J. Bradford DeLong December 1999 An extended passage from William Poundstone's (1992) marvelous
book
What Poundstone means is that, since both players know that
the supergame is going to last for 100 periods, there is no reason
for people to cooperate in round 100 to induce subsequent cooperation.
Hence--whatever else people do--the Nash equilibrium strategy
But once you know that the other player will defect in period
100 no matter what you do, the same argument applies to period
99: whatever else people do, the Nash equilibrium strategy Thus the situation "unravels." As long as there is a known, certain last period the only Nash equilibrium is to defect, always, from the first period. And real people don't do that--at least not unless they are John von Neumann or John Nash.
Alchian wound up with +40 Williams wound up with +63 The full 100-round C,C outcome is +50, +100; the full 100-round D,D outcome is 0, +50. So even though we identify with Williams--as the smart one, the one trying to induce cooperation, the one understnading that it was the two of them playing the umpire--nevertheless, Alchian "won" in that he got much closer to the total possible value of the game for his payoff matrix... |

I did indeed find the passage that you quoted from _Prisoners' Dilemma_ to be hilarious. All the more so because Armen Alchian was one of the players. One wonders what a game between, say, Milton Friedman and Paul Samuelson would've been like. Axelrod's computer tournaments were fascinating to read about, but it was even more fun to read about those human players.

Contributed by Mike Tamada <tamada@oxy.edu> on March 27, 2001

It would be interesting to know why Flood and Dresher chose the asymmetrical pay-off that they did, for this introduces an extraneous complication into the game. JW appears to be trying to optimise the joint pay-off, using a strategy somewhat akin to tit-for-tat. On the other hand, AA appears to be trying to maximise his comparative score against JW. A CD outcome puts AA 3 units behind JW, and occurs 7 times. However a CD outcome puts AA 2 units ahead (and is the only outcome that is better for AA than for JW). This occurs 18 times, including games 99 and 100, where it follows a long string of CC outcomes. It could be argued therefore that the two are actually playing different games, and indeed, in one sense, it could be argued that AA won the game that he was playing, even if it wasn't the game that JW was playing.

Of course the Nash equilibrium is still a poor outcome for both games, but Nash's analysis assumes that the players do not co-operate. When game is iterated, there is room for the players to test for co-operation, as has happened here.

There is plenty more on this subject at http://directory.google.com/Top/Computers/Artificial_Life/Iterated_Prisoner_Dilemma/.Contributed by Anonymous on January 1, 2001.

I thought that Flood and Dresher were trying to test John Nash's belief that the equilibrium should be invariant to monotonic transformations--that what mattered was the best-response nature of the equilibrium strategy, and not the values of alternative payoffs.

But I don't know. It's just a guess.

Contributed by Brad DeLong (delong@econ.berkeley.edu) on January 1, 2001.