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?

(In reply to I beg to differ by pleasance)

In truth, this is accurate to an extent. Follow these scenarios using similar logic.

1 pirate. (A) He takes all.

2 pirates. (a,b) B goes first and takes all. A doesn't have a neccesary majority to kill him off.

3 pirates. (a,b,c) C will offer A all of the coins just so A won't kill him of since B would have C killed anyways. A gets all the money here, or close to it depending on how brave C is.

4 pirates (a,b,c,d) D will offer B most the coins since B holds D's life. B will come out ahead here.
5 pirates. (a,b,c,d,e) E will offer C and D half the coins a peice to save his life. Since neither has a more favorable position later on they will both except this. C has 250 and D has @50 leaving A, B, and E pennyless.

Posted by Jon on 2003-04-30 08:37:35