With appropriate assumptions, the most logical solution I can think of is:
The fiercest pirate will keep 951 coins for himself, and offer 1 coin to the third fiercest, 1 to the fifth, 1 to the seventh, etc, and 1 to the ninety-ninth.
I'll edit in the proof in a bit, I just have to work on the wording...
OK, for the purposes of argument, I make one major assumption: all pirates are greedy, ruthless, untrustworthy beings, who will always act in their own immediate best interest.
The motivation for this assumption is that it makes the puzzle actually solveable.
Consider the 3-pirate case. Call the pirates 1, 2 and 3, from most to least fiercest. Consider if 2 and 3 make an alliance such that they accept 1's offer, but throw him overboard anyway and split his share. That is to say that if 1's offer is a gold to 1, b gold to 2 and c gold to 3, both 2 and 3 will vote against it and 1 will be thrown overboard. 2 will then offer b+a/2 gold to 2 and c+a/2 gold to 3, who will vote for it.
In such a situation, 1 has no good strategy. Regardless of what he offers, 2 and 3 can do better by throwing 1 overboard. 1's optimum strategy is to claim no gold for himself, giving it all to 2 and 3, and hope in the benevolence of 2 and 3 to not throw him overboard anyway.
This is not necessarily 2 and 3's optimum strategy, but it seems fair to assume that whatever their optimum strategy is, it will leave 1 without a good strategy.
This makes for a less interesting logic puzzle (since the puzzle asks what 1's strategy should be), hence the assumption, which changes the puzzle parameters to be one that has an interesting solution.
With the assumption, if 2 and 3 make this alliance, and throw 1 overboard, 2 (being the ruthless untrustworthy pirate that he is) will then backstab 3 and take all the gold for himself (he can do this, it being the 2-pirate case). 3, being perfectly logical, knows 2 will do this, and never enter the alliance.
The upshot of all this is that a pirate will vote for an offer if it gives them more gold than they would get if they vote against it. If the amount of gold is the same either way, their vote cannot be trusted. Thus, to ensure a pirate's vote, you must simply offer more them gold than they could otherwise recieve. If the next pirate is guaranteed to win if they get to make an offer (which you assume as part of induction) then all you have to do to guarantee a vote is offer more than the next pirate in line would. Guarantee enough votes, and you guarantee a win, which finishes the inductive step.
So, now we start the inductive process to find this strategy.
The 1-person case is a no-brainer. The pirate keeps all 1000 for him/herself.
The 2-person case is also a no-brainer. The fiercer pirate keeps all 1000.
The 3-person case is simple, and described by others above. The fiercest keeps 999, and offers 1 to the weakest. The weakest will then vote for it, knowing that if he/she votes against it, the middle pirate will take all 1000.
In the 4-person case, the fiercest will take 999, and offer 1 to the third-fiercest. The third-fiercest knows that if he votes the offer down, he'll get nothing because he'll become the second-fiercest of the 3-person case, so he'll vote for it.
In the 5-person case, the fiercest needs another two votes to win, and there are two people who will get offered 0 gold if he's thrown overboard and it becomes the 4-person case. So he/she offers 1 gold to each of them, securing their votes.
And so on. In general terms:
In the n-person case, the fiercest will offer 1 gold to each of the third-, fifth-, seventh-, etc, fiercest pirates, who know that if he is thrown overboard it will become the (n-1)-person case, and they'll be the even-numbered pirates and recieve nothing.
By mathematical induction, this strategy will always ensure you get all the votes you need, and you will only pay 1 gold for each vote.
Last edited by phlip
on Tue Oct 10, 2006 5:45 am UTC, edited 1 time in total.
While no one overhear you quickly tell me not cow cow.
but how about watch phone?