In class we briefly examined “The Ultimatum Game” during the discussion of power in networks. The game set-up was that two people had to decide how to split, say, 100 dollars. One person got to propose a deal, and the other could either accept or reject the terms. If rejected, both would recieve 0; if accepted, each got the amount specified in the deal. This set-up gives the first person most of the power, so the deals observed when this experiment was/is run tend to benefit the person to propose the split. The only really interesting thing to notice was that people actually have some value of ‘fairness.’ In other words, below a certain amount, the second person will reject the deal, despite the payout of 0, because they feel cheated.

But what if it is more than 2 people. This game, with a few small additions, actually extends to an infinite number of players, but let’s examine just 5. This is known as the ‘Pirate Game,’ since it is most easily imagined with a group of pirates. Say 5 pirates, P1 through P5, have plunder of 100 coins of equal value). The pirates have a naturally decided hierarchy, so P1 outranks P2 who outranks P3 and so on until P5, who is at the bottom. The game starts with P1 proposing a 5 way split of the coins. The pirates then all vote on the deal. If majority accepts, then the split is made. If there is a tie, then the ranking pirate, P1 decides the outcome. If majority rejects the split, then P1 is thrown overboard and dies, and the game starts over with the remaining pirates, highest ranking pirate offering a split. The knee-jerk reaction to this game is that P1 would have to give himself very little in order to ensure he is not killed, but this is not the case. There are 3 forces at work for each pirate: desire to maximize own take; desire to live; and desire to kill another pirate, all other factors equal. Lets assume these are smart pirates well versed in game theory, and that they don’t care at all about fairness.

It is easiest to see the equilibrium outcome by starting backward. If just P4 and P5 remain, then P4 will suggest he get all the coins, they will vote, tie, and P4 force an accept on the deal. So P3, come his turn, knows that P4 will vote against him. If he gives P5 1 coin, P5 will vote for him, since P4 wont give him anything. This leaves P3 with 99 coins. So P2 knows P3 will vote against him since he’s a round away from 99 coins. If he sends 1 coin P4’s way, P4 will vote for him, because P4 knows P3 can pass his deal and leave P4 nothing. At worst P2 gets a tie and can force his deal through. So P1 can be sure P2 will vote against him. P2’s deal leaves P3 and P5 with nothing, so if P1 gives P3 and P5 a coin each, they will both vote for him, and his deal passes. The equilibrium distribution is therefore [98, 0, 1, 0, 1]. Despite there being a group, rather than an individual, working against the deal proposer, and despite the fact the group can kill the proposer, the first pirate still ends up have most of the power.

It seems reasonable that fairness would be a more complicated issue in this game. Most of the pirates are leveraged against several times, and choosing to reject based on fairness can have more implications than just a payout of 0.  Included here is a link to a site that analyzes this game in the context of the bank robbery scene of The Dark Knight:

http://mindyourdecisions.com/blog/2008/08/19/game-theory-in-the-dark-knight-a-critical-review-of-the-opening-scene-spoilers/