I'm stuck with understanding the way of finding mixed strategy Nash equilibrium for non-square matrices and want to explain my difficulties with the help of the following example.
Let the following table describe a strictly competitive game:
\begin{matrix}
& B_1 & B_2 \\
A_1 & 7 & -1 \\
A_2 & 5 & 4 \\
A_3 & 1 & 5 \\
A_4 & 3 & -2 \\
A_5 & 2 & 1
\end{matrix}
Matrix cells contain player $A$ payoffs $u_1(A_i, B_j)$ which are opposite to player $B$ payoffs $u_2(A_i, B_j)$ for each eligible pair $(i, j)$.
I have to find a Nash equilibrium for this game, so firstly I tried to find it in pure strategies. The following figure shows the best responses for player $A$: \begin{matrix} & B_1 & B_2 \\ A_1 & \underline{7} & -1 \\ A_2 & 5 & 4 \\ A_3 & 1 & \underline{5} \\ A_4 & 3 & -2 \\ A_5 & 2 & 1 \end{matrix}
and player $B$:
\begin{matrix}
& B_1 & B_2 \\
A_1 & 7 & -\underline{1} \\
A_2 & 5 & \underline{4} \\
A_3 & \underline{1} & 5 \\
A_4 & 3 & -\underline{2} \\
A_5 & 2 & \underline{1}
\end{matrix}
(remember that player $A$ tends to choose a maximum value from each column, while player $B$ - the minimum from each row).
As we have no profile (pair of some $A_i$ and $B_j$) which is the best response for both players, this problem has no solution in pure strategies (correct me if I'm wrong).
Thus, we have to look for some mixed strategy. First of all, I remove rows $4$ and $5$, as options $A_4$ and $A_5$ always give less payoff for player $A$ than the option $A_2$.
After that we have the following matrix which can't be simplified by the same way (removing majorated rows or minorated columns):
\begin{matrix}
& B_1 & B_2 \\
A_1 & 7 & -1 \\
A_2 & 5 & 4 \\
A_3 & 1 & 5 \\
\end{matrix}
Let $\alpha_i$ and $\beta_j$ be the probabilities of choosing option $A_i$ by player $A$ and option $B_j$ by player $B$ correspondently. There is a theorem that states:
Every action in the support of any player's equilibrium mixed strategy yields that player the same payoff.
For player $A$ it means:
$A_1$ payoff: $7\beta_1 - \beta_2$
$A_2$ payoff: $5\beta_1 + 4\beta_2$
$A_3$ payoff: $\beta_1 + 5\beta_2$
and all these expressions should be equal to each other.
Substituting $\beta_2 = 1 - \beta_1$, we got:
$$8\beta_1 - 1 = \beta_1 + 4 = 5 - 4\beta_1$$
This system obviously has no solutions, as equating the first expression to the second we get $\beta_1 = \frac{5}{7}$, the second to the third - $\beta_1 = \frac{1}{5}$, the first to the third - $\beta_1 = \frac{1}{2}$.
I wouldn't like to go on with my reasoning, as I wasn't able to find $\beta_1$ from these equations and so far I can't find a mixed equilibrium.
What am I doing wrong? Thanks in advance for help.