This is exercise 3.13 from wackerly's book (7th ed.)
Problem:
Refer to the coin-tossing game in Exercise 3.2. Calculat the mean and variance of Y, your winnings on a single play of the game. How much should you pay to play this game if your net winnings, the difference between the payoff and cost of playing, are to be mean 0?
3.2:
You and a friend play a game where each toss a balanced coin. If the upper faces on the coins are both tails, you win $1$ dollar, if both heads you win $2$ dollars otherwise you lose $1$ dollar, -1. Give the probability distribution for your winnings of Y, on a single play of the game.
Probability distribution:
$$ \begin{array}{c|lcr} y & -1 & 1 & 2 \\ \hline p(y) & 1/2 & 1/4 & 1/4 \\ \end{array} $$
The mean is $ E(Y) = \sum_{y} yp(y) $ which gives -1/4 since this is an accurate representation of the probability frequency distribution.
Which means on average I'd lose $0.25 playing. So, how would I find the number of times to have a mean of zero?
Your calculation of the expected value is incorrect. It should be
$$E(Y)=-\frac{1}{2}+ \frac{1}{4}+\frac{2}{4}=\frac{1}{4}.$$
The question now asks what you should pay so that your expected winnings are expected to be zero. Well, without you paying anything, your expected winnings are $\frac{1}{4}$, so you should pay the same amount.
Can you calculate the variance?