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:

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

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

\begin{equation} \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} \end{equation}

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


1 Answer 1


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.

  • $\begingroup$ $Ll \succeq Lr$ is not true, it is the other way around? $Lr$ itself, however, is weakly dominant. $\endgroup$
    – Giskard
    Commented May 17 at 10:10
  • 1
    $\begingroup$ You're quite right, I'll edit. $\endgroup$ Commented May 18 at 5:58

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