Problem 20 · 1997 AJHSME
Stretch
Counting & Probability
casework
A pair of 8-sided dice have sides numbered 1 through 8, each equally likely. What is the probability that the product of the two numbers facing up exceeds 36?
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Answer: A — 5/32.
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Hint 1 of 2
A product over 36 is rare — both dice have to be large. So don't list all 64 outcomes; only the top rolls matter. Sweep the first die from high down and ask 'how big must the other be?'
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Hint 2 of 2
When few outcomes qualify, count the SUCCESSES directly by organized cases instead of computing every possibility.
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Approach: case on the first die, find the threshold for the second
- For each value of the first die, find the smallest second value that pushes the product past 36: a 1–4 can't reach it; a 5 needs 8 (1 way); a 6 needs 7 or 8 (2); a 7 needs 6, 7, 8 (3); an 8 needs 5, 6, 7, 8 (4).
- Counts 1 + 2 + 3 + 4 = 10 ordered pairs out of 8 × 8 = 64 total.
- Probability = 10/64 = 5/32.
- Nice pattern: the winning counts run 1, 2, 3, 4 as the first die climbs 5→8 — a neat staircase that signals you've been systematic, not just guessing.
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