Probability two dice sum to a multiple of three

Question

You roll two fair six-sided dice. What is the probability the total is a multiple of 3 (that is, 3, 6, 9, or 12)?

Accepted Answer

Work on the 36-cell grid of ordered outcomes. Use the known counts for each total (which follow 6 - s-7): - sum 3: 2 outcomes, - sum 6: 5 outcomes, - sum 9: 4 outcomes, - sum 12: 1 outcome. Adding the favorable outcomes: 2 + 5 + 4 + 1 = 12. So P(	ext{sum divisible by } 3) = \frac{12}{36} = \frac{1}{3}. A slick check using remainders: a total is divisible by 3 exactly when the two dice's remainders modulo 3 add to 0 mod 3. Each die is uniform over remainders \{0,1,2\} (values \{3,6\},\{1,4\},\{2,5\}), so the pair of remainders is uniform over 9 equally likely combinations, and exactly 3 of th…

Probability two dice sum to a multiple of three

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