# Pirate's Golden Throw

Five pirates are parting ways after finding a treasure of 100 pieces of gold. The pirates decide to split it based on a vote. Each pirate, from oldest to youngest, gets to propose a plan on how to split the gold.

If at least 50 percent of the other remaining pirates agree on the plan, that is how they will split the gold. If less than 50 percent of the pirates agree, the pirate who came up with the plan will be thrown overboard. Each pirate is smart, greedy, and wants to throw as many others overboard as possible without reducing the amount of gold they get.

What plan can the first (oldest) pirate propose to live and get as much gold as possible?

• Though this question does not (as far as I can tell) appear on this site, a variant of it does here puzzling.stackexchange.com/questions/13206/… which contains a link to the original well-known puzzle and answer Jun 20, 2016 at 9:46
• @GordonK Isn't this the general version of this puzzle? Jun 20, 2016 at 9:49

Let's call the pirates Albrecht, Beatrix, Carla, Daniel and Emily, oldest to youngest. The puzzle doesn't specify that pirates want to stay on board more than they want to throw people overboard. While it doesn't affect the end result, it does affect a couple of steps along the way. I'll write a solution for both cases.

Solution for wanting to stay on the boat:

Start by reducing the problem to the final pirate alone, since the pirates consider what happen if the current suggestion doesn't pass.
$One\ pirate:$ Emily suggests 100 to herself and votes yes, obviously.
$Two\ pirates:$ Doesn't matter what Daniel suggests. Emily gets 100 coins if she's alone and she wins the vote if it's 50/50, which also lets her throw Daniel overboard. She votes no, Daniel gets thrown overboard.
$Three\ pirates:$ Carla suggests 100,0,0. Emily votes no since both one and two pirates left results in 100 coins, which is better than 0 coins. Daniel votes yes because a no from him would result in getting thrown overboard. 0 coins is better than getting thrown overboard. Carla obviously votes yes to her own suggestion. 2 vs 1, Carla gets all the coins.
$Four\ pirates:$ Beatrix suggests 98,0,1,1. Emily votes yes because she gets 1 coin while she gets 0 coins if Beatrix loses the vote. Daniel votes yes because he gets 1 coin, which is more than the 0 he gets from voting down Beatrix. Carla votes no since she gets 100 coins if Beatrix's suggestions fails. Beatrix obviously votes yes. 3 vs. 1, vote passes.
$Five\ pirates:$ Albrecht suggests 97,0,1,0,2. Emily votes yes because she gets 2 coins instead of 1 coin. Daniel votes no because he gets 0 coins instead of 1 coin. Carla votes yes because she gets 1 coin instead of 0 coins. Beatrix votes no because she gets 0 coins instead of 98. Albrecht obviously votes yes. 3 vs 2, vote passes. This is the solution

Solution for wanting to throw people off more than wanting to stay on.

$One\ pirate:$ Emily suggests 100 to herself and the vote passes.
$Two\ pirates:$ Daniel gets voted off since Emily gets all the coins anyway and she wins the 1 vs 1 vote.
$Three\ pirates:$ Carla suggests 99,1,0. Emily votes no because she gets all the coins if Carla gets thrown off. Daniel votes yes because he gets no coins if Carla is thrown overboard. If Carla had given him no coins, there would be no monetary loss for him to throw her off the boat, but since he wants to throw her off more than he wants to stay, the 100,0,0 split would get Carla thrown off. Carla votes yes. Vote passes.
$Four\ pirates:$ Beatrix suggests 97,0,2,1. Emily votes yes because she gets 1 coin instead of 0 coins. Daniel votes yes because he gets 2 coins instead of 1 coin. Carla votes no because she would get 0 coins instead of 99 coins. Beatrix votes yes, 3 vs 1 votes, vote passes.
$Five\ pirates:$ Albrecht suggests 97,0,1,0,2. Emily votes yes because she gets 2 coins instead of 1 coin. Daniel votes no because he gets 0 coins instead of 2 coins. Carla votes yes because she gets 1 coin instead of 0 coins. Beatrix votes no because she gets 0 coins instead of 97 coins. Albrecht votes yes. 3 vs 2, vote passes.

This is similar a very famous problem called Pirate Game.

The solution for this is

97,0,1,0,2 in the same order of priority. Here is the full solution: If only two pirates are left D will be thrown out as he will never have majority. So C can keep 100 gold with him and still D votes for him. So B can give 1 gold to D and 1 gold to E to ensure they get more than before and keep the rest 98. Likewise A needs two votes so he can give 1 to C who will get nothing in case B comes into power and 2 to any of D or E keeping 97 with him.

• I believe that the last pirate would not agree to only getting 1 gold piece. Jun 20, 2016 at 10:23
• He will because if he doesn't and the eldest person A is killed. He is going to end up with nothing as B will keep 99 and give 1 to D and both will vote. Jun 20, 2016 at 10:28
• Here is the full solution: Lets suppose A B C are thrown out so D will keep 100 gold with him. So C needs to give only 1 gold to E to ensure he votes for him. Similarly B can offer D one to sway him in his favor as D knows he will get nothing from C. Finally, A has to give 1 gold to C and D as they were getting nothing from B to have their vote. Jun 20, 2016 at 10:29
• But 99, 0, 1, 0 with B and D agreeing is not a solution for 4 pirates, because at least half of the other remaining pirates must agree (the one who proposes can't vote). This should be the solution for n remaining pirates assuming that all pirates have the same priority of live > get more gold > kill others (only the oldest pirate wants to explicitly survive in the op): 1: 100 2: E votes for death of D, gets 100 again. 3: 100, 0, 0 (2nd pirate votes yes to save his life) 4: 98, 0, 1, 1 5: 97, 0, 1, 0, 2 OR 97, 0, 1, 2, 0 Jun 20, 2016 at 10:33
• I see your point I thought this one is similar to the old one wait I am editing my solution. Jun 20, 2016 at 10:41