A pure test of backward induction, and a good one because forward reasoning is hopeless.
The setup
Five pirates, ranked A (most senior) to E, divide 100 coins. The most senior proposes a split; everyone votes; if the proposal gets at least half the votes it passes, otherwise the proposer is thrown overboard and the next most senior proposes.
Pirates are perfectly rational, value survival first, then coins, and prefer fewer rivals if otherwise indifferent.
Solve backwards
Two pirates (D, E): D proposes, and his own vote is half of two, so it passes regardless. D takes 100, E gets 0. E hates this outcome.
Three (C, D, E): C needs one more vote. E currently faces getting 0 if C dies, so E accepts 1 coin. C proposes 99, 0, 1.
Four (B, C, D, E): B needs one more vote. If B dies, D gets 0 under C's plan - so D accepts 1 coin. B proposes 99, 0, 1, 0.
Five (A, B, C, D, E): A needs two more votes. If A dies, B's plan gives C 0 and E 0. So C and E each accept 1 coin.
A proposes 98, 0, 1, 0, 1. It passes 3-2.
The principle
Each pirate compares your offer to what they get if you die, not to what feels fair. Buy the votes that are cheapest - the ones who do badly in the next scenario down.
Note the alternation: the pirates who get bribed at each stage flip, because the "next scenario" flips.
Common follow-ups
What if there are 200 pirates and 100 coins? The structure changes once there are more pirates than coins can bribe. Proposers start needing to survive on votes from pirates who get nothing but prefer the proposer to live - and some proposers cannot survive at all. The pattern involves powers of 2.
What if ties fail instead of pass? Every answer shifts, because the proposer's own vote no longer counts as half.
Why it is asked
Backward induction is exactly the reasoning used in sequential games and in thinking about what a counterparty will do next. Getting it right requires resisting the urge to reason forward from five pirates, which never works.
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