Sorry for my English, I am French but i couldn't find help on the French website (so I am here).
I have a question about this two-player game:
$$ \begin{array}{c|cc} & y_1 & y_2 \\ \hline x_1 & a,0 & a,0 \\ x_2 & 1,-1 & -1,1 \\ x_3 & -1,1 & 1,-1 \end{array} $$
When $a<0$ and $a>1$ it's easy because using domination we can find the Nash equilibrium (mixed), but when $0 \leq a \leq 1$ this is more difficult.
If $(p_1, p_2, 1-p_1-p_2)$ is the strategy of the player $1$ and $(q, 1-q)$ the strategy of the player $2$, by the indifference theorem for player $1$ I have $$ E(x_1) = E(x_2) = E(x_3) \iff a = 2q-1 = -2q+1 $$ and for player $2$ I have $$ E(y_1) = E(y_2) \iff -p_2 + (1-p_1-p_2) = p_2 - (1-p_1-p_2) $$ What is the Nash equilibrium (mixed and pure) of this game for $0 \leq a \leq 1$? I can't solve this equation...
For $a \gt 0$, try Player 1 playing $x_1$ all the time and Player 2 playing $y_1$ and $y_2$ with equal probability