Junior Quant Trader · Interview · Game theory · reported by a candidate
Five pirates of strictly ranked seniority divide 100 gold coins.
The report doesn't give the full rules. Standard version: the most senior pirate proposes a split; everyone votes; it passes with at least half the votes (the proposer votes too); otherwise the proposer is thrown overboard and the next most senior proposes. Pirates are perfectly rational, want gold first, and given equal gold would rather throw someone overboard.
- (a) How many coins does the most senior pirate keep?
Answer and worked solution
Answer: 98
Key idea. Backward induction: solve for 2 pirates, then 3, and so on. Each proposer buys the cheapest votes.
- 2 pirates: the proposer's own vote is half, so the proposal passes: 100, 0.
- 3: needs one more vote. Buy the pirate who'd get 0 with two left, for 1 coin: 99, 0, 1.
- 4: needs one more vote. Buy the pirate who'd get 0 in the three-pirate split: 99, 0, 1, 0.
- 5: needs two more votes. Buy the two who'd get 0 in the four-pirate split: 98, 0, 1, 0, 1.
Check. Each bought pirate is strictly better off than in the next round down, so they vote yes.
Trap. Offering a coin to the pirates who'd get 1 next round: they'd need 2.
- (b) Follow-up (reported): with seven pirates?
Answer and worked solution
Answer: 97
Key idea. Continue the induction to seven pirates.
- 6 pirates: 98, 0, 1, 0, 1, 0 (two extra votes needed).
- 7 pirates: three extra votes needed. Buy the three pirates with 0 in the six-pirate split, 1 coin each.
- The proposer keeps 97.
Check. The pattern: with n pirates the proposer keeps 100 − ⌈n/2⌉ + 1 (until the coins run out, past 200 pirates).
Interviewers may push further: What changes when there are more than 200 pirates and not enough coins to buy votes?
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