# Nash equilibrium: mixed strategy

I'm struggling with basic Nash equilibrium question. There is a game

I have to find Nash equilibria. I know that (a,w) and (c,y) are pure strategy Nash equilibria. However, I'm confused when looking at the answer about mixed strategy NE. I don't know how to calculate probabilities. I saw in the answer: player 2 plays w with probability p, and y with probability 1-p. It has to be that case that 3p=3(1-p). Therefore p=0.5. Player 2 playes with probability q and 1-q. Therefore 3q=2q+2(1-q). Could somebody explain me why we calculate the probabilities like that?

First, do iterated deletion of weakly dominated strategies. If a strategies is weakly dominated, there is some other strategy that gives the player at least as high a payoff in every scenario that might occur, so you might as well use the dominating strategy whenever you would have used the dominated one.

Second, choose the row player's strategy to make the column player indifferent over her remaining pure strategies, and choose the column player's strategy to make the row player indifferent over his remaining pure strategies. In order for a player to be willing to randomize, they must be indifferent, so their opponent must be randomizing to keep them uncertain; think about rock-paper-scissors, where you play 1/3-1/3-1/3. You're doing it to be unpredictable, to make your opponent uncertain of what to do, and they respond with a similar kind of randomness. That's the essence of a mixed strategy Nash eqm.

In this problem, there are no weakly dominated strategies, so there is a mixed Nash eqm where each player mixes over all four, but whatever.

If you restrict attention to the strategies a, w, c, and y, what happens? Well, column wants to make row indifferent, so if $$p$$ is the probability of $$w$$ and $$1-p$$ is the probability of $$y$$, row is indifferent between $$a$$ and $$c$$ if $$3p + (1-p)0 = 0p + (1-p)3$$ or $$3p = 3 - 3p$$ or $$p^* = 1/2$$.

For column to be indifferent between $$w$$ and $$y$$, row must pick $$q$$ so that $$3q + (1-q) 0 = 2(1-q) + 2q$$ or $$3q = 2$$ or $$q^* = 2/3$$.

Does that help, or is it computing the expected utilities that is the problem?

• Thank you so much for your deep explanation. Now I got it Jun 30, 2020 at 15:50