Five pirates have found a treasure chest containing one hundred coins, and are now in the process of dividing the treasure amongst themselves. In their usual democratic fashion they decide that the oldest of them will decide on how split up the treasure. After he has made his decision they will all vote to accept or reject his plan.
If at least half of them vote to reject the plan, the oldest pirate gets thrown overboard, and the process is restarted with the next-to-oldest coming up with the plan, and so on.
Given that the pirates are all trying maximise the amount of treasure they get, how should the oldest pirate propose they share it?
Keep in mind that they, as proper pirates, are all perfectly logical and rational. And getting no coins is just as bad as getting thrown overboard.
Answer
One solution is to work our way from smaller groups to increasingly larger ones. If there’s only one pirate he will obviously get all the coins himself.
If there are two pirates then the younger pirate can always vote to throw the older one overboard. The youngest ends up with all the coins.
If there are three pirates, the next-oldest knows that if the oldest pirate’s proposal is rejected, he will end up with nothing, as in the previous example. The oldest pirate knows this as well, so he gives one coin to the next-oldest, and 99 to himself. The youngest will get nothing and vote no, but he will be in a minority.
With four pirates, the youngest knows that if he doesn’t accept the oldest’s proposal, he will get nothing. The oldest needs at least three pirates to vote with him, so he gives one to the youngest and two to the the next-youngest. The division would be, from oldest to youngest: 97, 0, 2, 1. (If he were to give just one to the next-youngest, the next-youngest could just as well reject as accept the plan).
This can be continued indefinitely. In a group of five the division of coins would be: 97, 0, 1, 0, 2.