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 Posted: Sun Dec 21, 2008 6:10 pm
First off I don't know if this is actually accepted as mathematics....so if it's not close enough to what's intended to be here, then by all means Admins, move it to wherever it's supposed to be.
Anyway, is it possible to create a solution with no Nash equilibrium? Where one strategy always beats another, sort of like rock-paper-scissors? And, if so, how would that be represented? I haven't done a lot of work in game theory, so all I know is the usual two person matrix...
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Posted: Wed Dec 31, 2008 11:03 am
Pick up a book on game theory. You'll run into tons of examples of games without pure Nash equilibria, like rock paper scissors. As far as I know, most games do have mixed Nash equilibria. Mixed equilibria occur when players choose some of their moves at random; the mixed Nash equilibria are then the equilibrium choices of probability distributions across moves, rather than of the moves themselves.
In the rock-paper-scissors example, there is no pure Nash equilibrium, since having any definitive strategy would make you too easy to beat. However, if both people choose to play each of rock, paper, and scissors at random, with each move having equal probability, you get is a mixed equilibrium. Given that the other player chooses this probability distribution, there is no choice of moves for you (including any other probability distribution) that allows you to win more often on average.
I hope that helped a bit, but seriously pick up a game theory book if you can. There are a ton out there that are pretty easy reads.
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