How to compute a mixed Nash Equilibrium where only one payoff is given. Let's say I have this: 
$$
        \begin{matrix}
         & A & B \\
        X & 1 & 2 \\
        Y & 2 & 1 \\
        \end{matrix}
$$
That is the payoff for me if I make those moves where I am the row player. How would I compute the mixed Nash Equilibrium for me if I only know this? The method I know involves using the opponent's payoff to calculate so I'm not sure what to do here.
 A: If it is a zero-sum game, then the equilibrium is $(X,Y) = (0.5,0.5)$. That is, the row player should choose the rows with equal probability. Doing so will guarantee expected winnings of $\frac{1+2}2 = 1.5$ regardless of the column player's strategy. Similarly, the column player can guarantee that the expected winnings for the row player are no more than $1.5$ by choosing the columns with equal probability.
Suppose the row player does not adopt this strategy; say he or she chooses $X$ more often than $Y$. Then the column player will always play $A$, and as a result, the row player will win $1$ more often than $2$, and the expected winnings will be less than $1.5$. Analogous comments hold for the column player. 
If you want a more technical approach, suppose the row player plays $X$ with probability $p$ and $Y$ with probability $1-p$, and similarly the column player plays $A$ and $B$ with probabilities $q$ and $1-q$. Then the row player's expected winnings are:
$$\begin{align}w &= pq + 2(1-p)q + 2p(1-q) + (1-p)(1-q)\\
&= 1 + p + q - 2pq.
\end{align}$$
Taking partial derivatives,
$$\frac{\partial w}{\partial p} = 1 - 2q\\
\frac{\partial w}{\partial q} = 1 - 2p$$
which are equal to $0$ when $p = q = \frac12$.
