# Nash equilibrium for 2 players' game

Consider a game with two players P1 and P2. For P1 the set of strategies is $x_1,...,x_m$ and $y_1,...,y_n$ for P2, gains are $f_1(x_i,y_j)$ for P1 and $f_2(x_i,y_j)$ for P2. Define mixed strategies by $\mathbf{p}_1$ and $\mathbf{p}_2$ where $$\mathbf{p}_1 = (p_{11},...,p_{1m})\quad\text{and}\quad \mathbf{p}_2 = (p_{21},...,p_{2n}) .$$

Then expected gains are given by $$g_1(\mathbf{p}_1,\mathbf{p}_2) = \sum\limits_{i=1}^m\sum\limits_{j=1}^n f_1(x_i,y_j)p_{1i}\,p_{2j}$$ and $$g_2(\mathbf{p}_1,\mathbf{p}_2) = \sum\limits_{i=1}^m\sum\limits_{j=1}^n f_2(x_i,y_j)p_{1i}\,p_{2j}.$$

If $\displaystyle{\frac{\partial g_1}{\partial p_{1i}}>0}$ then by increasing $p_{1i}$ we increase $g_1$, so in Nash equilibrium have to hold $$\frac{\partial g_1}{\partial p_{1i}} = 0\text{ and }\frac{\partial g_2}{\partial p_{2j}} = 0$$ for all $i = 1,...,m$ and $j = 1,...,n$. These are systems of linear equations. The first system contains $m$ equations on $n$ variables and the second contains $n$ equations on $m$ variables. It can be the case that there is no solution if $m\neq n$, but Nash equilibrium has to exist.

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But the functions $g_1$ and $g_2$ are not a priori known. – GWu Apr 13 '11 at 14:40
@GWu: Why not? The gains are known (they define the game), and the probabilities are the variables -- in what sense are the functions not known? – joriki Apr 13 '11 at 14:45
What do you mean? they are given as expectations. – Ilya Apr 13 '11 at 14:45

You didn't take into account the constraints: $\sum p_{1i}=\sum p_{2j}=1$ and $p_{1i},p_{2j}>0$. If there's no equilibrium in the interior, it has to occur on the boundary.
Did you mean that the sum is equal to $1$? – Ilya Apr 13 '11 at 14:46
But at the time e.g. only one $p_{1k}$ can be $1$. Can we use it? – Ilya Apr 13 '11 at 14:56