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URL:  https://boards.fool.com/a-long-attempt-this-first-part-assumes-that-every-16640532.aspx

Subject:  Re: The Pirate Puzzle Date:  2/5/2002  12:43 PM
Author:  galagan Number:  141 of 297

A long attempt:

This first part assumes that every pirate, including the proposer, gets a vote.

As has been said, #10 will vote against everything, because he might be the last alive and get everything.

#9 knows he'll lose if he and #10 are the last, unless he gives #10 everything and #10 has mercy. So he'd accept any offer from #8, even if it gives him nothing, because it gives him certainty of his life and leaves him free from the whims of #10. He'd also accept an earlier offer, but only if it gives him something, since he knows he can always wait for #8, accept his offer, and take nothing but his life. In fact, since #9 has effective swing-vote power, he might take a chance and try to extract a lot in offers from 1-7.

#8 knows he's in control if it's 8-10 left, so he'll vote against everything.

#7 knows he's doomed if 7-10 are left. So he'll accept any offer from #6, and any earlier offer that gives him something, with the same swing-vote extraction possibilities as mentioned for #9.

Similarly, #6, #4, and #2 are in control and vote against everything, while #5 and #3 try to use their swing-vote power to extract things from earlier bidders, or suck it up and vote for a plan that gives them nothing if they'd be up next).

So #1 looks doomed. The only chance is to figure out the deal that #2 would give, and give it himself. That means give #2 most of it, give tokens to the odd numbers, and keep nothing for himself.

Why does #2 vote for this instead of killing off #1 and proposing it himself? No economic reason.

This next part assumes that the proposing pirate does not get a vote.

#10 votes against everything, since he gets it all if he's left last.

#9 knows he'll lose, so he votes for anything 7 proposes. But for people before 7, he'll want something in return for his support (swing-vote deal).

#8 knows he'll lose (1-1 tie will be bad), so he votes for anything, with the same swing-vote deal before 7.

#7 knows he's got it locked up, so he gives everything to himself and wins 2-1. He thus votes against anything before him.

#6 knows he's doomed on a 2-2 tie, so he votes for anything 5 says, and uses swing-vote power before 5.

#5 knows he needs 6, 8, and 9. So he needs to give 8 and 9 something. If he thinks he'll succeed, he'll vote against everyone before him to get to that point.

#4 knows he's doomed, so he votes for anything 3 says, and tries to use swing-vote power before that.

#3 needs 4, 6, 8, and 9, so he'll give 6, 8, and 9 something, and vote against 1 and 2 if he thinks he'll get away with it.

#2 knows he's doomed, so he votes for anything 1 says. Except that if #2 comes up, he can try to offer the same deal that #3 would propose, and hope that #3 will have mercy on him and accept it rather than killing #2 off and proposing the same deal himself. Still, #2 would accept a #1 offer since it guarantees his life rather than leaving it to the whim of #3.

#1 needs 2, 4, 6, 8, and 9, so he'll give 4, 6, 8, and 9 something. He doesn't need to give 2 anything. How 1, 4, 6, 8, and 9 split the money depends on how effective each is in using their swing-vote power. Since 9 has the most chances to escape eventual death, a deal would probably give 9 the most, 8 the next most, 6 the next most, 4 the next most, and 1 the least.

dan
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