nash equilibirum help! seems tricky

Question

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 & A & B & C& D \\ \hline A & (8,8)& (0,10) & (0,5) & (0,0) \\ \hline B&(10,0)& (1,1)& (-9,5)&(-9,0)\\ \hline C&(5,0)& (5,-9) & (-5,-5)& (-15,0)\\ \hline D&(0,0)& (0,-9)& (0,-15)&(-10,10)\\ \hline \end{array}

Answer 1

After eliminating D by weakly domination, observe that some mixture of A and B dominates C. It is the case. The rest is straightforward.

Answer 2

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 & B & C\\ \hline A & (0,10) & (0,5)\\ \hline C& (5,-9) & (-5,-5)\\ \hline D& (0,-9)& (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 & A & B & C\\ \hline A & (8,8)& (0,10) & (0,5)\\ \hline B&(10,0)& (1,1)& (-9,5)\\ \hline C&(5,0)& (5,-9) & (-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 & A & C\\ \hline A & (8,8)& (0,5)\\ \hline B&(10,0)&(-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 $C$.