Consider the game below. Consider $p$ the probability of Player A playing U, and $(1-p)$ the probability of Player A playing D. Similarly, $q$ and $(1-q)$ for Player B playing L and R.

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My goal is to find the mixed-strategy Nash equilibrium. So, I compute the expected values and the respective $q$ and $p$ as below.

Player A

$E_{U} = q + (1-q)*3 = 3 - 2q$

$E_{D} = (1-q)*4 = 4 - 4q$

$E_{U} = E_{D}, q = 1/2$

Player B

$E_{L} = p + (1-p)*3 = 3 - 2p$

$E_{R} = 2p + (1-p)*4 = 4 - 2p$

$E_{L} = E_{R}, 3 - 2p = 4 - 2p, 3 = 4$, invalid solution!

So, I would conclude this game has no mixed-strategy Nash equilibrium. However, there is a claim saying: "Every finite game has a mixed strategy Nash equilibrium". I'm struggling with this impasse. Can anyone help me?


1 Answer 1


A pure strategy equilibrium is a mixed strategy equilibrium. In this game, R strictly dominates L, and there is a unique Nash equilibrium (D,R). The reason for the explicit mention of "mixed" in the statement of Nash's theorem is that there is not always a pure equilibrium so strict mixing can be necessary (e.g., in matching pennies).


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