nash equilibirum help! seems tricky Any advice for finding all nash equilibrium for this symmetric game? (B,B) looks like one but I feel like there are more. I tried looking for strictly dominant strategies, but only A weakly dominates D.
\begin{array}{|c|c|c|c|c|}
\hline &amp; A &amp; B &amp; C&amp; D \\
\hline  A &amp; (8,8)&amp; (0,10) &amp; (0,5) &amp; (0,0) \\ 
\hline B&amp;(10,0)&amp; (1,1)&amp; (-9,5)&amp;(-9,0)\\ 
\hline C&amp;(5,0)&amp; (5,-9) &amp; (-5,-5)&amp; (-15,0)\\ 
\hline D&amp;(0,0)&amp; (0,-9)&amp; (0,-15)&amp;(-10,10)\\ 
\hline 
\end{array}
As CommonerG noted, $D$ is worse than $A$ for the row player unless the column player mixes only $B$ and $C$.

So first assume that the column player mixes only $B$ and $C$. Then $C$ strictly dominates $B$ for the row player. That leaves us with

\begin{array}{|c|c|c|}
\hline &amp; B &amp; C\\
\hline  A &amp; (0,10) &amp; (0,5)\\ 
\hline C&amp; (5,-9) &amp; (-5,-5)\\ 
\hline D&amp; (0,-9)&amp; (0,-15)\\ 
\hline 
\end{array}

If the column player puts higher weight on $B$, the row player must choose $C$, but then the column player would choose $C$. On the other hand, if the column player puts higher weight on $C$, the row player must choose a mixture of $A$ and $D$, but then the column player would choose $B$. Thus there is no Nash equilibrium in this case.

It follows that neither player mixes only $B$ and $C$, and hence $D$ can be eliminated, leaving us with

\begin{array}{|c|c|c|c|}
\hline &amp; A &amp; B &amp; C\\
\hline  A &amp; (8,8)&amp; (0,10) &amp; (0,5)\\ 
\hline B&amp;(10,0)&amp; (1,1)&amp; (-9,5)\\ 
\hline C&amp;(5,0)&amp; (5,-9) &amp; (-5,-5)\\ 
\hline 
\end{array}

If the column player mixes only $A$ and $B$, then $B$ strictly dominates $A$ for the row player, who thus mixes only $B$ and $C$. Since we've excluded that, it follows that neither player mixes only $A$ and $B$.

If the column player mixes only $A$ and $C$, then $A$ strictly dominates $C$ for the row player, leaving us with

\begin{array}{|c|c|c|}
\hline &amp; A &amp; C\\
\hline  A &amp; (8,8)&amp; (0,5)\\ 
\hline B&amp;(10,0)&amp;(-9,5)\\ 
\hline 
\end{array}

If the row player puts any weight on $A$, the column player chooses $A$, but then the row player must choose $B$. On the other hand, if the row player chooses $B$, the column player chooses $C$, but then the row player must choose $A$. Thus there is no Nash equilibrium in this case.

Now we're left only with the case where both players mix all strategies $A$, $B$ and $C$. That implies that both players can make all three strategies yield the same payoff for the other player, yielding a system of linear equations: $8w_1+0w_2+0w_3=10w_1+1w_2-9w_3=5w_1+5w_2-5w_3$ with $w_1+w_2+w_3=1$.

The solution can be obtained by producing two homogeneous equations by subtraction, e.g. $2w_1+1w_2-9w_3=0$ and $-3w_1+5w_2-5w_3=0$, taking the cross product to find $w\propto(40,37,13)$ and normalizing to find that the only Nash equilibrium is the one in which both players apply the mixed strategy with weights $4/9$, $37/90$ and $13/90$ for strategies $A$, $B$ and $C$, respectively. The value of the game for both players is $8w_1=32/9$.

Note that both players playing $B$ is not a Nash equilibrium, since both players would switch to $C$.
After eliminating D by weakly domination, observe that some mixture of A and B dominates C. It is the case.
The rest is straightforward.