The Fundamental Theorem of Matrix Games, and the "indifference" method of solving games In the following we will consider two-person zero-sum games. Let $A = (a_{ij})$ be the payoff-matrix of such a game. In this book the fundamental theorem of such games is states as:
Theorem: Given any $m \times n$ matrix
$$
 A = (a_{ij})
$$
there exists vectors $X = (x_1, \ldots, x_m), Y = (y_1, \ldots, y_n)$
with $y_i \ge 0, x_i \ge 0$ and $x_1 + \ldots x_m = y_1 + \ldots y_n = 1$ and a real number $v$ such that:


*

*$x_1 a_{1j} + \ldots + x_m a_{mj} \ge v$ for $j = 1,\ldots n$

*$a_{i1} y_1 + \ldots + a_{in} y_n \le v$ for $i = 1, \ldots m$.


The number $v$ is called the value of the game. The vectors $X$ and $Y$ are the optimal strategies for player 1 and player 2 respectively.
Lets consider player 1. A common method to find an optimal strategy for him is to find a probability distribution $X = (x_1, \ldots x_n)$ such that player 2 is indifferent regarding his pure strategies, meaning we are going to solve the equations
$$
 \sum_{i=1}^m x_i a_{i1} = \sum_{i=1}^m x_i a_{i2} = \ldots = \sum_{i=1}^m x_i a_{in}
$$
and $x_1 + \ldots + x_n = 1$, $x_i \ge 0$.
This is done for example in this question here on stackexchange.
Lets denote by $E(X,j) = \sum_{i=1}^m x_i a_{ij}$ the expectation for player 1 if player 2 plays his pure strategy $j$ (i.e. chooses the $j$-th column of the payoff-matrix).
Then as I see it the fundamental theorem stated above says that for the value $v$ of the game we have $E(X, j) \ge v$ for each pure strategy $j$ of player 2. But the above method of solving a game uses, I guess, a much stronger claim, namely that $$v = E(X, 1) = E(X,2) = \ldots = E(X,n)$$ if $X$ is optimal? (This is the condition we solve for). So here are my question.


*

*Am I right in my objection, and if not

*Can you give me an example of a game such that $E(X, j) > v$ for some optimal strategy $X$ of player 1 and some pure strategy $j$ of player 2 (i.e. not equality with the value for the pure strategy)?


Thank you!
 A: First: the players must be indifferent only between strategies that they assign positive probability. What is really hard about finding equilibria is to find which strategies receive positive prob.
Second: the method of equating payoffs (of strategies that have positive payoff) does not use $E[X,j]=v$ since (accordingly to the question) these are the payoffs of 1 given a pure strategy of 2. If you are going to look at the payoffs of 1 then you must consider the pure strategies of 1 given the mixed strategy of 2 -- maybe you want to call it $E[j,Y]$? Because this is a zero-sum game it really doesn't matter, see point forth. But it matters for general games.
Third: In a zero-sum (or constant) bi-matrix game: you will attain the value with equality.
Forth: Let $(X,Y)$ be a mixed strategy Nash eq. of zero-sum game then $u_2(X,s_j)=c$ for all $s_j$ such that $y_j>0$ and constant $c$. Since it is zero sum game $u_1(X,s_j)=-c$. So yes: the payoff of player 1 when 1 mixes accordingly to $X$  is constant for any strategy that 2 plays with positive prob.
Unrelated but of interest: In zero-sum games, equilibria are interchangeable: if $(X,Y)$ and $(W,Z)$ are equilibria then $(X,Z)$ and $(W,Y)$ are also equilibria. 
A: If I understand your question right, I think you can look at Rock, Paper, Scissors. v is going to be 0, which both players can achieve if they play each with probability 1/3.
Suppose instead Y chooses 1, 0, 0, and X anticipates this.
X plays 0, 1, 0, and his payoff is 1, and Y's payo0ff is -1.
We have to distinguish an equilibrium from an optimum. The equilibrium just means that we've found a point where neither player wants to switch strategies. The question is, how do we get to an equilibrium in the first place? Maybe we can eliminate dominated strategies and get there. But maybe not. Not in Rock, Paper, Scissors. In repeated plays, we might evolve there. But we don't get there just by calculating and saying, this is the equilibrium, therefore it is "optimal", because it's only optimal under assumptions about the opponent's strategy we may not be in a position to make.
