Game Theory

Game Theory Parables for Game Designers

A Pirate Solution (read A Pirate Puzzle First!)

Be sure to read the original puzzle here first!

To find the solution, you actually need to start at the end, at the very last scenario where a decision has to be made.  This is when there are only two pirates left, D and E, and D is proposing a split of the 100 gold pieces.

With only two pirates left, the vote will be a tie, but D holds the tiebreaker.  Thus, they can offer 100 gold for D, and 0 for E.  D will vote in favour, E will vote against, and it will pass.

Now take one step back, to when C is proposing a split amongst C, D, and E.  C knows that if their proposal is rejected, D will propose 100 for D and 0 for E (C having been thrown overboard).  So C has to get E on board with their proposal, and can do this by offering E anything more than 0 gold pieces.  C can therefore propose 99 gold for C, 0 gold for D, and 1 gold for E.  E will actually vote in favour of this proposal, because if it is rejected they will get 0 gold pieces instead of the 1 that they will get if it is accepted.  D will vote against it, while C will vote in favour: the end result is 2 votes in favour, 1 opposed, so this proposal is accepted.

Now take another step back, to when B is proposing a split amongst B, C, D, and E.  B knows that if their proposal is rejected, the (99, 0, 1) proposal will be made and accepted when C is in charge.  As above, B needs to win over at least one more voter to ensure their proposal passes.  Pirate D is an obvious target, because they are getting 0 gold pieces under C’s proposal.  So B has to offer D at least one gold to get their vote: a split of 99 for B, 0 for C, 1 for D, and 0 for E will garner 2 votes in favour, 2 opposed.  As B holds the tiebreaker, and they voted in favour, this proposal will be accepted.

Finally, stepping all the way back to when A is proposing a split amongst A, B, C, D, and E.  A knows that if their proposal is rejected, the (99, 0, 1, 0) proposal will be made and accepted when B is in charge.  As above, A needs to win over two voters, and C and E are the ideal targets.  Those two pirates would get 0 gold if B was in charge, so A needs to offer then at least one gold each to get their votes.  A can propose (98, 0, 1, 0, 1) to get the votes of C and E, which would make 3 votes in favour, 2 opposed, and the proposal passes.

Thus, the solution that perfectly rational pirates would come up with is that A will propose 98 gold for A, 0 for B and D, and 1 each for C and E.  A, C, and E will vote in favour, B and D will oppose, and the proposal will be accepted.

This solution is counter-intuitive: most people are drawn to a “fair” distribution of 20 gold coins each, while some may offer more to the captain, while others even offer more to the lower pirates to ensure their votes.

This counter-intuitiveness should be no surprise if you’ve followed other examples on this site.  It is arrived at by a process called Backwards Induction: “look ahead, reason back.”  This process starts at the very last decision(s) possible in a sequential game like this, determines what a rational player would do when faced with that decision, and then moves back a step.  Each step back, the decision at the next step in the future is known (by this analysis), which means that this decision can be analyzed as well.  Trace this all the way back to the very first decision, and a solution of sorts is found.

However, there’s good reason to think that there is a problem with Backwards Induction…

Sam Hillier