2006 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2006 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10A solutions, or check the answer key.

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Concepts:expected valuedice (probability)

Difficulty rating: 1390

13.

A player pays $5\$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\$12

$30\$30

$50\$50

$60\$60

$100\$100

Solution:

The player wins only if the first roll is even (probability 12\frac12) and the second roll matches it (probability 16\frac16), so P(win)=1216=112.P(\text{win}) = \frac12 \cdot \frac16 = \frac{1}{12}.

For a fair game, 112x=5,\frac{1}{12} x = 5, so x=60.x = 60.

Thus, the correct answer is D.

Problem 13 in Other Years