Four dice are thrown together and the values that turn up are noted. What is…

2025

Four dice are thrown together and the values that turn up are noted. What is the probability that their sum is 20?

  1. A.

    17/648

  2. B.

    35/1296

  3. C.

    11/432

  4. D.

    1/54

Attempted by 7 students.

Show answer & explanation

Correct answer: B

Concept

For n independent dice each with s faces, every ordered sequence of face values is equally likely, so P(event) = (number of ordered rolls giving the event) / sn. Counting the favourable rolls for a fixed sum means listing every distinct combination of dice values (a value-combination) that adds to the target sum, counting how many ordered arrangements each combination has, and adding those counts.

Application to this roll

Total ordered outcomes: each of the four dice independently shows one of 6 faces, so the sample space size is 64 = 1296.

The maximum possible sum with four dice is 6+6+6+6 = 24, so a sum of 20 is exactly 4 short of the maximum. Writing each die's shortfall from 6 as d = 6 minus its value (0 to 5), the four shortfalls must add to 24 minus 20 = 4.

Shortfall pattern

Dice values

Arrangements

4, 0, 0, 0

2, 6, 6, 6

4!/3! = 4

3, 1, 0, 0

3, 5, 6, 6

4!/2! = 12

2, 2, 0, 0

4, 4, 6, 6

4!/(2!2!) = 6

2, 1, 1, 0

4, 5, 5, 6

4!/2! = 12

1, 1, 1, 1

5, 5, 5, 5

4!/4! = 1

Adding the arrangement counts: 4 + 12 + 6 + 12 + 1 = 35 favourable ordered rolls, so the probability is 35/1296.

Cross-check

Replacing every die value v by 7 minus v turns a roll summing to 20 into one summing to 4×7 minus 20 = 8, and this substitution pairs up the two sets of rolls one-to-one, so the count of rolls summing to 20 must equal the count summing to 8. Directly: the equation x1+x2+x3+x4 = 8 minus 4 = 4 with each 0 <= xi <= 5 has C(7,3) = 35 non-negative solutions (no upper-bound correction is needed since 4 is less than 5), confirming 35 favourable rolls independently.

Explore the full course: Tcs Live Preparation