# Mixed Strategy Nash Equilibrium for finite Game

Consider the following Game, where player 1 has strategies $A,B$ and player 2 has strategies $X,Y,Z$: \begin{array}{c|c|c|c} & X & Y & Z \\ \hline A & 4,3 & 6,7 & 0,4 \\ \hline B & 5,5 & 5,-1 & -4,2 \end{array}

$Z$ is strictly dominated by the mixed strategy $\hat{\sigma}_2=\left(\frac{2}{3},\frac{1}{3},0\right)$, since \begin{align} v_2(Z,\sigma_1=(p,1-p)) &= 4p+2(1-p) = 2p+2 \\\ v_2\left(\sigma_2=\hat{\sigma}_2,\sigma_1=(p,1-p)\right) &= \frac{2}{3}\cdot p\cdot 3 + \frac{1}{3}\cdot p\cdot 7 + 0\cdot p\cdot 4 \\\ &\quad + \frac{2}{3}\cdot (1-p)\cdot 5 + \frac{1}{3}\cdot (1-p)\cdot (-1) + 0\cdot (1-p)\cdot 2 \\\ &= \frac{4}{3}p + 3 > 2p + 2 = v_2(Z,\sigma_1=(p,1-p)) \quad \text{for } 0\leq p\leq 1 \end{align}

Since this is a finite game, $Z$ can eliminated without "losing" any Nash Equilibria. So there should not be a NE, that contains a positive probability of playing $Z$.

However a calculation without the elimination yields:

\begin{align} v_2(X,\sigma_1=(p,1-p)) = v_{2,X} &= 3p+5(1-p) = 5-2p \\\ v_2(Y,\sigma_1=(p,1-p)) = v_{2,Y} &= 7p-(1-p) = 8p-1 \\\ v_2(Z,\sigma_1=(p,1-p)) = v_{2,Z} &= 4p+2(1-p) = 2p+2 \end{align} equating these: \begin{align} v_{2,X} = v_{2,Y} &\Rightarrow p=0.6 \\\ v_{2,X} = v_{2,Z} &\Rightarrow p=0.75 \\\ v_{2,Y} = v_{2,Z} &\Rightarrow p=0.5 \end{align} Since these three numbers are different from each other there cannot be a NE in which player 2 plays all three strategies with positive probability. Thus \begin{align} v_1(A,\sigma_2=(q,1-q,0)) &= v_{1,AXY} = 4q+6(1-q) = 6-2q \\\ v_1(B,\sigma_2=(q,1-q,0)) &= v_{1,BXY} = 5q+5(1-q) = 5 \\\ {} \\\ v_1(A,\sigma_2=(q,0,1-q)) &= v_{1,AXZ} = 4q+0(1-q) = 4q \\\ v_1(B,\sigma_2=(q,0,1-q)) &= v_{1,BXZ} = 5q-4(1-q) = 9q-4 \\\ {} \\\ v_1(A,\sigma_2=(0,q,1-q)) &= v_{1,AYZ} = 6q+0(1-q) = 6q \\\ v_1(B,\sigma_2=(0,q,1-q)) &= v_{1,BYZ} = 5q-4(1-q) = 9q-4 \end{align} are all possibilities. Equating yields: \begin{align} v_{1,AXY} &= v_{1,BXY} \Rightarrow q=0.5 \\\ v_{1,AXZ} &= v_{1,BXZ} \Rightarrow q=0.8 \\\ v_{1,AYZ} &= v_{1,BYZ} \Rightarrow q=1.5>1 \end{align} So the NE are given by \begin{align} \sigma^* = \left(\left(\sigma_1^*=\left(\frac{6}{10},\frac{4}{10}\right),\sigma_2^*=\left(\frac{1}{2},\frac{1}{2},0\right)\right),\left(\sigma_1^*=\left(\frac{3}{4},\frac{1}{4}\right),\sigma_2^*=\left(\frac{4}{5},0,\frac{1}{5}\right)\right)\right) \end{align}

But one of the equilibria contains $Z$ with positive probability!

Where is my mistake here?

Let $S_i$ be the set of pure and $\Delta S_i$ the set of mixed strategies of player $i$. If \begin{align} \sigma=(\sigma_i,\sigma_{-i})\in\Delta S:=\Delta S_1 \times\dots\times\Delta S_n \end{align} is a NE of a mixed extension of a game, then for all $s_i,s_i^\prime \in \text{support}(\sigma_i)$: \begin{align} v_i(s_i,\sigma_{-i}) = v_i(s_i^\prime,\sigma_{-i}) \quad (*). \end{align} Here $\text{support}(\sigma_i)=\{s_i\in S_i: \sigma_i(s_i)>0\}$ and $v_i$ is the expected payoff for player $i$.
Now the point is that I used Eq. $(*)$ to derive two strategy profiles. Unfortunately Eq. $(*)$ is a necessary but not sufficient condition. So what one has to do after finding these profiles is check whether they really only contain best responses. And that is simply not the case for the second profile:
One can easily see that playing $Y$ if player 1 plays $\left(\frac{3}{4},\frac{1}{4}\right)$ yields a payoff $v_2=5$ whereas playing $\left(\frac{4}{5},0,\frac{1}{5}\right)$ only yields a payoff of $v_2=3.5$.