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The pirate game is a simple mathematical game. It is a multi-player version of the ultimatum game.
There are five rational pirates (in strict order of seniority A, B, C, D and E) who found 100 gold coins. They must decide how to distribute them.
The pirate world's rules of distribution say that the most senior pirate first proposes a plan of distribution. The pirates, including the proposer, then vote on whether to accept this distribution. If the majority accepts the plan, the coins are dispersed and the game ends. In case of a tie vote, the proposer has the casting vote. If the majority rejects the plan, 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. The process repeats until a plan is accepted or if there is one pirate left.
Pirates base their decisions on four factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins he receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.And finally, 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.
To increase the chance of his plan being accepted, one might expect that Pirate A will have to offer the other pirates most of the gold. However, this is far from the theoretical result. When each of the pirates votes, they won't just be thinking about the current proposal, but also other outcomes down the line. In addition, the order of seniority is known in advance so each of them can accurately predict how the others might vote in any scenario. This becomes apparent if we work backwards.
The final possible scenario would have all the pirates except D and E thrown overboard. Since D is senior to E, he has the casting vote; so, D would propose to keep 100 for himself and 0 for E.
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 one coin in this round to win E's vote. Therefore, when only three are left the allocation is C:99, D:0, E:1.
If B, C, D and E remain, B can offer 1 to D; because B has the casting vote, only D's vote is required. Thus, B proposes B:99, C:0, D:1, E:0.
(In the previous round, 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.)
With this knowledge, A can count on C and E's support for the following allocation, which is the final solution:
(Note: 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.)
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 + 1 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. For instance, with 100 gold pieces and 500 pirates, pirates #500 through #457 die, and then #456 survives (as 456 = 200 + 28) as he has the 128 guaranteed self-preservation votes of pirates #329 through #456, plus 100 votes from the pirates he bribes, making up the 228 votes that he needs. The numbers of pirates past #200 who can guarantee their survival as captain with 100 gold pieces are #201, #202, #204, #208, #216, #232, #264, #328, #456, #712, etc.: they are separated by longer and longer strings of pirates who are doomed no matter what division they propose.
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