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2006 AMC 10A Problem 13

Problem 13 of 25IntermediateProbability & Statistics

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.)

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Solution

The player wins only if the first roll is even (probability 12\frac12) and the second roll matches it (probability 16\frac16), so the win probability is 12⋅16=112.\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.
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Tagged: expected value · dice (probability)

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