Five pirates of different ages want to divide a treasure of 100 gold coins. They agree that the oldest among them will propose how to divide the coins, and then all the pirates will vote on the proposal. If more than half the pirates support it, it will be adopted and the coins distributed accordingly. Otherwise the proposer will be thrown overboard and the process repeated with the remaining pirates.
If all five pirates are smart, rational, greedy, and fearful of death, what will happen?
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Work backward:
- If there are only two pirates, the elder will need to give all 100 coins to the younger or he’ll go to the sharks.
- If there’s an intermediate pirate, though, the eldest can pay him 1 coin to gain his cooperation. The youngest, outvoted, must then accept nothing. So, youngest to oldest, the three would receive 0, 1, and 99 gold pieces.
- If there are four pirates, the division is 1, 2, 0, 97. The two youngest pirates must be bribed just enough to oppose throwing the proposer overboard and arriving at the three-pirate disposition immediately above.
- Similarly, with five pirates the eldest should propose the division 2, 0, 1, 0, 97, youngest to oldest. By spending just 3 coins judiciously, the eldest pirate can get the majority he needs.
From the Wikibooks puzzle collection. Somewhat related: How can three people divide a cake amenably?
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