I am trying to understand how to compute all Nash equilibria in a 2 player game, but I fail when there are more than 2 possible options to play. Could somebody explain to me how to calculate a matrix like this (without computer) \begin{matrix} 1,1 & 10,0 & -10,1 \\ 0,10 & 1,1 & 10,1 \\ 1,-10 & 1,10 & 1,1 \end{matrix}

I tried it with the support concept, but I don't get it...


A Nash equilibrium is a profile of strategies $(s_1,s_2)$ such that the strategies are best responses to each other, i.e., no player can do strictly better by deviating. This helps us to find the (pure strategy) Nash equilibria.

To start, we find the best response for player 1 for each of the strategies player 2 can play. I will demonstrate this by underlining the best responses: \begin{matrix} & A &B&C \\ A&\underline{1},1 & \underline{10},0 & -10,1 \\ B&0,10 & 1,1 & \underline{10},1 \\ C&\underline{1},-10 & 1,10 & 1,1 \end{matrix} Player 1 is the row player, player 2 is the column player. If 2 plays column A, then player 1's best response is to play either row $A$ or $C$, which gives him 1 rather than 0 as payoff. Similarly, the best response to column $B$ is row $A$, and to column $C$ it is row $B$.

Now we do the same for player 2 by underlining the best responses of the column player: \begin{matrix} & A &B&C \\ A&\underline{1},\underline{1} & \underline{10},0 & -10,\underline{1} \\ B&0,\underline{10} & 1,1 & \underline{10},1 \\ C&\underline{1},-10 & 1,\underline{10} & 1,1 \end{matrix} So, if player 1 plays row $A$ then player 2 best responds either with column $A$ or column $C$, giving him 1 rather than 0. We also find the best responses for row $B$ and $C$.

Now a pure strategy Nash equilibrium is a cell where both payoffs are underlined, i.e., where both strategies are best responses to each other. In the example, the unique pure strategy equilibrium is $(A,A)$. (There may also be mixed strategy equilibria.) In all other cells, at least one player has an incentive to deviate (because it gives him a higher payoff).

EDIT: How to compute mixed strategy equilibria in discrete games?

In a mixed Nash strategy equilibrium, each of the players must be indifferent between any of the pure strategies played with positive probability. If this were not the case, then there is a profitable deviation (play the pure strategy with higher payoff with higher probability).

Consider player 2. He plays column $A$ with probability $p$, $B$ with probability $q$, and $C$ with probability $1-p-q$. We need to find $p,q$ such that player 1 is indifferent between his pure strategies $A,B,C$. He is indifferent between row $A$ (left hand side) and row $B$ (right hand side) if $p,q$ are such that $$p+10q-10(1-q-p)=q+10(1-p-q).$$ He is indifferent between $B$ and $C$ if $$q+10(1-p-q)=p+q+1-q-p=1.$$ You just have to solve the first condition for $q$ as function of $p$, substitute $q$ in the second condition and you have $p$. Inserting $p$ again in the first gives you $q$.

Now we do the same with strategies for player 1 such that player 2 is indifferent. Player 1 plays $A$ with probability $x$, $B$ with probability $y$ and $C$ with probability $1-x-y$. The two conditions that follow are \begin{equation} 1x+10y-10(1-x-y)=x+10(1-x-y) \\ x+10(1-x-y)=1 \end{equation} Solve this again to find $x,y$. This is a mixed-strategy equilibrium, because neither player has a profitable deviation. Remember, we constructed the profile $(x,y;p,q)$ such that the other player is indifferent between his pure strategies. So, no matter how the other player unilaterally deviates, his expected payoff will be identical to that in equilibrium $(x,y;p,q)$. In general, depending on the solutions $x,y,p,q$, there may be infinitely many Nash equilibria, or none. The more pure strategies there are, the more tedious it is to compute mixed strategy equilibria, since we solve for $N-1$ variables for each player ($N$ being the number of pure strategies of the other player).  

  • $\begingroup$ Thank you very much for your answer, unfortunately my problem is not to find the pure strategy equilibriums, but the mixed. I am sorry that i was not clear enough in my question, I was looking for a method to compute all nash equilibriums, and i want to be sure after that i have all. Do you know any algorithm(that is calculable by hand) for it? $\endgroup$ – user106371 Nov 12 '13 at 19:06
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    $\begingroup$ I included an edit that described how to find the mixed strategy equilibria. $\endgroup$ – Nameless Nov 12 '13 at 19:34
  • $\begingroup$ Good answer! But why $(B, B)$ and $(C, C)$ are not a pure strategies Nash equilibrium ? $\endgroup$ – npisinp Apr 27 '14 at 18:16
  • $\begingroup$ Because if player 2 plays $B$, then player 1 wants to play $A$ (giving him 10) rather than $B$ (giving him 1). Similarly, if 1 plays $B$, player 2 wants to play $A$ rather than $B$. By looking at the underlined best responses, you find similar profitable deviations for $(C,C)$. $\endgroup$ – Nameless Apr 29 '14 at 19:18
  • $\begingroup$ Dear @Nameless, you wrote: ''We need to find $p,q$ such that player $1$ is indifferent between his pure strategies $A,B,C$.'' In this case, are you supposing that $A$, $B$ and $C$ are in the support of $s_1$? $\endgroup$ – rgm May 17 '17 at 21:41

I think I finally understood how to get all equilibrias by the support concept.

Given payoff matrix A for Player 1 and B for Player 2. That a point $(x,y)=(x_1,x_2,x_3,y_1,y_2,y_3)$ is a Nash equilibria with supp $x =I$ and supp $ y =J$ (which means that $x_i>0 $ for $i \in I$ and $ x_i=0 $ for $ i \not\in I$) is equivalent to

$ 1) \forall i,k \in I : (Ay)_i = (Ay)_k \\ 2) \forall i \in I,\forall k \not\in I : (Ay)_k ≤ (Ay)_i \\ 3) \forall j,l \in J : (Bx)_j = (Bx)_l \\ 4) \forall j \in J,\forall l \not\in J : (Bx)_l ≤ (Bx)_j $

So to get all Nash equilibrias, I have to check for every support combination that is possible, if all equalities and inequalities hold on. If yes the equations tell for what points or intervalls there are Nash equilibrias. In the example above the equations of Nameless are the equations for the case $I=\{1,2,3\}=J$. All in all there are to check 49 equation systems, which is a lot of work...

Thank you for your answer Nameless, it helped me a lot to understand the main principle.


@user106371. Your answer is exactly true. Assigning probabilities to each strategy and solving the system of equations is not a complete solution, only one case. In order to calculate number of equation systems need to be solved, you can use the following formula:

$$\sum_{k=1}^m \sum_{p=1}^n \binom{m}{k} \binom{n}{p}$$

where $m$ and $n$ are the number of strategies of the players.


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