[_D&L_]

A nash equillibrium is simply where each player plays with the best response to his opponents strategy. Its an equillibrium, because no player has an incentive to deviate.


If you play rock, i play paper
If you play scissor, i play rock

goes around in circles....


but if you play 1/3 rock, 1/3 paper, 1/3 scissor
the best i can do is copy you.

Once each of us starts playing this strategy, neither of us has an incentive to deviate. Rock, paper, scissors is misleading because people may think Nash EQ means both players play the same strategy. This only occurs in Rock, Paper, Scissors because both players are acting simultaneously, with the same strategy set, and no assymetric information. They are "symmetric players" (my term) hence their strategies are symmetric.

Pocker is not so simple as rock, paper, scissors. There are rotating assymetries, assymetric information, and plenty of non-obvious dominated strategies (strategies that should never be played). Thus, even though the nash EQ breaks even with all other strategies in rock, paper, scissors, this does not happen in poker - far from it.

A nash EQ in a zero sum game will beat or break even with all other strategies (i mean after a full rotation at a table, their are positional advantages and a Nash EQ does not mean your just as strong in every position) . In poker, my own calcs is that a nash eq. will beat virtually all of the opposing strategy space; your chance of playing by luck a non-nash strategy that breaks even is virtual zero. You'll have to just take my word on that. I can prove it, but i'm not showing those proofs (or going to take the time to formally prove it).