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

Solution:

The player wins only if the first roll is even (probability $\frac12$ ) and the second roll matches it (probability $\frac16$ ), so $P(\text{win}) = \frac12 \cdot \frac16 = \frac{1}{12}.$

For a fair game, $\frac{1}{12} x = 5,$ so $x = 60.$

Thus, the correct answer is D.

2006 AMC 10A Problem 13

Answer: D

Difficulty rating: 1390

Solution:

Problem 13 in Other Years