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.
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?
Best Answer
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).