If four dice are thrown and the values are noted. What is the probability of…

Concept: To count how many ways several dice (each restricted to a range of face values) can sum to a target, count the integer solutions of x1 + x2 + x3 + x4 = target with each xi restricted to its allowed range -- by shifting each variable to start at 0 and applying the standard stars-and-bars formula for the number of non-negative integer solutions.

Application:

1. Each of the four dice can independently show any value from 1 to 6, so the total number of equally likely outcomes is 64 = 1296.

2. Let x1, x2, x3, x4 be the values shown by the four dice, so x1 + x2 + x3 + x4 = 20 with each xi between 1 and 6.

3. Substitute zi = 6 - xi for each die, so each zi lies between 0 and 5, and the equation becomes z1 + z2 + z3 + z4 = 24 - 20 = 4.

4. Since the four zi values must add up to only 4, no single zi can ever exceed 5, so the upper bound never gets violated and no correction is needed.

5. The number of non-negative integer solutions of z1 + z2 + z3 + z4 = 4 is given by the stars-and-bars formula C(4 + 4 - 1, 4 - 1) = C(7, 3) = 35.

6. So the probability of the sum being 20 is 35 / 1296.

Cross-check: Cross-check by directly listing the partitions of 20 into four dice values from 1 to 6, and counting the number of orderings (arrangements) of each partition:

• {6, 6, 6, 2}: 4 arrangements

• {6, 6, 5, 3}: 12 arrangements

• {6, 6, 4, 4}: 6 arrangements

• {6, 5, 5, 4}: 12 arrangements

• {5, 5, 5, 5}: 1 arrangement

Total = 4 + 12 + 6 + 12 + 1 = 35, which matches the stars-and-bars result and confirms the probability is 35/1296.