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 :)


1 Answer 1


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.


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