# One pure Nash equilibrium, no mixed equilibrium?

I'm given a 4x4 payoff matrix

$$\begin{bmatrix}(0,7)&(2,5)&(7,0)&(0,1)\\(5,2)&(3,3)&(5,2)&(0,1)\\(7,0)&(2,5)&(0,7)&(0,1)\\(0,0)&(0,-2)&(0,0)&(10,-1)\end{bmatrix}$$

There isn't any strategy that is dominated for both players. As a combination of best responses, I see that (3,3) is a pure strategy Nash equilibrium. I tried to compute mixed strategy Nash equilibrium by setting probabilities of row player's actions as p,q,r,s and equating the payoffs for the column player, which yields p= -1/3

Does this mean there is no mixed strategy Nash equilibrium? Thus only one Nash equilibrium exists?

• It only means that there is no equilibrium in which the row player plays each strategy with a positive probability. – Michael Greinecker Oct 7 '20 at 9:57
• So if I limit the strategies, there can be mixed equilibriums? But then how do I know which strategy to discard? – ph8ndstne Oct 7 '20 at 10:01
• In principle, you have to do work through every subset of strategies for each player. But here the problem is a bit simpler. The fourth strategy of the column player, for example, is strictly dominated by a mixed strategy. – Michael Greinecker Oct 7 '20 at 10:07
• Oh I understand now. So to reduce a strategy, I need to find distributions of other strategies that dominate it right? – ph8ndstne Oct 7 '20 at 10:09
• If that is possible, the strategy will certainly never be played with positive probability. So you can discard the strategy when looking for Nash equilibria. – Michael Greinecker Oct 7 '20 at 10:11