After a long season of plunder, a pirate team of five Prudent Pirates has amassed a booty of 500 golden coins. Before they part their ways, the five decide to divide the treasure.

They that they will each propose a division strategy in order of their seniority: first the oldest pirate will propose the strategy for the division of coins. All five will then vote on it, and if at least half vote "Yes", the strategy will be used to divide the coins. If the majority rejects the plan however, the oldest pirate will be killed, and the whole process will be repeated with the remaining pirates, with the second oldest proposing his strategy.

Since all the pirates are very prudent, each one will want to claim as many coins for himself without getting killed. Given this, how many coins will each of the pirates (5 - 1, with 5 being the oldest) get, and why? What strategy will the oldest pirate propose?

This solution assumes the pirates will vote YES to break even. They will not vote NO  unless they can get more gold.

P1 will always loose all the gold to P2.
P2  always votes NO to P3.
P3 can bribe P1 with one coin.
P4 can bribe P2 with one coin.
P5 can bribe P1 and P2 with one coin each.

P5 = 498
P4 = 0
P3 = 0
P2 = 1
P1 = 1

Once again the CEO is the only one who gets a rase!
Oops, Fashed on Time Bandits there for a second.

Posted by Axorion on 2007-11-20 03:58:40