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Given a ($n$-player) symmetric game and two equilibriums $s_1,s_2$, is it true that if the support (the set of strategies with positive probabilities) of $s_1$ is identical to the support of $s_2$ then $\frac{s_1+s_2}{2}$ is also a symmetric equilibrium?
By $\frac{s_1+s_2}{2}$ I mean that a strategy $i$ will be selected by a player with a probability which equals the average of the probability of $i$ under $s_1$ and the probability of $i$ under $s_2$.
This seems obvious. If $\sigma_1,\sigma_2$ are in the joint support of $s_1,s_2$ and $\tau$ is not, then (writing $P$ for the first player's payoff) you need $$P(\tau,s)\le P(\sigma_1,s)=P(\sigma_2,s)$$
where $s$ is a convex combination of the $s_i$. But if you replace $s$ with $s_1$ or $s_2$, this holds, so it still holds after you average over $s_1$ and $s_2$.
What am I missing?
