Nash equilibria
The game theory section is quite confused, which is amusing given how straightforward the analysis claims to be.

The article can't decide whether it's talking about a single-round Nash equilibrium, or the Nash equilibrium of a repeated game.  It talks about repetition, and a threatened "retaliation" the Predictor, but it describes an equilibrium which doesn't need retaliation or a repeated game at all.  The equilibrium {choose both boxes, predict both boxes} is quite stable for the single-round game, and is the only Nash equilibrium.

Both players choosing "one box" is also a Nash equilibrium for the repeated game.  However, when we have a repeated game, another Nash equilibrium is possible.  If the Predictor adopts a strategy "always choose one box unless the Chooser chose two boxes last round" then the Chooser can also always choose one box and have no incentive to deviate.  Deviating would net an extra 1,000 for that round, but would cap the maximum payoff at 1,000 in the next round, where it would have been 1,000,000.  Hence, the deviation is discouraged and the equilibrium is stable.

The real question is, is any of this germane to the article's topic?  If this section could be made concise and simple, I could see including it.  I don't see a way to make it concise and still correct unless you want to dismiss the possibility of playing repeatedly.  02:41, 12 Aug 2004 (UTC)