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I am trying to solve the following problem.

Find the value of G, where G is a 2-person zero sum game described by an $n$ x $n $ square matrix A such that:

$ -a_{ii} = \sum_{j \neq i} a_{ij} \\ a_{ij} \geq 0 $

and $[A]_{ij} = a_{ij} \space \space i \neq j $

and $ [A]_{ii} = - a_{ii}$

That is, the sum of all diagonal entries are non-positive, the sum of all non-diagonal entries are non-negative and the sum of the entries in each row is zero where the diagonal entry is equal to the negative of the sum of all other entries in its row.

I believe that value is zero. I have verified this for $2$ x $2$ examples. Also, by giving each column equal weight player-2 can guarantee not paying more than $0$.

any help is welcome, thanks.

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by giving each column equal weight player-2 can guarantee not paying more than $0$.

Thus, in order to show that the value of the game is $0$, by the well-known theorem, it remains to show that Player 2 cannot guarantee not to pay less than $0$. Let in a mixed strategy Player 2 chooses column $i$ with probability $q_i$. Since $\sum q_i=1$, there exists $i$ with $q_i\le 1/n$. Then it is easy to check that when Player 1 chooses $i$-th row his expected gain is at least $0$.

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