Number of unique faces throwing a dice We select with replacement $N$ times a random number with $M$ options where $M\gg N$. How do I calculate the probability of having $n$ unique numbers $n\ge N$ per selection?
For instance, throwing three times a six-faced dice means $N=3$ and $M=6$. What is the probability of having three unique faces $P(n=3)$ (e.g. $\{3,4,5\},\ \{6,2,4\}$), or having two unique faces $P(n=2)$ (e.g. $\{3,4,3\},\ \{6,2,2\}$) or having one unique face $P(n=1)$ (e.g. $\{3,3,3\},\ \{5,5,5\}$)?

This is related to Stirling numbers of the second kind and the probability you are looking for is $$\dfrac{\frac{M!}{(M-n)!}S_2(N,n)}{M^N }$$
So in your example where $M=6$ and $N=3$ and $S_2(3,1)=1,S_2(3,2)=3,S_2(3,3)=1$, you would have 


*

*$\Pr(n=1)=\dfrac{\frac{6!}{5!}\times 1}{6^3} = \dfrac{1}{36}$

*$\Pr(n=2)=\dfrac{\frac{6!}{4!}\times 3}{6^3} = \dfrac{5}{12}$

*$\Pr(n=3)=\dfrac{\frac{6!}{3!}\times 1}{6^3} = \dfrac{5}{9}$


and as you would hope, these sum to $1$
