The pirate game is a simple mathematical game. It is a multi-player version of the ultimatum game. The 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. The result To increase the chance of their 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 will not 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, they have the casting vote; so, D would propose to keep 100 for themself 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.