<p>Consider the game of Battle of Sexes, where
<span class="math-container">$$\begin{matrix}
&amp;&amp;Women\\
 &amp;&amp; Football &amp; Resaurant\\
Men&amp;Football&amp;2,1 &amp; 0,0  \\
&amp;Restaurant&amp;0,0 &amp; 1,2
\end{matrix}$$</span></p>
<p>Then the Nash Equilibriums will be: <span class="math-container">$$NE_1=\{F,F\}\;\; NE_2 = \{R,R\}$$</span></p>
<p>However, if the game is like that:
<span class="math-container">$$\begin{matrix}
&amp;&amp;Women\\
 &amp;&amp; Football &amp; Resaurant\\
Men&amp;Football&amp;2,1 &amp; 1,1  \\
&amp;Restaurant&amp;1,1 &amp; 1,2
\end{matrix}$$</span>
Can we eliminate the choice of restaurant for men and choice of football for women, which makes game only solvable with mixed-strategy Nash Equilibrium. Since, from the perspective of men, if women chooses restaurant, he is indifferent between choosing restaurant and football, so he will basically choose football. So if the women knows the men to choose football, she will be indifferent between choosing football and restaurant. Thus they cannot reach a Nash Equilibrium. So in this case should we use mixed-strategy to find Nash Equilibrium or rather than eliminating choices, as in the previous case, should we assume that in case of men choose football because women will be indifferent, then it will stick to football as well?</p>

CC BY-SA 4.0
2023-11-17T20:02:54.777Z
"he is indifferent between choosing restaurant and football, so he will basically choose football." If he is indifferent, he can choose both with any probability.
Michael Greinecker
CC BY-SA 4.0
2023-11-18T18:10:46.880Z
In this case, can we eliminate the choice of restaurant for men and football for women, which leads to mixed strategy method? Or as in the first case, the pure strategy Nash Equilibriums will be same?
Tunay Sabri Yüksel
CC BY-SA 4.0
2023-11-18T19:43:08.307Z
There is a continuum of Nash equilibria. For example, every strategy profile in which one player goes to their "preferred" location and the other plays any mixed strategy is a Nash equilibrium,
Michael Greinecker
<game-theory><nash-equilibrium>
Mixed Strategy Nash Equilibrium
Tunay Sabri Yüksel
# Mixed Strategy Nash Equilibrium

(Asked by: username_0 on 2023-11-17)

Consider the game of Battle of Sexes, where
$$\begin{matrix}
&&Women\\
&& Football & Resaurant\\
Men&Football&2,1 & 0,0 \\
&Restaurant&0,0 & 1,2
\end{matrix}$$

Then the Nash Equilibriums will be: $$NE\_1=\{F,F\}\;\; NE\_2 = \{R,R\}$$

However, if the game is like that:
$$\begin{matrix}
&&Women\\
&& Football & Resaurant\\
Men&Football&2,1 & 1,1 \\
&Restaurant&1,1 & 1,2
\end{matrix}$$
Can we eliminate the choice of restaurant for men and choice of football for women, which makes game only solvable with mixed-strategy Nash Equilibrium. Since, from the perspective of men, if women chooses restaurant, he is indifferent between choosing restaurant and football, so he will basically choose football. So if the women knows the men to choose football, she will be indifferent between choosing football and restaurant. Thus they cannot reach a Nash Equilibrium. So in this case should we use mixed-strategy to find Nash Equilibrium or rather than eliminating choices, as in the previous case, should we assume that in case of men choose football because women will be indifferent, then it will stick to football as well?

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: "he is indifferent between choosing restaurant and football, so he will basically choose football." If he is indifferent, he can choose both with any probability.  
> **username_0**: In this case, can we eliminate the choice of restaurant for men and football for women, which leads to mixed strategy method? Or as in the first case, the pure strategy Nash Equilibriums will be same?  
> **username_1**: There is a continuum of Nash equilibria. For example, every strategy profile in which one player goes to their "preferred" location and the other plays any mixed strategy is a Nash equilibrium,  

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## 0 answers

