[Guest]

Ten pirates plundered a ship and stole 100 gold coins. They decided to divide the coins by following these rules. The pirates were ranked: the captain was number 10, the first mate was number 9, down to the lowliest pirate at number 1. Pirate 10 would propose a plan to divide up the coins, and all 10 pirates would vote on the plan. If the plan got at least 50% of the votes, it would pass. If it got less than 50%, pirate 10 would be killed, and pirate 9 would get a chance to propose a plan. Again the plan would be voted on, with the same results. This would continue until a plan got at least 50% of the votes.

Assuming all pirates are perfectly logical (i.e., they can instantly deduce the logic of their situation and all possible situations), all pirates know that all other pirates are perfectly logical, and all pirates will act to guarantee that they greedily receive the maximum amount of gold for themselves, what happens in this situation? What about for N pirates?

Remember: Each plan only needs 50% of the votes to pass, and all pirates vote; that is, the pirate who proposed the plan votes on his own plan.

[Guest]

this has been posted before

[Guest]

this has been posted before

Damn, really? I searched the forums, I guess I missed it. Sorry.

[Guest]

i saw this one a while ago, it took me like an hour to solve and explain it properly.

the youngest gets 6, second youngest gets 4, third youngest gets 2, and 88 to the eldest.

explanation: if there were only two pirates, the older one will propose that all the coins will go to him, and vote for it.. which will make the vote pass with a 50%. this case does not satisfy the younger of the two. but it will not reach here. if there were three pirates, the eldest will want to win the youngest pirate's vote, he knows that the youngest will not get any gold if only two pirates are left, so he would propose 99 pieces for himself and one for the youngest, the youngest will accept this IF there were 3 pirates, but this does not satisfy his greed. if there were four pirates, the eldest would propose a bigger share (2 coins) to the youngest than what the third would propose, and 0 coins to the second since the eldest needs only one vote along with his to pass the proposal. the rest of the coins would go to him. if there were 5 pirates, the eldest would want to win the votes of two pirates, therefore, he would give 3 coins to the youngest and one coin to the second. both of them would favor this choice than what would come from the pirate younger than current eldest. if there were 6 pirates, to prevent the last scenario from happening (because it would mean his death), the eldest would give even a bigger share to the first two, and 0 to the third since his vote is redundant. so it would be, 4 to the youngest, 2 to the second youngest, and the rest for him. 7 pirates: the seventh will give 5 to the youngest, 3 to the second and 1 to the third to secure a 50% vote. 8 pirates: gives 6 to the youngest, 4 to the second, and 2 to the third, he gets 88 coins, a 50% vote, and his life.

pattern: the pattern here is that the pirate younger than the mid-aged one gets a single coin (if the total number of pirates is even), or 2 coins (the total pirates is odd), then moving towards the youngest pirate, we add 2 coins to each pirate..

if we continue down the line, we will reach a point where the eldest will distribute all the coins in a manner where he would try to save his life and get no coins. this is reached when there are 22 pirates, the eldest will distribute gold from youngest to the 10th in this way: 19, 17, 15, 13, 11, 9, 7, 5, 3, 1. they will all vote in favor plus him will be 11 votes...

Edited March 15, 2011 by hussein

[k-man]

i saw this one a while ago, it took me like an hour to solve and explain it properly.

the youngest gets 6, second youngest gets 4, third youngest gets 2, and 88 to the eldest.

explanation: if there were only two pirates, the older one will propose that all the coins will go to him, and vote for it.. which will make the vote pass with a 50%. this case does not satisfy the younger of the two. but it will not reach here. if there were three pirates, the eldest will want to win the youngest pirate's vote, he knows that the youngest will not get any gold if only two pirates are left, so he would propose 99 pieces for himself and one for the youngest, the youngest will accept this IF there were 3 pirates, but this does not satisfy his greed. if there were four pirates, the eldest would propose a bigger share (2 coins) to the youngest than what the third would propose, and 0 coins to the second since the eldest needs only one vote along with his to pass the proposal. the rest of the coins would go to him. if there were 5 pirates, the eldest would want to win the votes of two pirates, therefore, he would give 3 coins to the youngest and one coin to the second. both of them would favor this choice than what would come from the pirate younger than current eldest. if there were 6 pirates, to prevent the last scenario from happening (because it would mean his death), the eldest would give even a bigger share to the first two, and 0 to the third since his vote is redundant. so it would be, 4 to the youngest, 2 to the second youngest, and the rest for him. 7 pirates: the seventh will give 5 to the youngest, 3 to the second and 1 to the third to secure a 50% vote. 8 pirates: gives 6 to the youngest, 4 to the second, and 2 to the third, he gets 88 coins, a 50% vote, and his life.

pattern: the pattern here is that the pirate younger than the mid-aged one gets a single coin (if the total number of pirates is even), or 2 coins (the total pirates is odd), then moving towards the youngest pirate, we add 2 coins to each pirate..

if we continue down the line, we will reach a point where the eldest will distribute all the coins in a manner where he would try to save his life and get no coins. this is reached when there are 22 pirates, the eldest will distribute gold from youngest to the 10th in this way: 19, 17, 15, 13, 11, 9, 7, 5, 3, 1. they will all vote in favor plus him will be 11 votes...

There is a small flaw in your logic that doesn't provide for the maximum gain to the captain.

I will not go into the complete solution for 10 pirates, but for 4 pirates you can give only 1 coin to the second pirate to win his vote as he gets nothing if you get killed. The same way for 5 pirates, you can get 2 needed votes by giving 1 coin to the lowest and 1 coin to the 3rd pirate as they lose out if you get killed. You can keep going in this manner without ever giving more than 1 coin to anybody, but the key is to give that coin to those who would get nothing in the next iteration.

[Guest]

There is a small flaw in your logic that doesn't provide for the maximum gain to the captain.

I will not go into the complete solution for 10 pirates, but for 4 pirates you can give only 1 coin to the second pirate to win his vote as he gets nothing if you get killed. The same way for 5 pirates, you can get 2 needed votes by giving 1 coin to the lowest and 1 coin to the 3rd pirate as they lose out if you get killed. You can keep going in this manner without ever giving more than 1 coin to anybody, but the key is to give that coin to those who would get nothing in the next iteration.

wow! u're right!i would've never guessed it. ty

[Guest]

k-man, you're correct! For 10 pirates, pirate 10 gets 96 pieces of gold, and pirates 8, 6, 4, and 2 each get 1 piece.

For N pirates, every other pirate, starting with pirate N-2, gets 1 piece of gold, and pirate N gets the rest.