The pirate game produces a unique result. There are 5 pirates and 100 gold peices and they have to distribute the gold. They are ranked by strength, 5 is the strongest, 1 is the weakest. The strongest proposes a way of distributing the money and they vote on it. Any vote with 50% or more wins. If the vote is less than 50%, the proposing pirate is killed. What’s the final distribution?

Pirate 5 gets 98 gold coins, Pirate 4 gets 0, Pirate 3 gets 1, Pirate 2 gets 0, Pirate 1 gets 1. How?

You go to the end and work backwards

P1 and P2 are the last two voters. P2 can take 100 gold and threaten to kill P1. P2 wins the vote.
Add in P3. P3 can bribe P1 with 1 gold coin and take 99 gold for himself. P1 will accept any gold. P2 loses the vote.
Add in P4. P4 takes 99 coins and bribes P2 with 1 coin to win.
Add in P5. P5 proposes that he takes 98 coins, and bribes P1 and P3 for their votes with 1 gold each. This is the best deal P1 and P3 can get, so they vote for it.

Asymmetrical bargaining power, even with a democratic vote, leads to a concentration of wealth.
This pattern holds for 10 pirates. Pirate 10 takes 96 coins and bribes 4 others with 1 coin each.

This continues up to 200 pirates with 100 gold. The strongest pirate’s share is slowly diluted because he has to bribe a very large number of voters with 1 gold each. In the end, he only has one coin himself. Half the other pirates get nothing at all.

After 200 though, the game gets funny. They run out of money to bribe voters. The stongest pirate must take nothing and bribe half the other pirates so they don’t kill him. The Pirate Holocaust begins at player 203. He can’t bribe the half the voters even if he takes no money at all.

Pirate mass murder occurs when gold is scarce. Where gold is abundant, the strongest pirate always takes the lion’s share even with a democratic vote.

If there are 500 pirates with 10,000 gold, the strongest pirate will take 9,751 for himself and bribes 249 others with 1 gold each.

Suck it Pirate #2!

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