Finding the expected value of a card game with replacement

Question

A player chooses two cards with replacement from a standard deck. If both of the cards are hearts he wins 25 dollars. If he draws only one heart, he wins 5 dollars. If he draws no hearts he gets no prize. The cost of the game is 4 dollars. Find the expected value of the game for the player.

Answer 1

Let $ X $ be the numbers of hearts drawn. There are three outcomes: 1) two hearts drawn $(X=2)$, 2) one heart drawn $(X=1)$, 3) zero hearts drawn $(X=0)$.

What is the probability of each? Since you are drawing with replacement, then $ X \sim Binomial(n=2,p=13/52=1/4) $.

$ P(X=2) = {{2}\choose{2}} \big(\frac{1}{4}\big)^2 \big(1-\frac{1}{4}\big)^{2-2} = \big(\frac{1}{4}\big)^2 $

$ P(X=1) = {{2}\choose{1}} \big(\frac{1}{4}\big)^1 \big(1-\frac{1}{4}\big)^{2-1} = 2 \big(\frac{1}{4}\big) \big(\frac{3}{4}\big) $

$ P(X=0) = {{2}\choose{0}} \big(\frac{1}{4}\big)^0 \big(1-\frac{1}{4}\big)^{2-0} = \big(\frac{3}{4}\big)^2 $

Let $ W(X=i) $ denote the winnings from drawing $ i $ hearts, i.e. $ W(X=2) = 25 $ , $ W(X=1) = 5 $ , $ W(X=0) = 0 $. You can now use this information to compute the expected winnings as follows:

\begin{equation} E(W) = \sum_{i=0}^{i=2} P(X=i) \times W(X=i) \\ = P(X=0)W(X=0) + P(X=1)W(X=1) + P(X=2)W(X=2) \\ = \Big(\frac{3}{4}\Big)^2(0) + 2 \Big(\frac{1}{4}\Big) \Big(\frac{3}{4}\Big)(5) + \Big(\frac{1}{4}\Big)^2(25) \end{equation}

The expected earning from playing this game once is simply the expected winnings from one game minus the cost of one game, i.e. $$ E(P) = E(W) - 4 $$