The Pirate Gold-Division Game

Five pirates, ranked by seniority, must split 100 gold coins. The most senior proposes a split; all pirates, including the proposer, vote; if at least half accept, it happens; if not, the proposer is thrown overboard and the next most senior proposes instead. Pirates are perfectly rational, care about gold first and survival second, and would rather see a rival thrown overboard than not, all else equal. How should the most senior pirate split the gold?

With only 2 pirates left, the senior one proposes keeping all 100 for themselves — they need only their own vote (a 1-of-2 tie counts as acceptance), so the junior pirate gets 0 regardless. Add a third pirate: the proposer needs one more vote, and the cheapest way to buy pirate 3 (the most junior)'s vote is 1 coin, since pirate 3 would otherwise get 0 if it falls to the 2-pirate case. So with 3 pirates, the split is 99/0/1. With 4 pirates, the proposer needs one more vote; the pirate who'd get 0 in the 3-pirate outcome (pirate 2) will accept just 1 coin, giving 99/0/1/0. With all 5, the proposer needs two votes; the two pirates who'd get 0 in the 4-pirate case are pirates 2 and 4, so 1 coin each buys them off: the answer is 98/0/1/0/1.

The puzzle is really testing whether a candidate can hold a chain of "what happens if this proposal fails" scenarios in their head and reason from the base case outward, which is the same discipline used in pricing American options or any other backward-induction problem.