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2006 AMC 10A Problem 13

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2006 AMC 10A Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Expected Value · difficulty 2 of 5 · about 2 min · position P11-15

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.)

  1. A)

    $12

  2. B)

    $30

  3. C)

    $50

  4. D)

    $60

  5. E)

    $100

Answer

D

Key insight

Winning needs an even first roll (1/2) and a matching second roll (1/6), so P(win) = 1/12 and the fair prize is 5 × 12 = 60.

Solution

To win, two things must happen in order. First, the initial roll must be even: probability 36=12\tfrac{3}{6} = \tfrac12. Second, the next roll must show that same number: whatever the first number was, there is 11 matching face out of 66, probability 16\tfrac16.

Because the second roll is independent of the first,
P(win)=12⋅16=112.P(\text{win}) = \frac12 \cdot \frac16 = \frac{1}{12}.

The game is fair when (probability of winning) ×\times (prize) equals the $5\$5 entry fee:
112⋅W=5⟹W=60.\frac{1}{12} \cdot W = 5 \quad\Longrightarrow\quad W = 60.

The answer is (D) $60\boxed{\textbf{(D)}\ \$60}.

Why this works

A multi-stage win condition has probability equal to the product of the stage probabilities, as long as each stage's probability is taken given the earlier stages happened. "Fair" is an expected-value statement: the expected winnings must equal the cost, so the prize is the cost divided by the win probability.

The trap

Using 1/6 alone for the winning probability (prize 30, choice (B)) by forgetting the first roll must be even, or using 1/2 for 'any even' on the second roll.

Common mistakes

  • Using 16\tfrac16 alone for the winning probability (prize 3030, choice (B)) by forgetting the first roll must be even, or using 12\tfrac12 for "any even" on the second roll.
  • Computing 5⋅12=605 \cdot 12 = 60 correctly but then subtracting the $5\$5 fee, thinking the prize should be net profit.

Techniques

Set up the equation/formula and compute; no special trick needed