I am wondering for a two person zero sum symmetric game, is it true that, for any fixed mixed strategy $$p = [p_1, ... , p_n]^T$$ of player 1, is it true that $$max_q\space \pi_2(p,q)\geq0$$ where$$q=[q_1, ... , q_n]^T$$ denotes a mixed strategy for player 2, and$$max_q\space \pi_2(p,q)$$denotes the payoff function of player 2.
2026-05-14 06:18:05.1778739485
Game Theory: players payoff in symmetric zero sum game
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Yes. Note that if a zero-sum game $(A,B)=(A,-A)$ is symmetric, then $A$ is skew-symmetric:
$$A^\top = -A,$$
so $a_{ij} = -a_{ji}$ for all $i$ and $j$. Now fix $p$ as a mixed strategy of the row player as in your question and consider what the column player gets by playing $p$ too:
$$p^\top A p = \sum \sum p_i a_{ij} p_j = \sum \sum p_i (-a_{ji}) p_j = - - \sum \sum p_i a_{ji} p_j = -p^\top A p,$$
so $p^\top A p = 0$. Thus indeed we have that $\max_q \pi_2(p,q) \ge 0$.
BTW, actually a matrix $A$ is skew-symmetric if and only if $x^\top Ax = 0$ for all $x$.