I came across the following game:

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The question is to find potential equilibria in mixed and pure strategies. The solution says that there is an equilibrium in pure strategies (B,N), but none in mixed strategies. Mathematically this makes sense to me, since if you solve for the mixed strategy equilibrium, you get the solution that player 2 would have to play strategy L "-100%" of the time, in order to make player 1 indifferent between strategy A and strategy B. What I don't comprehend however, is how this result is to be reconciled with Nash's theorem, which states that every game with a finite number of players and a finite number of pure strategies has at least one equilibrium in mixed strategies. In the case at hand, we do in fact have a finite number of players and pure strategies. So how is it possible that there is no mixed strategy equilibrium?

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    $\begingroup$ There is no such theorem for proper mixed strategies. For example the prisoner's dilemma has a unique equilibrium in pure strategies. In your interpretation of Nash's theorem you have to interpret pure strategies as a degenerate form of mixed strategy where one strategy is played with probability 1 and all others with probability 0. $\endgroup$ – Bayesian Dec 23 '20 at 12:26

As I said in the comment, Nash's theorem shows the existence of a Nash equilibrium (possibly but not necessarily in mixed strategies). If you are interested in Nash equilibria is proper mixed strategies, i.e., NE in which all players play at least two actions with positive probability, you can easily show the following impossibility result: No such NE can exist if one player $i$ has a stricly dominant strategy, i.e., a strategy that gives a higher payoff regardless of the other players' strategy. Since the strictly dominant strategy is $i$'s unique best response against all strategies of the others, this pure strategy must be a part of each equilibrium. Each other strategy by $i$ can be improved upon by deviating to the dominant strategy. In your case 1 has a strictly dominant strategy, $B$ - so even if 2 always plays $L$, 1 would not be indifferent between $A$ and $B$, but always choose $B, 9>8$. Without indifference, mixing is never optimal and therefore never part of a NE.

Even if $B$ was only weakly dominant, it would always be the unique best response to every strategy in which player 2 plays $L$ and $N$ with positive probability.


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