I'm new to game theory, so I apologize if I get some terms wrong.
I have a zero-sum game with payoff matrix $A$ for the row player. Mixed strategies are allowed, and so I understand that there is an optimal strategy for each player at the Nash equilibrium. This optimal strategy has an expected payoff for the row player, which I will call $E(A)$.
Computing $A$ is computationally intensive, and so I would like to avoid it. I can more easily calculate $A_{max}$ and $A_{min}$, such that $A_{min_{ij}} \le A_{ij} \le A_{max_{ij}}$ for each $i,j$ in bounds of the payoff matrix. Intuitively, it seems to me that the expected payoff of the game $A$ must be between the expected payoffs of games $A_{min}$ and $A_{max}$, or in other words that $E(A_{min}) \le E(A) \le E(A_{max})$. If this is the case, I can skip games with poor payoffs. Is my intuition correct that $E(A_{min}) \le E(A) \le E(A_{max})$?
Yes. You can consider payoff as a random variable, where sample space are payoff matrix cells, probability of cell is probability of playing it in optimal strategies (=product of row and column probability), and value of this variable at given cell is simple value in the cell.
Now, general result - if you have random variable $\xi$ that is a.s. in interval $[x, y]$, then it's expectation is also in this interval.
For discrete case, it's really simple: $\mathbb E \xi = \sum_{i, j} p_{ij} A_{ij} \leq \sum_{i,j} p_{ij} A_{\max} = A_\max$, and same for $\min$.