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Pirate 5's optimum strategy allows him to keep 497 coins and live to tell about it.
As we have seen, if it's 2's turn, he will die and 1 will get all the money.
Therefore, everybody knows that if 3 goes overboard, so does 2. If it's 3's turn, he can exploit 2's predicament by keeping all of the money for himself. Pirate 2 will still vote for him and he'll have a majority.
Therefore, if 4 goes over, 1 and 2 won't get any money. Pirate 4 can bribe them each with a coin, and keep 498 for himself. Pirate 3 gets no money.
Therefore, if 5 goes over, 3 will get no money and 1 and 2 will get a coin each. Pirate 5 needs two votes, besides himself, to win. So he bribes pirate 3 with a coin and one of Pirates 1 and 2 with two coins. He keeps 497 coins for himself!
It is worth noting that although "they are graduates of CSU" may seem anecdotal and trivial it is actually crucial to the problem. If the pirates had graduated from Colorado School of Mines, they would have all worked this solution out before agreeing to the rather unusual method of distributing the coins. They would realize that nobody was going to get thrown overboard and that there was a good chance they'd end up with nothing or almost nothing.
Of course the mathematician tells us that the expected payoff of the above solution is 100 coins which is just as good as any other solution that doesn't involve killing people. However, the economist would tell us that this is really not a problem of expected value but rather of expected utility. Because marginal utility diminishes as quantity increases it is intuitively clear that a gauranteed 100 gold coins is much better than the 497/5 + 1/5 + 2/5 + 0/5 + 0/5 expected value offered by the above solution. They therefore would have come up with some other method of dividing up the money (perhaps an even split or maybe random plank walkings determined by straw drawings?)