Three pirates (in order of seniority A, B, C) find a treasure chest containing 100 (indivisible)
coins. They have the following rules regarding the distribution of treasure. The most senior
pirate on the ship proposes a plan of how to distribute the coins, and everyone takes a vote on
the plan. If there are at least as many votes in favor as against, the vote passes and distribution
is done accordingly. If the majority votes against, the proposer is thrown overboard, after
which the now most senior pirate makes a proposal. Pirates prefer more coins to less. If a
pirate is indifferent between voting for or against in terms of coins, he prefers throwing the
proposer overboard. Find the sub-game perfect Nash equilibrium of this game. Hint: use
backward induction and read carefully.
The perfect nash equilibrium of the game if found as follows:
The oldest pirate will propose a 98:0:1:0:1 split, which means that the oldest pirate gets 98 coins, the middle pirate gets 1 coin and the youngest pirate gets 1 coin.
We can name the pirates from oldest to youngest as follows:(Ali,Baily,Cosy,Dan, Else).
Working backwards:
2 pirates: Dan splits the coins 100:0 giving himself all the gold. His 50 % vote is enough to ensure the deal.
3 pirates: cosy splits the coins 99:0:1 Else will accept the deal, getting only one coin because h there will be only two pirates left and he knows that if he rejects the deal, he gets nothing.
4 pirates: Baily splits the coins 99:0:1:0. By the same reasoning as before, Dan will support the deal. Baily would not waste a spare coin on Cosy, because Cosy knows that if he rejects the proposal, he will pocked 99 coins once Baily is thrown overboard. Baily would also not give a coin to Else, because Else knows that if he rejects the proposal he will receive a coin from Cosy in the next round anyway.
5 pirates: Ali splits the coins 98:0:1:0:1 . By offering a gold coin to Cosy, who would otherwise get nothing, he is assured of a deal.
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