# Finding the mixed-strategy Nash equilibrium in this game

Find the pure and mixed-strategy Nash equilibrium in this game:

$$\begin{array}{c|cccc} P_1 \text{/} P_2 & \text{Ll} & \text{Lr} & \text{Rl} & \text{Rr} \\ \hline \text{T} & (3,2) & (3,2) & (1,1) & (1,1) \\ \text{B} & (4,3) & (2,4) & (4,3) & (2,4) \\ \end{array}$$

By using the definition of pure strategy Nash equilibria, I found these two Nash equilibria:

$$$$\text{Pure-strategy N.E.} = \begin{cases} ([T],[Lr]) \\ ([B],[Rr]) \end{cases}$$$$

For the mixed-strategy Nash equilibria, it appears that the answer is:

$$$$\text{Mixed-strategy N.E.} = \begin{cases} ([T], q[Ll] + (1-q)[Lr]) & \text{if } q < \frac{1}{2} \\ ([B], q[Lr] + (1-q)[Rr]) & \text{if } q < \frac{1}{2} \end{cases}$$$$

How would one derive the two mixed-strategy Nash equilibria?

First, note that there is a lot of weak domination: $$Lr \succeq Ll, ~ Lr \succeq Rl, ~ Rr \succeq Rl, Ll \succeq Rl.$$

Consequently, $$\nexists$$ MSNE in which both players mix at the same time. If $$P_1$$ mixed, then $$P_2$$ would strictly prefer to play a pure strategy.

Consider then, as the answer suggests, a candidate equilibrium in which $$P_1$$ plays a pure strategy, whilst $$P_2$$ mixes.

If $$P_1$$ plays $$T$$, then $$P_2$$ is indifferent between $$Ll$$ and $$Lr$$.

$$\implies$$ $$P_1 \to T, ~P_2 \to q(Ll) + (1 - q)Lr$$.
$$P_2$$ is indifferent by construction, so we need merely show that $$P_1$$ prefers $$T$$ to $$B$$: \begin{align} 3 & \geq 4q + 2(1 - q) \\ & \geq 2 + 2q \\ 1 &\geq 2q \implies 1/2 \geq q \end{align}

Likewise, if $$P_1 \to T$$, then it must be that $$P_2 \to q(Lr) + (1-q)Rr$$. Again, only need to show $$P_1$$ prefers $$B$$ to $$T$$: \begin{align} 2 & \geq 3q + 1(1 - q) \\ 1/2 & \geq q \end{align}

This fully describes the Nash Equilibrium.

• $Ll \succeq Lr$ is not true, it is the other way around? $Lr$ itself, however, is weakly dominant. May 17 at 10:10
• You're quite right, I'll edit. May 18 at 5:58
• @matthewoulton Thank you May 18 at 12:09