Does introducing mixed strategy Nash Equilibria override the pure strategy Nash Equilibria?

Question

I am a bit confused as to why pure strategy Nash Equilibria are often used in analyzing a game alongside mixed strategy Nash Equilibria. While I understand that pure strategy Nash Equilibria can be thought of as special cases of mixed strategy Nash Equilibria, the pure strategy Nash Equilibria of a game typically do not pop out of the computation used to find mixed strategy Nash Equilibria. This leads me to believe that pure strategy Nash Equilibira are generally valid only when mixing is not allowed.

Is this correct? For example, in the Battle of the Sexes game, do the pure strategy Nash Equilibria go away when mixing is allowed, or should players continue to consider pure strategy options (assuming no alterations to the game such as randomness or burning money)? Should the player always prefer to used the Mixed Strategy Nash Equilibirum over the pure strategy Nash Equilibria as long as mixing is possible? Does this result generalize to all games?

Answer 1

No, this is merely an artifact of a method of calculating equilibria in mixed strategies.

Formally, a Nash equilibrium is defined in terms of inequalities. These inequalities state that the expected payoff of the (possibly pure, degenerate) equilibrium mixed strategy is at least as large as that of any other mixed strategy given, the mixed strategies of the other players.

This gives us infinitely many inequalities, but one can show that it is enough to check that the expected payoff of the equilibrium strategy is at least as high as any alternative pure strategy. The linearity of expected payoffs in probabilities implies that the set of (possibly pure) mixed best replies to a profile of strategies of the others is the set of mixed strategies that play pure best replies with positive probability.

If it is optimal to mix between two pure strategies, playing any of the two pure strategies is optimal too. Proper mixed best replies are never strict. This means that when we look at equilibria in which a player mixes two pure strategies, we have to specify the strategies of the other player so that both pure strategies give the same payoff. Here, we work with equalities instead of inequalities. But that is only so because we are looking for a player who actually mixes.

There are many settings in which mixed strategy equilibria are less reasonable than pure strategy equilibria. Consider the game of choosing on which side of the street to drive, left or right- modeled as a pure coordination game.

$$\begin{array} {|r|r|}\hline & L & R \\ \hline L & 1,1 & 0,0 \\ \hline R & 0,0 & 1,1 \\ \hline \end{array}$$

If everyone drives on the left, driving on the left is optimal. If everyone drives on the right, driving on the right is optimal. Frontal crashes into other cars are bad. There is also a mixed strategy equilibrium in which everyone randomizes with probability $1/2$ for both pure strategies. We would not expect that equilibrium to be a good predictor of what happens in reality (though the Harsanyi-Selten theory of equilibrium selection would actually choose it). Indeed, if we think of equilibria as the outcome of some learning process, we would expect only the equilibria in pure strategies to persist. If people are slightly more likely to drive on the left, driving on the left is optimal, and vice versa. In the long run, the behavior should converge to everyone driving on the same side.