Determine the value(s) of a so that if the game is played many times it will be fair.
$\begin{bmatrix}16 & -2\\-8 & a\end{bmatrix}$
The game is in terms of winnings for row where + is a gain and - is a loss. I understand that a game is considered fair if the value of the game is 0. (No advantage for row or column). In order to achieve this value I was thinking of having a equal to 1? using the formula v=ad-bc/a-b-c+d I would obtain 0. Does this make sense?
How do I go about finding all possible values of a that will make the game fair?
Clearly if $a < 0$, the game won't be fair. If $a \ge 0$ there is no saddle point, so the optimal strategies are mixed. The column player's optimal strategy will have to make rows $1$ and $2$ have equal expected values, and that expected value is the value of the game. Thus if the column player's optimal mixed strategy chooses column $1$ with probability $p$ and $2$ with $1-p$, for the game to be fair we must have $$ \eqalign{16 p - 2 (1-p) &= 0\cr -8p + a(1-p) &= 0\cr}$$ From the first equation $p = 2/18 = 1/9$, and from the second $a = (8/9)/(8/9) = 1$.