Five Pirates, 100 Coins

Question

Five perfectly rational pirates (seniority A > B > C > D > E) split 100 gold coins. The most senior proposes a division; all pirates including the proposer vote; a strict majority, make it: at least half, accepts it. If rejected, the proposer is thrown overboard and the next most senior proposes. Pirates maximize coins, prefer fewer deaths only as tiebreaker (and prefer throwing others overboard i

Accepted Answer

Backward induction: the only way to reason about sequential rational play. Two pirates (D, E): D proposes 100 for himself, 0 for E; D's own vote makes the 1-of-2 tie, which passes. D: 100, E: 0. Three (C, D, E): E gets 0 in the 2-pirate world, so E accepts any positive offer. C proposes (C: 99, D: 0, E: 1): C and E vote yes (2 of 3). D's vote is unbuyable for less than 101. Four (B, C, D, E): B needs 2 votes including his own, buy the cheapest pirate. D gets 0 in the 3-pirate world → offer D 1: (B: 99, C: 0, D: 1, E: 0). B + D = 2 of 4 = tie → passes. Five (A, …, E): A needs 3 votes (≥ half of 5 means 3, since 2.5 rounds up). In the 4-pirate world C gets 0 and E gets 0 → buy both for 1 each: boxedA: 98,quad B: 0,quad C: 1,quad D: 0,quad E: 1 A, C, E vote yes, 3 of 5. Why it feels wrong but isn't: the proposer's power comes from everyone's outside option being computed in the continuation game, not from fairness intuitions. The interview follow-ups probe whether your induction is mechanical or understood, change the voting rule (strict majority) or let pirates be bloodthirsty, and the bought pirates and amounts shift. The transferable skill: price each vote at (continuation value +

Five Pirates, 100 Coins

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