Variance when playing a game with a fair coin I am having a hard time with this question for some reason. 
You and a friend play a game where you each toss a balanced coin. If the upper faces on
the coins are both tails, you win \$1; if the faces are both heads, you win \$2; if the coins
do not match (one shows head and the other tail), you lose \$1. 
Calculate the expected value and standard deviation for your total winnings from this
game if you play 50 times.
PMF Values:
\begin{array}{c|c}
 $& p\\\hline
+$1  &  .25\\
+$2  & .25\\
-$1  & .50
\end{array}
I have calculated the expectation as $$1(.25)+2(.25)+(-1)(.5) = .25,$$ so $$E(50X) = 50\cdot.25 = \$12.5,$$ which I have confirmed is correct.
I know I need to get $\operatorname{Var}(50X)$, but doing a standard variance calculation and then using the formula $a^2\operatorname{Var}(X)$ is not giving me the correct value.
What step am I missing?

Variance is the mean of the squares minus the square of the mean.
$$
0.25\cdot1^2+0.25\cdot2^2+0.5\cdot(-1)^2-0.25^2=1.6875
$$
For independent events, the variance of a sum is the sum of the variances, so the variance for $50$ events is
$$
50\cdot1.6875=84.375
$$
