# 5 Negotiating Pirates Puzzle

5 pirates of different ages have a treasure of 100 gold coins.

On their ship, they decide to split the coins using this scheme:

The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it.

If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.

As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.

But before the oldest pirate makes his proposition, the other pirates can publicly announce their voting policy, in turn, for the propositions that the oldest pirate could make. They take turns updating their policies, but each pirate gets no more than 200 turns per eldest's proposition (to ensure they do not take infinitely long to state/update their policies).

It is common knowledge that pirates keep to their word - once they announce their policy, they will stick to it, and it is common knowledge that they will stick to it.

Assuming that all 5 pirates are intelligent, rational, greedy, and above all they do not wish to die, (and are rather good at math for pirates) what will happen?

• Could you elaborate on "they take turns updating their policies"? How do they take turns, who begins? Can pirates pass their turn? – Magma Jul 17 '19 at 11:05
• They go from the 2nd oldest (i.e. the oldest pirate, the one that will be making the proposition, does not need to announce his policy) to the youngest 200 times. Pirates can announce a new policy or keep their previously announced policy. They cannot skip their turn. – rinspy Jul 17 '19 at 11:15
• Can you define "voting policy"? – LeppyR64 Jul 17 '19 at 11:44
• @LeppyR64 A statement of the form "I will vote 'Yes' if allocated >= N coins, otherwise I will vote 'No'". – rinspy Jul 17 '19 at 11:46

Working backwards as in the classic... A,B,C,D,E

Two pirates remaining:

D takes 100, E takes 0.

Three Pirates Remaining: C,D,E

D states his usual intention of voting No for all amounts, since he's guaranteed 100.
E proposes that he will take all 100 of the coins, saving C's life.
D noticing that he now gets 0 coins, then proposes he will vote yes for 99 coins
E noticing that he now gets 0 coins, then proposes he will vote yes for 98 coins
So on and so forth until...
D proposes that he will accept 1 coin.

Since they are all perfect logicians, if D initially proposes his usual No, then E would immediately propose to take 1 coin. Therefore they skip all the banter and D initially proposes to take 1 coin.

C takes 99, D takes 1, E takes 0

Four Pirates: B,C,D,E

The same process as with three will follow...
C proposes immediately to take 1 coin.

B takes 99, C takes 1, D takes 0, E takes 0

Five Pirates: A,B,C,D,E

A takes 98, B takes 1, C takes 1, D takes 0, E takes 0

• In the three pirate example, why is D first to act? Shouldn't C make a proposal first? – Trenin Jul 17 '19 at 18:03
• @trenin The proposal does not happen until after the voters have decided their policies. – LeppyR64 Jul 17 '19 at 18:07
• Which among the 4 lower pirates gets the payoffs is ambiguous. In the three pirate case, E has nothing to lose and a potential for gain to make the same deal as D. It is then up to C to decide who gets paid off. And similarly for higher numbers of pirates. – Paul Sinclair Jul 18 '19 at 1:02