Question

You draw a card from a deck. If you get a red card, you win nothing. If you got a spade, you win $5. For any club, you win $10 plus an extra $20 for the ace of clubs. Define random variable X as the amount you win at one game.a) What is the pmf of X?b) Find expected value and standard deviation of X.

Answer 1

• Probability of drawing a red card = 26/52 = 1/2
• Probability of drawing a spade = 13/52 = 1/4
• Probability of drawing a non-ace club = 12/52 = 3/13
• Probability of drawing the ace of clubs = 1/52

Then the PMF for your winnings $X$ is

$P(X=x)=\begin{cases}\frac12&\text{for }x=0\\\frac14&\text{for }x=5\\\frac3{13}&\text{for }x=10\\\frac1{52}&\text{for }x=30\\0&\text{otherwise}\end{cases}$

You can expect to win

$E[X]=\displaystyle\sum_xxP(X=x)=\frac02+\frac54+\frac{10\cdot3}{13}+\frac{30}{52}=\frac{215}{52}$

or about $4.13 per game.

Recall that variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\sum_xx^2P(X=x)=\frac{0^2}2+\frac{5^2}4+\frac{10^2\cdot3}{13}+\frac{30^2}{52}=\frac{2425}{52}$

so that the variance of your winnings is

$\mathrm{Var}[X]=\dfrac{2425}{52}-\left(\dfrac{215}{52}\right)^2=\dfrac{79875}{2704}$

or about 29.54 "squared" dollars. Standard deviation is the square root of variance, so the standard deviation of $X$ is approximately $5.44.