Question

4. Consider this game. Pick a card randomly from 52 playing cards. If

you get an Ace, you will win $x. Otherwise, you will lose $1. If the game is fair, what is the value of x?
A- 8
B- 10
C- 11
D-12

Answer 1

The expected value of the game is the sum of the products of the outcomes and their probabilities. In this case, we have two possible outcomes: winning x dollars with probability 4/52 (since there are 4 Aces in a deck of 52 cards) or losing 1 dollar with probability 48/52 (since there are 48 non-Aces in a deck of 52 cards). Therefore, the expected value of the game is:

E(x) = (4/52) * x + (48/52) * (-1)

Simplifying:

E(x) = (1/13) * x - (12/13)

For the game to be fair, the expected value must be zero. Therefore:

0 = (1/13) * x - (12/13)

(12/13) = (1/13) * x

x = 12

Therefore, the value of x that makes the game fair is $12.

So the answer is D - 12.