Answer: Tuesday Teaser #26
Whatever distribution Pirate A proposes, Pirates B-E will vote on it according to whether or not they can survive/get more money from the next distribution (and so on), so in order to think about this problem we need to work backwards.
Let’s consider the case that Pirates A, B and C have been thrown overboard and it’s Pirate D’s turn to propose a distribution. Then it is clear that Pirate D can propose all 100 coins to go to himself, vote in favour and keep all of the coins.
That means that all Pirate C has to do to gain the two votes he would need if it came to his turn, would be to propose to give 1 coin to Pirate E and 99 to himself. He could then be sure of Pirate E’s vote as well as his own.
Similarly if it came to Pirate B’s turn, he can propose to give 1 coin to Pirate D and 99 to himself. He could then be sure of Pirate D’s vote as well as his own.
So now Pirate A can be sure of the votes of Pirates C and E by offering them 1 coin each. Incredibly, this means he can achieve 98 of the gold coins for himself.
This puzzle is extremely interesting when the number of pirates is increased. If you’ve enjoyed this puzzle, I would recommend setting aside ten minutes to read Iain Stewart’s analysis of the problem, in which he concludes: “Truly, the meek shall inherit the worth.”
