A crew consisting of 3 democratic pirates finds a bounty of 100 coins.
The strongest pirate must propose a bounty split and all pirates vote. If the vote passes, meaning that it gets at least half the votes, his proposal is accepted. If the vote doesn’t pass, he is killed, and the process is repeated.
Knowing that all pirates possess perfect logic and their priorities are, in order to survive, then more bounty, then more deaths, how will the coins end up being split?
Introduction:
Let’s look at a couple of examples.
Consider the case when the strongest pirate is greedy and proposes that he gets all the bounty. Suppose his proposal is rejected and he is killed.
Now, the median pirate proposes 100-0 and it passes as it’s a tie at worst. This is a better outcome than the first for the median pirate, as he gets more money.
It is also a better result for the weakest pirate, as there are fewer pirates alive. In hindsight, for both of them, the vote on the first proposal was the right decision
As the strongest pirate has perfect logic and doesn’t want to die, he needs to do something else.
Let’s look at another option he was to try and survive, He could propose 0-0-100. This would pass as both he and the weakest player would vote (as he knows that he would get 0 otherwise) So, the strongest player has an option in which he stays alive.
But, does he have a better one?
Solutions:
Current Case
As we’ve seen before, if the strongest pirate dies, the median pirate will go for 100-0.
For the strongest one to survive, he needs another vote.
Getting a vote from the median pirate:
- As the strongest pirate can’t offer more than 100 coins, and in case of a tie, the median pirate still would rather see him dead, he won’t be able to persuade him.
Getting a vote from the weak pirate:
- In the other scenario, the weak pirate gets 0; given the “piratey” priorities, it’s enough to offer 1 coin to persuade him.
As the strongest pirate maximizes his loot, he will go for the 99-0-1 split. This is his best proposal and will pass with a 2v1 vote.
A More General Case
Let’s try and solve the problem in the case of 100 pirates and 100 coins.
For n pirates, let’s denote them 
and 
the optimal bounty split.
We know that:
In the case of 4 pirates, due to 
, we know that 
will accept 
and thus: 
.
Similarly:
Continuing (formally through mathematical induction), we get our final answer:
- 0’s for odd pirates between 1 and 99,
- 1’s for even pirates between 2 and 98,
- The remaining 51 coins are for the strongest one.