The pirate game is a simple mathematical game. It is a multi-player version of the ultimatum game.
Contents
The game
There are 5 rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.
The pirates have a strict order of seniority: A is senior to B, who is senior to C, who is senior to D, who is senior to E.
The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. In case of a tie vote, the proposer has the casting vote. If the distribution is accepted, the coins are disbursed and the game ends. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.
Pirates base their decisions on three factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins each receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal. The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.
The result
It might be expected intuitively that Pirate A will have to allocate little if any to A for fear of being voted off so that there are fewer pirates to share between. However, this is quite far from the theoretical result.
This is apparent if we work backwards: if all except D and E have been thrown overboard, D proposes 100 for D and 0 for E. D has the casting vote, and so this is the allocation.
If there are three left (C, D and E) C knows that D will offer E 0 in the next round; therefore, C has to offer E 1 coin in this round to win E's vote, and get C's allocation through. Therefore, when only three are left the allocation is C:99, D:0, E:1.
If B, C, D and E remain, B considers being thrown overboard when deciding. To avoid being thrown overboard, B can simply offer 1 to D. Because B has the casting vote, the support only by D is sufficient. Thus B proposes B:99, C:0, D:1, E:0. One might consider proposing B:99, C:0, D:0, E:1, as E knows it won't be possible to get more coins, if any, if E throws B overboard. But, as each pirate is eager to throw the others overboard, E would prefer to kill B, to get the same amount of gold from C.
Assuming A knows all these things, A can count on C and E's support for the following allocation, which is the final solution:
Also, A:98, B:0, C:0, D:1, E:1 or other variants are not good enough, as D would rather throw A overboard to get the same amount of gold from B.
Extension
The solution follows the same general pattern for other numbers of pirates and/or coins, however the game changes in character when it is extended beyond there being twice as many pirates as there are coins. Ian Stewart wrote about Steve Omohundro's extension to an arbitrary number of pirates in the May 1999 edition of Scientific American and described the rather intricate pattern that emerges in the solution.
Supposing there are just 100 gold pieces, then:
In general, if G is the number of gold pieces and N (> 2G) is the number of pirates, then
Another way to see this is to realize that every Mth pirate will have the vote of all the pirates from M/2 to M out of self preservation since their survival is secured only with the survival of the Mth pirate. Because the highest ranking pirate can break the tie, the captain only needs the votes of half of the pirates over 2G, which only happens each time (2G + a Power of 2) is reached.