The Pirate Puzzle!

As we all know, math can be applied to real life situations. Even if you are a pirate.

The puzzle: Five smart, greedy and mean pirates find themselves on a boat with 100 gold coins. The pirates are ranked according to seniority, from 1 to 5, the most senior being ranked 1. Here is the deal: the most senior pirate must make a proposal for splitting the coins among the pirates, including herself (no fractional coins!). If her proposal has at least 50 percent acceptance (she votes, too), she is fine; otherwise, the other four pirates throw her into the shark-infested sea. The remaining pirates play the same game, starting with a proposal from the next most senior pirate, that is, the pirate ranked 2.

What should pirate No. 1 propose?

The original solution: she should propose giving one coin to the last and third pirate, and keeping the remaining 98 coins for herself.

The way to solve this puzzle is to start from 2 pirates and add one pirate at a time until there are 5 pirates.

Before moving on, let us clarify what we mean by smart, greedy and mean. Smart means that all pirates are able to deduce the logical consequences of any proposal. Greedy means that they want to maximize their earnings. Mean means that they will throw someone off the boat if means more gold for them.

Let us also change the seniority ranking, meaning that, from 1 to 5, 5 is the most senior and 1 the junior pirate.

Back to our two pirates, P1 and P2. In this case, P2 proposes 100 coins for herself and zero for P2, gets 1 vote (her own) and is safe. 

With 3 pirates we get to the core of this puzzle. P3 needs 2 votes, her own plus one. She can accomplish this by giving P1 a single coin and keeping the rest for herself. Why would P1 accept a single coin? Because the alternative is worse. If P1 votes against the plan (and P3 is tossed to the sharks) then there are only two pirates, and P1 gets nothing. So P3 proposes 99/0/1.

Moving to 4 pirates, we see the emerging pattern: by looking at the previous proposal, P4 can propose 99/0/1/0 and get two votes (her own plus the pirate getting the single coin).

For 5 pirates P5 will need 3 votes — her own plus 2. She therefore proposes 98/0/1/0/1.