Solving this puzzle doesn’t require any particular leaps of imagination, just a systematic logical approach. The key is to work ‘back-to-front’; first consider what happens if all the splits are rejected and work back from there. Recalling that the hierarchy states that captain outranks master, outranks mate, outranks cabin boy, the argument runs as follows.
- CABIN BOY: If the captain, the master and the mate were all to lose the votes on their proposed splits, it would get down to the cabin boy, and he would get 100 gold pieces.
- MATE: Now consider the situation for the mate if the captain and master have already lost their votes. He needs the cabin boy to agree to his proposed split or he will be obliged to throw himself overboard (remember that a tied vote is not good enough for a split to be accepted). However, if he proposes any split in which the cabin boy gets less than 100 gold pieces, it will be rejected, because the cabin boy knows that he can then get all the money anyway. Therefore, the mate must offer the cabin boy everything and hope that doesn’t just throw him overboard for fun.
- MASTER: Now consider the situation for the master if the captain has been thrown overboard. He needs only one of the other two pirates to support his split in order to win the vote. However, as we have seen, the cabin boy knows that if the master’s split is rejected, then he will get all the money, so there is nothing that the master can do to ensure the cabin boy’s vote. Conversely however, the mate will get nothing if the vote is rejected, so the master need only offer him a single gold piece to ensure his cooperation. The master therefore proposes to keep 99 gold pieces, to give 1 gold piece to the mate and nothing to the cabin boy.
- CAPTAIN: As captain, you must get two other pirates to support your suggestion. We now know that if your proposal is rejected, the master will get 99 gold pieces, so his support could only be guaranteed by offering him all the money. However, the mate and the cabin boy are only in line to receive 1 and 0 gold pieces respectively, so their votes can be secured by offering them a much more reasonable 2 and 1 gold pieces. You should therefore propose to keep 97 gold pieces for yourself, offering nothing to the master, 2 to the mate and 1 to the cabin boy.
Although this puzzle is relatively straightforward, I think that the answer is quite interesting. It seems counter-intuitive that the solution would involve keeping such a huge share of the money for yourself, but that is what the logical argument tells us, given the assumptions about the pirates that were stated in the question.
In real life, as you might expect, people’s behaviour tends not to reflect that of these pirates. Firstly, people tend to offer more equitable divisions even when this reduces their own gains (as observed in the Dictator Game) and, secondly, people tend to reject splits that they do not perceive to be fair, even if doing so will result in a worse outcome for themselves (as observed in the Ultimatum Game).
To the best of my knowledge, however, no psychological studies exist in which one of the possible outcomes is being devoured by sharks, so there is insufficient evidence to draw firm conclusions over how people might react if genuinely forced to play the pirate division game.
P.S. After writing this, I discovered that the puzzle is actually quite a well-known mathematical game, which even has its own Wikipedia page! Since I only had fairly hazy memories of it from a good fifteen years ago, I had vaguely assumed that it would be languishing in some forgotten book of logic puzzles from the 1970s, probably set in a different context, but no. It is in fact a firmly established part of the canon of recreational mathematics, pirates and all.
Ian Stewart published an article on the puzzle in 1999, with a lovely extension up to very large numbers of pirates, which is worth checking out. Incidentally, the date on this article matches quite well with my recollection of when I first heard about it, so it may well be than my school maths teacher got it from this very source.
However, there are differences between the version of the puzzle that I posed and those given by the sources above. In those other versions, pirates explicitly prefer not to be thrown overboard (which admittedly makes sense) and, more importantly, only a tie is required for a vote to be carried. These differences change the solutions considerably, though not the methods used to obtain them. I am pretty certain that I heard the puzzle in the way that I have stated it, so perhaps it wasn’t taken from the Stewart article after all.
A quick think about the solution to my formulation of the problem for large numbers of pirates suggests that ambiguity sets in as soon as you have at least seven. Working the logic through, you find that the most senior pirate of seven has two possible ways to guarantee the required number of votes: one of the fourth and sixth most senior pirates will be offered 2 gold pieces and the other will be offered nothing. At that point, you either have to impose a rule about how pirates should choose between their colleagues in such circumstances or you have to decide whether pirates prefer to have 1 guaranteed gold piece or 2 possible gold pieces. These decisions will then affect the optimal behaviour and the takings of even more senior pirates.
Bonus points to anyone who can be bothered to work through all these possible versions of the puzzle to find the solutions for arbitrarily many pirates in each case.
Thomas Oléron Evans, 2014