By: Shawn

Shawn — Tue, 22 Nov 2011 03:28:27 +0000

The way to find a mixed-strategy that “cannot be exploited” is to find your mixture of strategies that will make your opponent indifferent between his or her pure strategies. In symmetric games such as rock-paper-scissors, it’s easy to see that the mixed-strategy Nash equilibrium occurs when each player chooses to play each pure strategy with 1/3 probability. However, since both players have the symmetric payoffs, it’s not obvious that this is because Player A is choosing a mixture that would make Player B “break even” (at least in expectation) regardless of what strategy he or she chooses. If a game starts out symmetric but then becomes lopsided, as in the mismatched pennies game, it is the player whose payoffs did not change who should adjust their mixture in order to fall back into equilibrium. Mixed-strategies are definitely one of the most difficult concepts in game theory, but I think this particular principle at least can be thought of relatively intuitively. If I know that my opponent has the same expected payoff from both of his or her strategies, he or she will likely choose to play the pure strategy that gives the highest expected payoff from the strategy that I play more often. For example, if Player n gets 2 when he chooses left and I choose up, 2 when he chooses right and I choose down, and 0 otherwise, he’s going to always play left if I choose up more than half the time and he will always play right if I play down more than half the time. To get him not to have a pure strategy be a strict best response, I have to mix evenly between up and down. In other words, the expected payoff from playing either pure strategy against my mixture should be equal. That is what allows a mixture to be a best response (albeit only weakly).

Comments on: CP. The Prisoners Dilemma

By: Shawn