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2006 AMC 10B Problem 21

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2006 AMC 10B Problem 21AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 3 of 5 · about 4 min · position P21-25

For a particular peculiar pair of dice, the probabilities of rolling 11 , 22 , 33 , 44 , 55 , and 66 , on each die are in the ratio 1:2:3:4:5:61:2:3:4:5:6 . What is the probability of rolling a total of 77 on the two dice?

  1. A)

    463\frac{4}{63}

  2. B)

    18\frac{1}{8}

  3. C)

    863\frac{8}{63}

  4. D)

    16\frac{1}{6}

  5. E)

    27\frac{2}{7}

Answer

C

Key insight

Face k has probability k/21, so a total of 7 has probability (1*6 + 2*5 + 3*4 + 4*3 + 5*2 + 6*1)/21^2 = 56/441.

Solution

The six probabilities are proportional to 1,2,…,61, 2, \ldots, 6, which add to 2121, so face kk comes up with probability k21\frac{k}{21} on each die.

A total of 77 arises from the ordered outcomes (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)(1,6), (2,5), (3,4), (4,3), (5,2), (6,1). The dice are independent, so the outcome (j,7−j)(j, 7-j) has probability j(7−j)212\frac{j(7-j)}{21^2}. Adding the six:
1⋅6+2⋅5+3⋅4+4⋅3+5⋅2+6⋅1441=6+10+12+12+10+6441=56441.\frac{1\cdot6 + 2\cdot5 + 3\cdot4 + 4\cdot3 + 5\cdot2 + 6\cdot1}{441} = \frac{6 + 10 + 12 + 12 + 10 + 6}{441} = \frac{56}{441}.
Since 56=7⋅856 = 7 \cdot 8 and 441=7⋅63441 = 7 \cdot 63, this reduces to 863\frac{8}{63}.

The answer is (C) 863\boxed{\textbf{(C)}\ \frac{8}{63}}.

Why this works

A ratio of probabilities becomes actual probabilities once you divide by the ratio total, because the probabilities must sum to 11. After that the problem is an ordinary "sum of two dice" count, except each ordered outcome carries its own weight. The mirror symmetry j↔7−jj \leftrightarrow 7 - j halves the work: compute three products and double.

The trap

Counting each unordered pair once (3 terms, 28/441 = 4/63) instead of the six ordered outcomes on two distinct dice.

Common mistakes

  • Counting each unordered pair once (3 terms, 28/441 = 4/63) instead of the six ordered outcomes on two distinct dice.
  • Using 16\frac{1}{6} for each face out of habit, which gives the fair-dice answer 16\frac{1}{6}, choice (D).

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects