Any mathemathically minded individual mind giving me a hand? Let's see if I even manage to explain my problem at all. Ask if you need clarification on anything.

I'm trying to find a formula that returns the probability of getting k amount of the same result with n amount of x-sided dice. So, if k=3, n=10 and x=6 for example, the probability of getting three ones, twos, threes, fours, fives or sixes with 10 6-sided dice. Thing is, the last time I've done any actual math was in upper secondary school, 3 years ago, and even then probability was definitely not one of my strong points. I looked over some of my old math books while visiting my parents earlier this week, and here's what I've got so far:

Using binomial probability, it's easy to calculate the probability for getting k amount of a certain result (for example k amount of ones) with n amount of x-sided dice:

k = amount of same value

n = amount of dice

p = probability of getting a certain value = 1/x

x = the amount of sides on the dice

q = the complement of p = 1-p

P=(n combinations k)p^{k}q^{n-k}

Basically, IIRC, if I want to expand this to cover any value as long as there are k amount of the same (for example looking for k amount of ones or twos or threes etc), I need to repeat this for all the different values. So, the earlier formula times the amount of faces on the dice. But if n≥2k, all of the repetitions contain a bunch of cases where I'd get the required amount of multiple different values. I'd need to remove these duplicates, which seems stupidly complicated.

So, I figured I'd try to do it another way. Basically, take the complement of the binomial probability earlier (1-(n combinations k)p^{k}q^{n-k}), multiply it by itself for the amount of faces in the dice to get the probability of not getting the required amount of ones or twos etc etc, and then take the complement of that to get the probability of getting k amount of any value.

P(k amount of a certain value) = (n combinations k)p^{k}q^{n-k}

P(NOT k amount of a certain value) = 1-((n combinations k)p^{k}q^{n-k})

P(NOT k amount of any value) = (1-((n combinations k)p^{k}q^{n-k}))^{x} <- This is probably wrong

P(k amount of any value) = 1-(1-((n combinations k)p^{k}q^{n-k}))^{x}

I have no idea if this logic is faulty at some point, though I suspect it is since I tried it with k=1, n=10, p=(1/10) and got 99,25%. I'm pretty sure that should be 100%. Of course, there could also be a rounding problem somewhere or I might have made a typo both in excel and doing the calculation on paper.

Soooo, anyone want to take a stab at this and tell me where I went wrong?

EDIT: actually, now that I think about it, 100% would definitely not be the right answer for k=1, n=10, p=(1/10). 99,25% seems pretty plausible to be honest. Still, I'd like comments on the logic. I'm totally not sure about this.