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:
$$SW(s_1,s_2) = \sum _{i\in [n]}\sum_{j\in [n]}SW(i,j)\Pr_{s_1}(i)\Pr_{s_2}(j)$$


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 $SW(s')\leq SW(s)$?**

> - **If the statement is true, does it extend to symmetric games with any number of players?**

The intuition is that one may always resort to playing the symmetric equilibrium, hence if the other player would make more in asymmetric equilibrium it has to benefit them both.

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For example, consider the following simple game, for some $x>0$:

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

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

$A^t=
 \left( \begin{array}{ccc}
0 & x \\
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 $1+x$.

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