*Technology Review*

*a few months ago) has a neat solution - if you assume the other players are perfectly logical. In reality, I can't help thinking that pirate #5 would vote no if he were offered only two coins, thinking that, if he can just kill all the other pirates, he'd get twenty. The fact that he'll get fewer coins from the next two proposals is irrelevant - his greed overrides his decision. After eight years or so of playing and watching poker, I feel confident in saying there is no shortage of people who will give up a certain small win to try for a near-impossible big payoff. Anyway, here is the modified problem:*

Five pirates are dividing a treasure of 100 gold coins. The pirates first draw lots to determine an order. The first pirate then submits a proposal for dividing the coins, and all of the pirates vote on the proposal. If the proposal does not receive a majority vote (a tie is not good enough), then the first pirate is killed, and 20 coins are paid to the executioner (who is not one of the pirates). The second pirate then submits a proposal to divide the remaining 80 coins. The process is repeated until a proposal is accepted, or there is only one pirate left. If a proposal maximizes a particular pirate’s share, in the sense that he gets MORE from this proposal then he'll get from ANY future proposal, that pirate votes “yes;” otherwise, he votes “no.”

For example, pirate #5 will vote “no” on any proposal which gives him fewer than 20 coins, because he can get 20 if he's the last one left. In fact, he'll vote against a proposal which gives him exactly 20 coins, because he can already get 20, plus the added bonus of (at least) one less pirate next time there's booty to be divvied up.

How should the first pirate propose to divide the coins in order to maximize his share?

If you read the first link, you'll see the solution to the original problem, which should make this one kind of easy. But I don't want to strain your brain just yet. I'll post the answer on Sunday; more and better problems are on their way, just as soon as I think of them.

## 1 comment:

Okay, working backwards:

If the first three pirates are killed, Pirate #4 has to offer 21 coins to #5 (keeping 19 for himself) if he wants to live.

Knowing this, if the first two pirates are killed, pirate #3 can survive by offering 20 coins to #4, keeping 40 for himself.

Pirate #2 needs two votes to survive. He won't want to offer 41 coins to #3, so he can offer 21 to #4 and 22 coins to #5, leaving himself 37 coins.

Finally, pirate #1 only needs two votes for a majority, so he can offer 22 coins to #4 and 23 coins to #5, keeping 55 for himself.

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