Finding the Nash Equilibrium p in mixed strategies of a 2-player, symmetric zero-sum game with 3 pure strategies can be done by solving LP:
max $(0, 0, 0, 1)^{T}(p_1, \ p_2,\ p_3,\ \epsilon)$
s.t. $A \geq b$
Where A is:
\begin{bmatrix} a_{11} & a_{12} & a_{13} & -1 \\ a_{21} & a_{22} & a_{23} & -1 \\ a_{31} & a_{32} & a_{33} & -1 \\ \end{bmatrix}
and $b = (0,0,0)$
i.e. We maximize our worse payoff.
Now suppose player 2, who plays $q = (q_1,\ q_2,\ q_3)$ is under the constraint $q_1 + q_2 = c$, for some $c$ $0 \leq c \leq 1$.
Can this be solved? Thanks.