since rejection costs them nothing and drowns the proposer. Offer them one coin and their second priority outranks the third.
What this puzzle is really testing
Backward induction: solving a sequential game from its last move to its first, reusing each solved subgame as a fact. That is dynamic programming over game states, and the table above is the memo. The memoisation versus tabulation deep-dive is the same computation with arrays instead of pirates, and the recursion-with-a-base-case habit is examined in recursion as induction. Interviewers also watch whether you keep the rules straight under pressure: forgetting that the proposer votes, or that a tie passes, changes every row of the table.
Follow-ups to expect
Ten pirates (the pattern continues: the senior pirate keeps 96, paying one coin to every second pirate below). What if a strict majority is required instead of half? Then the two-pirate game changes, D cannot pass anything, and the whole table re-solves with different zeros. And the famous extreme: 500 pirates and 100 coins, where the senior pirates run out of coins to buy votes and start going overboard, which is a genuinely subtle analysis worth trying once.