Question

In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing.
(a) Find the expected winnings for a single game.

(b) Find the standard deviation of the winnings.

(c) If the game costs $5 to play, what would be the expected value of the net profit (or loss)?

(Hint: profit = winnings - cost; X - 5)

(d) If the game costs $5 to play, what would be the standard deviation of the net profit (or loss)?

(e) If the game costs $5 to play, should you play this game?

Answer 1

There are $\binom{52}3=\frac{52!}{3!(52-3)!}$ (or "52 choose 3") ways of drawing any 3 cards from the deck.

There are 13 hearts in the deck, and 26 cards with a black suit. So there are $\binom{13}3$ and $\binom{26}3$ ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is

$P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}$

and the probability of drawing 3 black cards is

$P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}$

All other combinations can be drawn with probability $1-\frac{11}{850}-\frac2{17}=\frac{739}{850}$.

Let $W$ be the random variable for one's potential winnings from playing the game. Then

$P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\ 50\\\frac2{17}&\text{for }w=\ 25\\\frac{739}{850}&\text{otherwise}\end{cases}$

a. For a single game, one can expect to win

$E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\ 50\cdot11}{850}+\frac{\ 20\cdot2}{17}+\frac{\ 0\cdot739}{850}=\ 3$

b. For a single game, one's winnings have a variance of

$V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2$

where

$E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\ 50^2\cdot11}{850}+\frac{\ 20^2\cdot2}{17}+\frac{\ 0^2\cdot739}{850}=\ ^2\frac{1350}{17}\approx\ ^279.41$

so that $V[W]=\ ^2\frac{1197}{17}\approx\ ^270.41$. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.

c. With a $5 buy-in, the expected value of the game would be

$E[W-\ 5]=E[W]-\ 5=-\ 2$

i.e. a player can expect to lose $2 by playing the game (on average).

d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:

$V[W-\ 5]=V[W]$

so the standard deviation is the same, roughly $8.39.

e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.