A while ago we had a puzzle about five pirates dividing 100 gold coins between them, which you can read here. We now have a variation on that puzzle.
There are five pirates on a ship with 100 gold coins of booty, and they are trying to divide the coins between them. The method that they have agreed is that the oldest pirate proposes a distribution of the coins and the other pirates vote on the proposal. If 50% or more of the pirates agree with the proposal, it is accepted and the coins distributed. If not then the pirate that made the proposal is thrown overboard and the oldest remaining pirate proposes a distribution of coins.
The pirates don't like each other, so if a pirate would get the same number of coins however he votes, he will vote against to kill the proposing pirate.
If everyone votes logically, what happens?
Let's work backwards from two pirates.
In this case the oldest proposes that the other pirate receives all 100 coins otherwise the vote will fail and he will be killed.
With three pirates the eldest will propose a split of 99, 1 and 0. The youngest pirate will vote against this, because if he doesn't there will be only two pirates left, and he gets everything. But the second youngest votes for it, otherwise he gets nothing.
With four pirates the eldest will propose a split of 97, 0, 2 and 1. The two youngest will support this as they get more than they will with three pirates.
So with the initial five pirates a split of 97, 0, 1, 0 and 2 will be supported by three pirates, the eldest, and the two who receive coins who would get less on the four-pirate vote.