Question

A deck of cards contains $52$cards, of which $4$are aces. You are offered the following wager: Draw one card at random from the deck. You win $\backslash (10$if the card drawn is an ace. Otherwise, you lose $\backslash )1$. If you make this wager very many times, what will be the mean amount you win?

(a) About −$\backslash (1$, because you will lose most of the time.

(b) About $\backslash )9$, because you win $\backslash (10$but lose only $\backslash )1$.

(c) About −$\backslash (0.15$; that is, on average you lose about $15$cents. (d) About $\backslash )0.77$; that is, on average you win about $77$cents.

(e) About $\$0$, because the random draw gives you a fair bet.

Short answer

The correct option is (C) About $-\$0.15$; that is, on average you lose about $15$ cents.

Step-by-step solution

Given Information

Given in the question that a deck of cards contains $52$cards, of which $4$are aces. You are offered the following wager: Draw one card at random from the deck. You win $\$10$if the card drawn is an ace. Otherwise, you lose $\$1$.

Explanation

A deck of cards contains $4$aces out of $52$cards.

The number of positive outcomes divided by the total number of possible outcomes equals the probability:

localid="1649915202706" $$\begin{gathered} P\left(\text{win}\right)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}} \\ =\frac{4}{52} \\ =\frac{1}{13} \end{gathered}$$

localid="1649915218751" $$\begin{gathered} P\left(\text{loss}\right)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}} \\ =\frac{48}{52} \\ =\frac{12}{13} \end{gathered}$$
The expected value is the sum of the product of each possibility with its probability:

localid="1649915557204" $$\begin{gathered} \mu =\sum xP\left(x\right) \\ =\$10\times \frac{1}{13}+(-\$1)\times \frac{12}{13} \\ =-\$\frac{2}{13} \\ \approx -\$0.15 \end{gathered}$$