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2018 AMC 10B Problem 9

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2018 AMC 10B Problem 9AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 2 of 5 · about 2 min · position P1-10

The faces of each of 77 standard dice are labeled with the integers from 11 to 66 . Let pp be the probability that when all 77 dice are rolled, the sum of the numbers on the top faces is 1010 . What other sum occurs with the same probability pp ?

  1. A)

    1313

  2. B)

    2626

  3. C)

    3232

  4. D)

    3939

  5. E)

    4242

Answer

D

Key insight

Replacing each face k by 7 - k is a bijection, so a sum of 10 is exactly as likely as 49 - 10 = 39.

Solution

Pair every outcome with its "flipped" outcome, in which each die showing kk is replaced by a die showing 7−k7 - k. This is a one-to-one correspondence on the 676^7 equally likely outcomes, and it turns a total of SS into a total of
7⋅7−S=49−S.7 \cdot 7 - S = 49 - S.
So the number of outcomes with sum SS equals the number with sum 49−S49 - S, and the two sums are equally likely.

For S=10S = 10 the partner sum is 49−10=3949 - 10 = 39.

The answer is (D) 39\boxed{\textbf{(D)}\ 39}.

Why this works

The distribution of a dice total is symmetric about its midpoint: the minimum sum is 77, the maximum is 4242, and 1010 sits 33 above the minimum just as 3939 sits 33 below the maximum. Whenever a problem asks for "the other value with the same probability," look for a reflection that maps outcomes to outcomes.

Alternative approach

Sum 1010 with seven dice means the excess over the all-ones roll is 33, distributed among 77 dice. Sum 3939 means the shortfall below the all-sixes roll is 33. Both are the number of ways to distribute 33 units among 77 dice, so the counts match.

The trap

Mirroring about the wrong value, e.g. answering the maximum sum 42 = 6*7 (choice E) instead of 49 - 10, the reflection of 10 about the mean 24.5.

Common mistakes

  • Mirroring about the wrong value, e.g. answering the maximum sum 42 = 6*7 (choice E) instead of 49 - 10, the reflection of 10 about the mean 24.5.
  • Guessing 1313 because it is "10 plus a small number," without any symmetry reasoning.

Techniques

Map the objects to something easier to count · Exploit symmetry to reduce work or pair up objects