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2007 AMC 10B Problem 22

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2007 AMC 10B Problem 22AMC 10 Step by Step
Name:                             Date: September 13, 2026
Expected Value · difficulty 4 of 5 · about 4 min · position P21-25

A player chooses one of the numbers 11 through 44 . After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered 11 through 4.4. If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins 11 dollar. If the number chosen appears on the bottom of both of the dice, then the player wins 22 dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses 11 dollar. What is the expected return to the player, in dollars, for one roll of the dice?

  1. A)

    −18-\frac{1}{8}

  2. B)

    −116-\frac{1}{16}

  3. C)

    00

  4. D)

    116\frac{1}{16}

  5. E)

    18\frac{1}{8}

Answer

B

Key insight

Each die shows the chosen number with probability 1/4, so the outcomes have probabilities 9/16, 6/16, 1/16 and the payoff sum is (-9 + 6 + 2)/16.

Solution

Which number the player picks is irrelevant by symmetry; call it kk. Each die independently shows kk with probability 14\frac14 and something else with probability 34\frac34. Out of the 1616 equally likely rolls:

  • both dice show kk: 1⋅1=11 \cdot 1 = 1 roll, payoff +2+2;
  • exactly one shows kk: 1⋅3+3⋅1=61 \cdot 3 + 3 \cdot 1 = 6 rolls, payoff +1+1;
  • neither shows kk: 3⋅3=93 \cdot 3 = 9 rolls, payoff −1-1.

The expected return is
116(2)+616(1)+916(−1)=2+6−916=−116.\frac{1}{16}(2) + \frac{6}{16}(1) + \frac{9}{16}(-1) = \frac{2 + 6 - 9}{16} = -\frac{1}{16}.

The answer is (B) −116\boxed{\textbf{(B)}\ -\frac{1}{16}}.

Why this works

Expected value is a weighted sum of payoffs, and with two independent dice the weights come from a 4×44 \times 4 grid of outcomes. Listing the three outcome classes with their counts out of 1616 keeps the arithmetic exact; a check is that 1+6+9=161 + 6 + 9 = 16. The house edge here is tiny, so a sign error swings the answer across three choices.

Alternative approach

Linearity: let MM be the number of matching dice, with E[M]=2⋅14=12E[M] = 2 \cdot \frac14 = \frac12. The payoff is MM dollars when M≥1M \ge 1 and −1-1 when M=0M = 0, i.e. payoff =M−[M=0]= M - [M = 0]. So the expectation is 12−916=−116\frac12 - \frac{9}{16} = -\frac{1}{16}.

The trap

Counting 'exactly one die matches' as 3/16 instead of 6/16, forgetting that either die could be the matching one.

Common mistakes

  • Counting 'exactly one die matches' as 3/16 instead of 6/16, forgetting that either die could be the matching one.
  • Treating the three outcomes as equally likely, or using 14+14=12\frac14 + \frac14 = \frac12 as the chance of at least one match (double-counting the both-match roll).

Techniques

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