# 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…