Consider an arbitrary 2x2 simultaneous game with complete information. Say that the model has only one pure-strategy Nash equilibrium. For example (first pay-off refers to Player 1):
Player 2 +---+-------+-------+ | | A | B | +---+-------+-------+ Player 1 | A | (1,2) | (2,1) | | B | (3,3) | (1,1) | +---+-------+-------+
Here, (3,3) is the only pure-strategy Nash equilibrium. The reason there is just one is, apparently, because one of the players have a dominant strategy (Player 2 always prefers A). If we change this, we would get two pure-strategy equilibria.
Abstracting from the fact that the pure-strategy equilibrium is also a mixed strategy one with probability 0/100%, does any simple game with just one pure-strategy equilibrium have no mixed strategy equilibrium? Is there a formal proof of this?
I educated guess is that there is no mixed strategy equilibrium. The "proof" I can think of is a best response plot. Basically, one player has a dominant strategy which means that for any probability, her response is a vertical/horizontal line at 0 or 1 (depending how probability is defined). Thus, regardless of how the other player's best response line looks like (i.e. where the indifference probability is), it will only cross the other player's line in one point. That point is the pure-strategy equilibrium.
PS: the adjective "simple" is to avoid more complex game scenarios like repetition, cooperation, incomplete information, etc.