AMC 10 · 2009 · #22

Grade 7 probability
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
probability-basicsystematic-enumeration easier-related-problem ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
Take the numbers 1 through 6 off of two dice — twelve numbers total, two copies of each value 1–6. Shuffle all twelve in a bag and glue them back on at random, one number per face, six faces per die. Roll both dice and add the two numbers on top. Find the probability that this sum equals 7.

Pick an answer.

(A)
$\ \frac{1}{9}$
(B)
$\ \frac{1}{8}$
(C)
$\ \frac{1}{6}$
(D)
$\ \frac{2}{11}$
(E)
$\ \frac{1}{5}$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

The random gluing and the rolling look like a lot of moving parts, but they only decide which single number ends up on top of each die. Tool #9 (Solve an Easier Related Problem) strips the story down to its core: it is the same as drawing two numbers out of the bag, one for each die's top. Tool #4 (Introduce a Variable) names die 1's top number n, which instantly fixes the partner 7-n that die 2 needs. Tool #2 (Make a Systematic List) counts how many of the remaining numbers hit that target by walking through the sum-to-7 pairs {1,6},{2,5},{3,4}. The whole thing then collapses to one division.

1STEP 1

Strip away the rolling

The gluing and rolling only pick which number tops each die, so by symmetry it is just drawing two of the twelve numbers from the bag.

P(sum=7) from two draws (no replacement) out of the 12 numbers
2STEP 2

Name die 1's top number

Let n be die 1's top number. Then die 2 must show 7-n, and 7-n is never n itself since 2n=7 has no whole-number solution.

need die 2 top=7-n, 7-n ≠ n
3STEP 3

Count the matching numbers left

Die 1's top uses up one number, leaving 11. Since 7-n is never n, both copies of 7-n survive, so 2 of the 11 work.

copies of (7-n) left=2, numbers left=11
4STEP 4

Divide to get the probability

Die 2's top is equally likely to be any of the 11, and 2 of them work, so the probability is 211\frac{2}{11} — choice (D).

P(sum=7)=2/11 → (D)
Answer
2/11
Cross-check by counting unordered pairs. Treat the twelve numbers as distinct tokens; the two tops form an unordered pair, and all C(12, 2)=66 pairs are equally likely. A sum of 7 comes from {1,6},{2,5},{3,4}, and since each value has 2 copies, each combination gives 2×2=4 token pairs, for 3×4=12 winning pairs. That is 12/66=2/11 — the same result. Size check too: 2/11≈0.18, just a hair above the 1/6≈0.17 you get from two ordinary dice, which makes sense — a normal die has 1 partner out of 6 that completes a 7, and here it is 2 partners out of 11, essentially the same rate.
💡Key takeaway

Whatever the first die shows on top, exactly two of the eleven leftover numbers make the sum 7 — so the probability is just 2/11, and the whole gluing-and-rolling story never mattered.

  • Strip away the rolling
  • Name die 1's top number
  • Count the matching numbers left
  • Divide to get the probability

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