2006 AMC 10A Problem 13

Question

A player pays

\$5

to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

\$12

\$30

\$50

\$60

\$100

Answer: D

Concepts:expected value dice (probability)

Difficulty rating: 1390

Small hint:

Find the probability of winning: an even roll, then a matching roll

Big hint:

The win probability is

\frac12 \cdot \frac16 = \frac{1}{12};

set (probability)(prize) equal to

5

Solution:

The player wins only if the first roll is even (probability

\frac12

) and the second roll matches it (probability

\frac16

), so the win probability is

\frac12 \cdot \frac16 = \frac{1}{12}.

For a fair game,

\frac{1}{12} x = 5,

so

x = 60.

Thus, the correct answer is D.