Consider a 2-player symmetric game given by a payoff matrix $A\in [0,1]^{n,n}$ for the row player (i.e. the column player matrix is $A^t$).

Define the social welfare as the sum of payoffs for both players, i.e. $$SW(i,j)=A(i,j)+A(j,i)$$

Define the social-welfare of a (possibly mixed) equilibrium in a straight forward manner: $$SW(s_1,s_2) = \sum _{i\in [n]}\sum_{j\in [m]}SW(i,j)\Pr_{s_1}(i)\Pr_{s_2}(j)$$

Define $Sup(s)=Sup(s_1)\cup Sup(s_2)$, where $Sup(s_i)$ is the set of strategies in $s_i$ that has a strictly positive probability.

Let $s=<s_1,s_2>$ be an equilibrium for the game, and let $<s',s'>$ be a symmetric equilibrium. Is it true that $Sup(s)=Sup(s')$ imply $SW(s')\leq SW(s)$?

For example, consider the following simple game:

$A= \left( \begin{array}{ccc} 0 & 1 \\ 0.5 & 0 \\ \end{array} \right) $

And the column player profit, given by $A^t$ is:

$A^t= \left( \begin{array}{ccc} 0 & 0.5 \\ 1 & 0 \\ \end{array} \right) $

There exists a asymmetric equilibrium $s$ where some player plays strategy $a$ and the other plays $b$. This gives a social welfare of 1.5.

The symmetric equilibrium $s'$ is reached when both play strategy $a$ with probability $p_a=\frac{2}{3}$ and $b$ otherwise, giving a social welfare of: $$SW(s')=1\cdot(2p_ap_b)+0.5(2p_ap_b)=\frac{2}{3}<SW(s)$$