<p>First, note that there is a <em>lot</em> of weak domination:
<span class="math-container">$$Lr \succeq Ll, ~ Lr \succeq Rl, ~ Rr \succeq Rl, Ll \succeq Rl.$$</span></p>
<p>Consequently, <span class="math-container">$\nexists$</span> MSNE in which both players mix at the same time. If <span class="math-container">$P_1$</span> mixed, then <span class="math-container">$P_2$</span> would strictly prefer to play a pure strategy.</p>
<p>Consider then, as the answer suggests, a candidate equilibrium in which <span class="math-container">$P_1$</span> plays a pure strategy, whilst <span class="math-container">$P_2$</span> mixes.</p>
<p>If <span class="math-container">$P_1$</span> plays <span class="math-container">$T$</span>, then <span class="math-container">$P_2$</span> is indifferent between <span class="math-container">$Ll$</span> and <span class="math-container">$Lr$</span>.</p>
<p><span class="math-container">$\implies$</span> <span class="math-container">$P_1 \to T, ~P_2 \to q(Ll) + (1 - q)Lr$</span>. <br />
<span class="math-container">$P_2$</span> is indifferent by construction, so we need merely show that <span class="math-container">$P_1$</span> prefers <span class="math-container">$T$</span> to <span class="math-container">$B$</span>:
<span class="math-container">$$\begin{align} 3 &amp; \geq 4q + 2(1 - q) \\
&amp; \geq 2 + 2q \\
1 &amp;\geq 2q \implies 1/2 \geq q
\end{align}$$</span></p>
<p>Likewise, if <span class="math-container">$P_1 \to T$</span>, then it must be that <span class="math-container">$P_2 \to q(Lr) + (1-q)Rr$</span>.
Again, only need to show <span class="math-container">$P_1$</span> prefers <span class="math-container">$B$</span> to <span class="math-container">$T$</span>:
<span class="math-container">$$\begin{align} 2 &amp; \geq 3q + 1(1 - q) \\
1/2 &amp; \geq q
\end{align}$$</span></p>
<p>This fully describes the Nash Equilibrium.</p>

CC BY-SA 4.0
2024-05-17T10:10:49.733Z
$Ll \succeq Lr$ is not true, it is the other way around? $Lr$ itself, however, is weakly dominant.
Giskard
CC BY-SA 4.0
2024-05-18T05:58:28.087Z
You're quite right, I'll edit.
matthewoulton
CC BY-SA 4.0
2024-05-17T09:55:01.153Z
2024-05-18T06:00:07.210Z
2024-05-18T06:00:07.210Z
matthewoulton
<p>Find the pure and mixed-strategy Nash equilibrium in this game:</p>
<p><span class="math-container">\begin{array}{c|cccc}
P_1 \text{/} P_2 &amp; \text{Ll} &amp; \text{Lr} &amp; \text{Rl} &amp; \text{Rr} \\
\hline
\text{T} &amp; (3,2) &amp; (3,2) &amp; (1,1) &amp; (1,1) \\
\text{B} &amp; (4,3) &amp; (2,4) &amp; (4,3) &amp; (2,4) \\
\end{array}</span></p>
<hr />
<p>By using the definition of pure strategy Nash equilibria, I found these two Nash equilibria:</p>
<p><span class="math-container">\begin{equation}
\text{Pure-strategy N.E.} = 
\begin{cases} 
([T],[Lr]) \\
([B],[Rr])
\end{cases}
\end{equation}</span></p>
<p>For the mixed-strategy Nash equilibria, it appears that the answer is:</p>
<p><span class="math-container">\begin{equation}
\text{Mixed-strategy N.E.} = 
\begin{cases} 
([T], q[Ll] + (1-q)[Lr]) &amp; \text{if } q &lt; \frac{1}{2} \\
([B], q[Lr] + (1-q)[Rr]) &amp; \text{if } q &lt; \frac{1}{2}
\end{cases}
\end{equation}</span></p>
<p>How would one derive the two mixed-strategy Nash equilibria?</p>

<game-theory><nash-equilibrium>
Finding the mixed-strategy Nash equilibrium in this game
user9487
# Finding the mixed-strategy Nash equilibrium in this game

(Asked by: username_0 on 2024-05-15)

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?

**Tags:** game-theory, nash-equilibrium


---

## 1 answers

### Answer by username_1 on 2024-05-17. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_2**: $Ll \succeq Lr$ is not true, it is the other way around? $Lr$ itself, however, is weakly dominant.  
> **username_1**: You're quite right, I'll edit.  

---

