# Finding Nash's Equilibrium in mixed strategies [closed]

How to find the mixed strategy equilibrium in the following game:

                     P2

L            R

L      (3,1)         (0,1)
P1   M      (1,1)         (1,1)
R      (0,1)         (4,1)

• We can eliminate M for P1 but even without eliminating it we should be able to calculate the same mixed strategy equilibrium. How to do it ? Sep 27, 2015 at 19:13
• I'm voting to close this question as off-topic because it is a basic question and there is no effort shown at a solution. Because the site is not aimed at solving homework problems we try to avoid these questions. If you write down what you have tried so far the question may be reopened and you will probably get helpful feedback. Sep 27, 2015 at 20:32
• I have tried solving this question by eliminating the dominated strategy. For player P1, M is a strictly dominated strategy because if he randomizes between L and M with 0.5 probability associated with each then the expected payoffs for that mixed strategy will be (1.5,1) and (2,1) which are in any case better for P1 than what he gets by playing M. So, M can be eliminated. And then we get an infinite number of Nash's equilibria in the game. However, if I do not want to eliminate M beforehand but solve this game with this given structure only then I am unable to find the solution. Please help. Sep 27, 2015 at 20:39

As you have suggested in a comment, $M$ will not be played with positive probability in any Nash equilibrium. This is true for any strictly dominated action.
Given that $M$ has zero probability, the rest of your question is answered in another question of yours in another thread at How infinite Nash equilibria are possible in a game?