Then the PMF for your winnings [tex]X[/tex] is
[tex]P(X=x)=\begin{cases}\frac12&\text{for }x=0\\\frac14&\text{for }x=5\\\frac3{13}&\text{for }x=10\\\frac1{52}&\text{for }x=30\\0&\text{otherwise}\end{cases}[/tex]
You can expect to win
[tex]E[X]=\displaystyle\sum_xxP(X=x)=\frac02+\frac54+\frac{10\cdot3}{13}+\frac{30}{52}=\frac{215}{52}[/tex]
or about $4.13 per game.
Recall that variance is
[tex]\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
where
[tex]E[X^2]=\displaystyle\sum_xx^2P(X=x)=\frac{0^2}2+\frac{5^2}4+\frac{10^2\cdot3}{13}+\frac{30^2}{52}=\frac{2425}{52}[/tex]
so that the variance of your winnings is
[tex]\mathrm{Var}[X]=\dfrac{2425}{52}-\left(\dfrac{215}{52}\right)^2=\dfrac{79875}{2704}[/tex]
or about 29.54 "squared" dollars. Standard deviation is the square root of variance, so the standard deviation of [tex]X[/tex] is approximately $5.44.
