<p>Your calculations are correct. If we take this game: </p>

<p>\begin{array}{| r | r |}
  \hline 
\hline                       
  &amp; A&amp; B\\
  \hline                      
A&amp;3,2&amp;1,1\\
B&amp;0,0&amp;2,3\\
  \hline 
\end{array}</p>

<p>then the mixed strategy probability used by the column player must make the row player indifferent: $3q+1(1-q)=0q+2(q-1)$, implying $q=1/4$.</p>

<hr>

<p>From your question, I infer that you have a solution claiming that the answer should be $q=2/5$. In that case, it looks to me like there has been a typo somewhere. Indeed, if we take the following game:</p>

<p>\begin{array}{| r | r |}
  \hline 
\hline                       
  &amp; A&amp; B\\
  \hline                      
A&amp;3,2&amp;0,0\\
B&amp;0,0&amp;2,3\\
  \hline 
\end{array}</p>

<p>We find the mixed strategy equilibrium implies the column player using probability $q$, with $3q=2(1-q)$ so $q=2/5$.</p>

<p>Note that this second, modified game is symmetric (as textbook battle of the sexes games usually are), further strengthening my suspicion that the confusion has been caused by a typo in the description of the game.</p>

CC BY-SA 3.0
2015-09-01T10:09:04.790Z
2015-09-01T10:09:04.790Z
Ubiquitous
<p>0
down vote
favorite
In the following game, the mixed strategy equilibrium is found as player 2 playing B 3 out of 5 times and A 2 out of 5 times but I am not able to get it. Please find the mistake in the following calculation: 
                         Player 2 
                    A                      B</p>

<pre><code>         A     3,2                    1,1 
</code></pre>

<p>Player 1</p>

<pre><code>         B     0,0                    2,3  
</code></pre>

<p>Payoff for player 1 if he plays A: 3q+1(1-q)  assuming player 2 plays A with probability q. And if he plays B payoff is 0q + 2(1-q)
For an equilibrium both payoffs should equal so 3q+1(1-q)=2(1-q) which gives q=1/4 i.e. probability of player 2 playing A which is not correct.</p>

<p>Thanks in advance.</p>

CC BY-SA 3.0
2015-08-31T23:23:44.037Z
Hi. Welcome to Economics.SE. This question seems to be a homework question (and requests descriptive detail without providing us any details of what you have tried) and we have a policy against providing help for homework questions **unless** the OP shows their work and articulates the piece of the problem with which they are having difficulty. Please consider attempting this on your own and including the details (in [Latex](http://meta.economics.stackexchange.com/questions/1417/mathjax-basic-tutorial-and-quick-reference)) here.
cc7768
<game-theory>
Calculating mixed strategy equilibrium in battle of sexes
Abhinav Arya
# Calculating mixed strategy equilibrium in battle of sexes

(Asked by: username_0 on 2015-08-31)

0
down vote
favorite
In the following game, the mixed strategy equilibrium is found as player 2 playing B 3 out of 5 times and A 2 out of 5 times but I am not able to get it. Please find the mistake in the following calculation: 
                         Player 2 
                    A                      B
         A     3,2                    1,1 

Player 1
         B     0,0                    2,3  

Payoff for player 1 if he plays A: 3q+1(1-q)  assuming player 2 plays A with probability q. And if he plays B payoff is 0q + 2(1-q)
For an equilibrium both payoffs should equal so 3q+1(1-q)=2(1-q) which gives q=1/4 i.e. probability of player 2 playing A which is not correct.
Thanks in advance.

**Tags:** game-theory
> **Comments:**
> **username_1**: Hi. Welcome to Economics.SE. This question seems to be a homework question (and requests descriptive detail without providing us any details of what you have tried) and we have a policy against providing help for homework questions **unless** the OP shows their work and articulates the piece of the problem with which they are having difficulty. Please consider attempting this on your own and including the details (in [Latex](http://meta.economics.stackexchange.com/questions/1417/mathjax-basic-tutorial-and-quick-reference)) here.  

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## 1 answers

### Answer by username_2 on 2015-09-01. Score: 4

Your calculations are correct. If we take this game:

\begin{array}{| r | r |}
\hline
\hline
& A& B\\
\hline
A&3,2&1,1\\
B&0,0&2,3\\
\hline
\end{array}

then the mixed strategy probability used by the column player must make the row player indifferent: $3q+1(1-q)=0q+2(q-1)$, implying $q=1/4$.

---

From your question, I infer that you have a solution claiming that the answer should be $q=2/5$. In that case, it looks to me like there has been a typo somewhere. Indeed, if we take the following game:

\begin{array}{| r | r |}
\hline
\hline
& A& B\\
\hline
A&3,2&0,0\\
B&0,0&2,3\\
\hline
\end{array}

We find the mixed strategy equilibrium implies the column player using probability $q$, with $3q=2(1-q)$ so $q=2/5$.

Note that this second, modified game is symmetric (as textbook battle of the sexes games usually are), further strengthening my suspicion that the confusion has been caused by a typo in the description of the game.


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