Let say we have $N$ dice with 6 faces. I'm asking my self, what is the probability that the maximum number of dice with the same face is $k$?
In more precise terms, what is the size of this set?
\begin{equation} D_k = \{ \, (d_i)_{i = 1}^{N} \ | \ \max_{j = 1 \dots 6} \ | \small{\{} \, i \ | \ d_i = j \ \small{\}} | = k \ \} \end{equation}
My approch was to use partitions of integers. Lets define $\Lambda = (\lambda_i)_{i = 1}^6$ a non-increasing partition of $N$ where $\lambda_i$ can be $0$. For example $\Lambda = (0,1,2,0,1,1)$ is a partition on $5$ and it represents a throw of 5 dice where the result is like $$2 \ \ 3 \ \ 5 \ \ 3 \ \ 6$$
Now I defince the space
\begin{equation} D^{\Lambda} = \{ \, (d_i)_{i = 1}^{N} \ | \ \small{\#} \small{\{} \, i \ | \ d_i = j \ \small{\}} = \lambda_j \ , \ \forall j = 1 \dots 6 \} \end{equation} where $\Lambda = (\lambda_i)_{i = 1}^6$ is a partition like the one above.
I prooved and checked with python that $$\small{\#} D^{\Lambda} = \cfrac{N!}{\lambda_1 ! \ \cdot \ ... \ \cdot \ \lambda_6 !}$$
Clearly $$D_k = \bigcup_{\Lambda \in P_k} D^\Lambda$$ where $P_k$ is the set of all partitions $(\lambda_i)_{i = 1}^6$ of $N$ such that $\max(\lambda_i) = k$. Now I'm stucked. I think there is a better way to calculate this probability but this is my try. I enjoyed the road since now so if yuo want, try it.
