I first heard this 10 years ago when my dad told me it, and it’s still a favorite of mine…
5 very smart bloodthirsty pirates have 100 pieces of gold they must divide among them. They agree on the following plan: First, they draw lots to order the pirates 1 through 5. The first pirate must propose a way to divide the gold. 50% or more of the pirates must agree on the plan or else he is killed and then it moves on to pirate 2 to make a proposal. and so on…
If 50% or more are in agreement of the proposed plan, the gold is divided as such and everyone goes their way.
First pirate needs to buyout the last two pirates. Just offering 30 to pirates 2 and 3 is no good. They will be incentivized to say no and kill him regardless of what he offers them. Their share will go up to 50 from 33 if they just refuse the first offer. In fact I’d make a deal and give up to 50 and 50 to whomever would agree which would presumably be pirates 4-5. At least I’d stay alive.
Shares of 10 for everyone. Pirates would only agree to double shares for the captain. They would shoot you if you tried to divide it up any other way. Haven’t you read the Invisible Hook?
the first pirate proposes that lots 2 and 3 get 40% each and 1 gets 20%. number 4-5 get 0. this makes sense because this would be the higherst possible payout to 2 and 3 so it gets thier vote and 1 end up with his life and 20% so he votes for it
Ok… so this is another of those recursive questions where you must start from the end case and work backwards. This is my solution. I’m not 100% sure, but I can’t find any logical flaws so far…
Notation: a/b/c/d/e means treasure allocation from worst pirate to best.
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If there are 2 pirates: 0/100 (pirate 1 needs only 50% of the vote, so just takes all the money)
If there are 3 pirates: 1/0/99 (pay off the last pirate, who will get zero otherwise)
**From here on, just keep paying off the cheapest pirate(s) in the next round, such that votes including the main pirate’s are at least 50%.
If there are 4 pirates: 0/1/0/99
If there are 5 pirates: 1/0/1/0/98
So pirate 1 takes 98%, and pirates 5 and 3 each get 1%.
that does not make sense to me. 3 or 5 would not agree with this because by not agreeing 1 would be eleminated meaning that 2 who was going to get zero could propose an even split and everyone would be better off. 2 would at the minimum need to propose a value high enough to get two others to agee with him
but based on the solution above that would require 2 pirates to take 1%. The proposing pirate would be 1 but the other would be better off not agreeing and forcing the elimination of the proposing pirate. you solution only works when you have 3 pirates. that way you can give 99% to pirate 3 and the proposing pirate takes the 1%. In truth you would not even need to offer pirate 3 98% but just above 50% since, when there is 3 pirates, upon elimination of the proposing pirate the two remaining would only get 50% each…
explain this. in round 1 you need 3 to agree, right? you have 1 take 98% and 2 pirates take 1%? The pirate who decides takes 1%…he can do that since it is his choice. the pirate who gets 98% would vote for it. but why would the third who only gets 1% vote for it?
Because if either of Pirates 3 or 5 refuse the 1%, Pirate 1 gets killed and the game progresses to the next round. In the second round, the equilibrium is for both Pirates 3 and 5 to receive 0%. So, they will both accept 1% in the first round if it means that the game will not progress to the second round.
I take it nobody here remembered the dictator game. The assumption was that they were smart and bloodthirty, not necessarily rational. Offering 1/0/1/0/98 will still get you killed given piratenorms (yes, I made that word up).