Rolling 4 dice, probability of a sum=S

ondansetron

TS Contributor
#1
Working on a problem (not homework for me) regarding the probability of a particular sum achieved by rolling n fair 6-sided dice.

The specific problem required me to find the probability that four dice are rolled and the sum is 21.
I said 6^4 possibilities = denominator.
The only possible "sets" that would sum to 21 would be (3+6+6+6), (5+5+5+6), or (4+5+6+6).
For the first two sums there are 4 choose 1 ways= 4 (for Sum 1) + 4 (for Sum 2) = 8 ways for sums 1 and 2.
Sum 3 I said, 'fix' either 4 or 5 and the problem becomes 3 choose 2 and then 4 places/dice where the 4 could come up, so 12 ways to get sum 3.
Total favorable ways for S=21 is 20/6^4.

When I read the solution I ran into my issue of thinking, why don't we care about the "non-6" numbers in the calculation of N choose k where K is the number of 6's?
For example, 3+5+6+6 is not favorable (although it's equally likely as the 4 being there instead of 3). So what am I missing in my understanding?
 

Dason

Ambassador to the humans
#3
I'm still happy to help and I might know what you're trying to get at - but if you could just add a bit of clarifying details that would be appreciated. Thanks!