<p>In 2x2 games (where there are two players and each has two strategies), this is the way to find mixed strategies. So called 'generic' 2x2 games have at most one mixed equilibrium, while non-generic games like</p>
<pre><code>                    Player 2
                   A       B 
Player 1     1   (0,0)   (0,0)
             2   (0,0)   (0,0) 
</code></pre>
<p>have infinitely many. (In the above game any strategy profile is a mixed equilibrium.) Your payoff equivalence equations still hold.</p>
<p>In two player games where the players may have more than two strategies (matrix games with size <span class="math-container">$m \times n$</span>, the number of Nash-equilibria may be more than one but still finite. If you assume that the equilibrium is <em>completely mixed</em>, that is, all strategies are played with positive probability, the expected payoffs of a player are still the same for each pure strategy.</p>
<p>There is no easy algorithm for finding all Nash-equilibria; one can assume that some strategies are played with 0 probability and see if an equilibrium is possible with this assumption.</p>

CC BY-SA 4.0
2022-11-08T10:48:04.080Z
2022-11-08T10:48:04.080Z
Giskard
<p>So I have been taught how to find a single mixed strategy Nash equilibrium in a 2 player game by ensuring both players are indifferent to which strategy is played.</p>
<p>So for example:</p>
<pre><code>                    Player 2
                   x      1-x
                   A       B 
Player 1     1   (1,0)   (0,1)
             2   (0,0)   (3,3)
</code></pre>
<p>Where <span class="math-container">$x$</span> is probability of playing strategy A</p>
<p>Then for player 1 we would try find the <span class="math-container">$EU_i(X)$</span> (expected utility of player i playing strategy X):</p>
<p><span class="math-container">$EU_1(1) = x(1) + (1-x)0$</span></p>
<p><span class="math-container">$EU_1(2) = x(0) + (1-x)3$</span></p>
<p>Then want:</p>
<p><span class="math-container">$EU_1(1) = EU_1(2)$</span></p>
<p><span class="math-container">$x = 3-3x$</span></p>
<p><span class="math-container">$x = 3/4$</span></p>
<p>I was wondering how you go about finding multiple Mixed Strategy Nash Equilibria if they exist as well as showing that you have found all the existing ones? I can't seem to find many resources online.</p>
<p>I've only been studying game theory for a few weeks so please bear that in mind :)</p>

<game-theory><nash-equilibrium>
How to find all mixed strategy Nash equilibrium
pk00
# How to find all mixed strategy Nash equilibrium

(Asked by: username_0 on 2022-11-07)

So I have been taught how to find a single mixed strategy Nash equilibrium in a 2 player game by ensuring both players are indifferent to which strategy is played.
So for example:
                    Player 2
                   x      1-x
                   A       B 
Player 1     1   (1,0)   (0,1)
             2   (0,0)   (3,3)

Where $x$ is probability of playing strategy A
Then for player 1 we would try find the $EU_i(X)$ (expected utility of player i playing strategy X):
$EU_1(1) = x(1) + (1-x)0$
$EU_1(2) = x(0) + (1-x)3$
Then want:
$EU_1(1) = EU_1(2)$
$x = 3-3x$
$x = 3/4$
I was wondering how you go about finding multiple Mixed Strategy Nash Equilibria if they exist as well as showing that you have found all the existing ones? I can't seem to find many resources online.
I've only been studying game theory for a few weeks so please bear that in mind :)

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

---

## 1 answers

### Answer by username_1 on 2022-11-08. Score: 1 | Accepted answer

In 2x2 games (where there are two players and each has two strategies), this is the way to find mixed strategies. So called 'generic' 2x2 games have at most one mixed equilibrium, while non-generic games like
                    Player 2
                   A       B 
Player 1     1   (0,0)   (0,0)
             2   (0,0)   (0,0) 

have infinitely many. (In the above game any strategy profile is a mixed equilibrium.) Your payoff equivalence equations still hold.
In two player games where the players may have more than two strategies (matrix games with size $m \times n$, the number of Nash-equilibria may be more than one but still finite. If you assume that the equilibrium is completely mixed, that is, all strategies are played with positive probability, the expected payoffs of a player are still the same for each pure strategy.
There is no easy algorithm for finding all Nash-equilibria; one can assume that some strategies are played with 0 probability and see if an equilibrium is possible with this assumption.


---

