I came across the following game:
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?