\begin{bmatrix}16&-2\\c&d\end{bmatrix}
Find values for $c$ and $b$ that will make this payoff matrix fair.
I'm new to game theory- would love someone to hear out my approach and let me know if I'm on the right track.
So In order for a game to be fair I know the expected value for the row and value of the game would have to be 0 so my intuition is to set up this equation. This equation represents the optimal strategy for the column player.
\begin{equation*} 16q-2(1-q)= 0 \end{equation*} From this I get $q=1/9$.
q is the probability for column 1 and 1-q is the probability for column 2
Next I tried to solve for $d$ using this equation \begin{equation*} -2q+d(1-q)= 0 \end{equation*} From this I got $d=1/4$. First can my value for a zero-sum game be a fraction? Also does this approach make sense. If I'm on the right track from here I would create an equation to solve for $p$ and then $c$.
Yes, this is a good approach to get $q$. You are using the fact that for each player at equilibrium their payoff is independent of what the other plays and to be fair it must be zero. Now you can write $\frac c9+\frac{8d}9=0, c+8d=0$
Your equation to get $d$ is not correct. The $q$ in it should be $p$, the chance the row player plays the first row. You can use the fact that the row player has to be indifferent to the column player's choice to write $-2q+d(1-p)=0=16p+c(1-p)$. Now we have three equations in three unknowns, $c,d,p$.