<p>If you find the P1's utility of each pure strategy to P2 playing <span class="math-container">$(q_1,q_2,1-q_1-q_2)$</span>, then you have:
<span class="math-container">$$ u_1 (H,(q_1,q_2,1-q_1-q_2)) = 2q_1 - 2q_2, \quad u_1(T, (q_1,q_2,1-q_1-q_2)) = -2q_1 + 2q_2, \quad u_1(O, (q_1,q_2,1-q_1-q_2)) = 3q_1 + 3q_2 -2.$$</span>
Then setting all of these to be equal will give you <span class="math-container">$q_1 = q_2 = q_3 = \frac{1}{3}$</span>.</p>
<p>However, this doesn't give you all the NE. Notice that when P2 plays <span class="math-container">$O$</span>, P1 is indifferent between <span class="math-container">$H$</span> and <span class="math-container">$T$</span>. So you might have a Nash equilibrium where P1 plays <span class="math-container">$p_1 H + (1-p_1) T$</span> and P2 plays <span class="math-container">$O$</span>. We now to need to make sure that P2 playing <span class="math-container">$O$</span> is the best response to <span class="math-container">$p_1 H + (1-p_1) T$</span>.
<span class="math-container">$$ u_2 ((p_1,1-p_1,0),H) = 2-4p_1, \quad u_1((p_1,1-p_1,0),T) = 4p_1 - 2, \quad u_2((p_1,1-p_1,0),O) = 1.$$</span>
For <span class="math-container">$O$</span> to be the best response, we need <span class="math-container">$1 \ge 2-4p_1$</span> and <span class="math-container">$1 \ge 4p_1 - 2$</span>, which together gives you <span class="math-container">$p_1 \in [\frac{1}{4},\frac{3}{4}]$</span>.</p>
<p>So we have another set of equilibrium which look like <span class="math-container">$$\left\{ (p_1 H + (1-p_1)T,O), \frac{1}{4} \le p_1 \le \frac{3}{4} \right\}.$$</span></p>
<p>Apply the same logic to the other player to get <span class="math-container">$$\left\{ (O, p_2 H + (1-p_2)T), \frac{1}{4} \le p_2 \le \frac{3}{4} \right\}.$$</span></p>

CC BY-SA 4.0
2023-05-10T00:10:01.170Z
2023-05-10T00:10:01.170Z
Gary Liang
<p><span class="math-container">\begin{array}{|c|c|}\hline
&amp;H&amp;T&amp;O\\\hline
H&amp;2,-2&amp;-2,2&amp;0,1\\\hline
T&amp;-2,2&amp;2,-2&amp;0,1\\\hline
O&amp;1,0&amp;1,0&amp;-2,-2\\\hline
\end{array}</span></p>
<p>I want to find an unique Nash Equilibrium here. I know that there is pure nash equilibrium since <span class="math-container">$BR_1(H) = H$</span>, <span class="math-container">$BR_1(T) = T$</span>, <span class="math-container">$BR_1(O) = H$</span> or <span class="math-container">$T$</span>. Also, <span class="math-container">$BR_2(H) = T$</span>, <span class="math-container">$BR_2(T) = H$</span>, <span class="math-container">$BR_2(O) = H$</span> or <span class="math-container">$T$</span></p>
<p>To find a mixed strategy N.E., I can use the way that (<span class="math-container">$p_1, p_2, 1-p_1-p_2$</span>) and (<span class="math-container">$q_1, q_2, 1-q_1-q_2$</span>) and I will find <span class="math-container">$(1/3,1/3,1/3)$</span>. I would like to ask that is there easy way to find a mixed strategy N.E. here?</p>

CC BY-SA 4.0
2023-03-29T05:53:58.650Z
@HerrK. is there any dominated strategy here? because I could not find it
Steve Josh
CC BY-SA 4.0
2023-03-29T10:57:11.733Z
I guess A = H, B = T, O = C? Indifference of player 1 between H and T gives you $p_1=p_2$, then indifference between H and O gives you $p_2=p_3$. Analogously for player 2.
VARulle
CC BY-SA 4.0
2023-03-29T13:56:11.337Z
@SteveJosh: I was mistaken. Sorry.
Herr K.
<game-theory>
Mixed Strategy Game Theory
Steve Josh
# Mixed Strategy Game Theory

(Asked by: username_0 on 2023-03-28)

\begin{array}{|c|c|}\hline
&H&T&O\\\hline
H&2,-2&-2,2&0,1\\\hline
T&-2,2&2,-2&0,1\\\hline
O&1,0&1,0&-2,-2\\\hline
\end{array}

I want to find an unique Nash Equilibrium here. I know that there is pure nash equilibrium since $BR\_1(H) = H$, $BR\_1(T) = T$, $BR\_1(O) = H$ or $T$. Also, $BR\_2(H) = T$, $BR\_2(T) = H$, $BR\_2(O) = H$ or $T$

To find a mixed strategy N.E., I can use the way that ($p\_1, p\_2, 1-p\_1-p\_2$) and ($q\_1, q\_2, 1-q\_1-q\_2$) and I will find $(1/3,1/3,1/3)$. I would like to ask that is there easy way to find a mixed strategy N.E. here?

**Tags:** game-theory
> **Comments:**
> **username_0**: @HerrK. is there any dominated strategy here? because I could not find it  
> **username_1**: I guess A = H, B = T, O = C? Indifference of player 1 between H and T gives you $p_1=p_2$, then indifference between H and O gives you $p_2=p_3$. Analogously for player 2.  
> **username_2**: @SteveJosh: I was mistaken. Sorry.  

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## 1 answers

### Answer by username_3 on 2023-05-10. Score: 1

If you find the P1's utility of each pure strategy to P2 playing $(q\_1,q\_2,1-q\_1-q\_2)$, then you have:
$$ u\_1 (H,(q\_1,q\_2,1-q\_1-q\_2)) = 2q\_1 - 2q\_2, \quad u\_1(T, (q\_1,q\_2,1-q\_1-q\_2)) = -2q\_1 + 2q\_2, \quad u\_1(O, (q\_1,q\_2,1-q\_1-q\_2)) = 3q\_1 + 3q\_2 -2.$$
Then setting all of these to be equal will give you $q\_1 = q\_2 = q\_3 = \frac{1}{3}$.

However, this doesn't give you all the NE. Notice that when P2 plays $O$, P1 is indifferent between $H$ and $T$. So you might have a Nash equilibrium where P1 plays $p\_1 H + (1-p\_1) T$ and P2 plays $O$. We now to need to make sure that P2 playing $O$ is the best response to $p\_1 H + (1-p\_1) T$.
$$ u\_2 ((p\_1,1-p\_1,0),H) = 2-4p\_1, \quad u\_1((p\_1,1-p\_1,0),T) = 4p\_1 - 2, \quad u\_2((p\_1,1-p\_1,0),O) = 1.$$
For $O$ to be the best response, we need $1 \ge 2-4p\_1$ and $1 \ge 4p\_1 - 2$, which together gives you $p\_1 \in [\frac{1}{4},\frac{3}{4}]$.

So we have another set of equilibrium which look like $$\left\{ (p\_1 H + (1-p\_1)T,O), \frac{1}{4} \le p\_1 \le \frac{3}{4} \right\}.$$

Apply the same logic to the other player to get $$\left\{ (O, p\_2 H + (1-p\_2)T), \frac{1}{4} \le p\_2 \le \frac{3}{4} \right\}.$$


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