Consider a static zero sum game with two players and a payoff matrix $U$ such that $\sum_{i\neq j} u_{i,j} =u_{i,i}$ for all $i \in \{1,...,n\}$. Player 1 and 2 choose numbers from 1 to $n$. Player 1 pays $u_{i,i}$ if both choose the same number $i$ and otherwise, if player 1 chooses $i$ and player 2 chooses $j$ then player 1 earns $u_{i,j}$. I need to compute the value of the game (or the expected payoff of the Nash equilibrium under mixed strategies, which is the same if I understand correctly).
Clearly if both choose to equally randomize (choose each number with probability $1/n$) then the expected payoff is $0$. Using the provided description of the entries of the matrix I can verify here player 1 doesn't has reason to unilaterally deviate, i.e. he always get $0$ as his expected payoff whichever strategy he chooses.
But as far as I understand to conclude that the value is $0$ it also needed to prove that player 2 also hasn't incentives to deviate, but I can't prove this. Can someone share some comments or hints? Thanks in advance!
Suppose that Player 1 equally randomises. Then if Player 2 plays $j \in \{1,\dots,n\}$, there is a $1/n$ chance of getting $u_{j,j}$, and then for each $j \ne i$, there is a $1/n$ chance of losing $u_{i,j}$. So Player 2's expected payoff is: $$ \frac{u_{j,j}}{n} - \sum_{i \ne j} \frac{ u_{i,j}}{n} = 0.$$ In other words, the expected payoff for every $j$ is $0$, so mixing is a best response to Player 1 equally randomising. In particular, Player 2 can also equally randomise.