A recent Wikipedia exploration of Arrow's Impossibility Theorem just led me to discover that there is a mathematical game called the pirate game. Here's the set-up:
And here's the perhaps counter-intuitive result:There are five rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them. The Pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.
The Pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates should then vote on whether to accept this distribution; the proposer is able to vote, and has the casting vote in the event of a tie. If the proposed allocation is approved by vote, it happens. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.
Pirates base their decisions on three factors. First of all, each pirate wants to survive. Secondly, each pirate wants to maximize the amount of gold coins he receives. Thirdly, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.
It might be expected intuitively that Pirate A will have to allocate little if any to himself for fear of being voted off so that there are fewer pirates to share between. However, this is as far from the theoretical result as is possible.
This is apparent if we work backwards: if all except D and E have been thrown overboard, D proposes 100 for himself and 0 for E. He has the casting vote, and so this is the allocation.
If there are three left (C, D and E) C knows that D will offer E 0 in the next round; therefore, C has to offer E 1 coin in this round to make E vote with him, and get his allocation through. Therefore, when only three are left the allocation is C:99, D:0, E:1.
If B, C, D and E remain, B knows this when he makes his decision. To avoid being thrown overboard, he can simply offer 1 to D. Because he has the casting vote, the support only by D is sufficient. Thus he proposes B:99, C:0, D:1, E:0. One might consider proposing B:99, C:0, D:0, E:1, as E knows he won't get more, if any, if he throws B overboard. But, as each pirate is eager to throw each other overboard, E would prefer to kill B, to get the same amount of gold from C.
Assuming A knows all these things, he can count on C and E's support for the following allocation, which is the final solution:
Also, A:98, B:0, C:0, D:1, E:1 or other variants are not good enough, as D would rather throw A overboard to get the same amount of gold from B.
A more in-depth explanation of the game (and an n=200+ pirates version) is available in Scientific American here. We've talked a lot about pirates' rationality (here and here) in the last few weeks; if nothing else, this game would indicate that avarice and violence are part of contemporary understandings of what it is to be a pirate.
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