Possibilities of rolling 15 with top 3 six-sided dice

Question

There are $7776$ possible outcomes of rolling $5$ six-sided dice, of this total there are $651$ possible outcomes where the $5$ dice equal exactly $15$.

How could you calculate the number of possible outcomes where the top three dice equal $15$?

Answer 1

There are three ways to roll 15 with three dice: 663, 654 and 555. I'll call this the greatest-sum-set (GSS); these have to be the greatest within the roll. I will split each case into three sub-cases, depending on how many times the smallest number in the GSS appears on the remaining two dice:

• 663|xx (x is 1 or 2, the x's can be different). 10 ways to place 6's, 3 ways to place 3 afterwards, the remaining spaces can be filled in 4 ways. Total 10 × 3 × 4 = 120 ways.
• 663|3x (one 3 on remaining dice). 10 ways to place 6's, 3 ways to place x, x can have 2 values. Total 10 × 3 × 2 = 60 ways.
• 663|33 (two 3's). Only 10 ways to place 6's and no choice after that.
• 654|xx (x < 4). 5 × 4 × 3 ways to place 654, remaining spaces can be filled in 9 ways. Total 540 ways.
• 654|4x. 5 × 4 × 3 ways to place 65x, x can have 3 values. Total 180 ways.
• 654|44. 5 × 4 ways to place 65, no choice afterwards. Total 20 ways.
• 555|xx (x < 5). 10 ways to place x's, which can assume 16 values. Total 160 ways.
• 555|5x. 5 ways to place x, which can assume 4 values. Total 20 ways.
• 555|55. Only one way for this.

Adding up all cases gives 1111 5-dice rolls where the top three numbers sum to 15.