Let's assume we throw a die $n$-times. What is the probability that all numbers appear with the same frequency if $n\to\infty$?
My approach:
Let be $P(A_n)$ the probability that each number appears $k$ times after $n$ throws. It's easy to see that $P(A_n)=\frac{n!}{(k!)^6 6^n}$ if $n\mod 6=0$ and $P(A_n)=0$ otherwise. The problem is to check the limit $\lim\limits_{n\to\infty}P(A_n)$. We only have to consider those $n$ with $n\mod 6=0$.
We know by Stirling approximation that $\lim\limits_{n\to\infty}\frac{n!}{\sqrt{2\pi n}~n^ne^{-n}}=1$. Also \begin{align*} &\frac{\sqrt{2\pi n}~n^ne^{-n}}{\left(\sqrt{2\pi k}~k^ke^{-k}\right)^6}=\frac{\sqrt{2\pi 6k}~(6k)^{6k}e^{-6k}}{8\pi^3 k^3~k^{6k}e^{-6k}}=\frac{\sqrt{2\pi 6k}}{8\pi^3 k^3}6^{6k}=\frac{\sqrt{2\pi 6k}}{8\pi^3 k^3}6^{n}\\ &\text{and } \lim\limits_{k\to\infty}\frac{\sqrt{2\pi 6k}}{8\pi^3 k^3}=0. \end{align*} Keeping this in mind we conclude: \begin{align*} &\frac{n!}{(k!)^6 6^n} \underset{=1}{\underbrace{\frac{\sqrt{2\pi n}~n^ne^{-n}}{\sqrt{2\pi n}~n^ne^{-n}}\left(\frac{\sqrt{2\pi k}~k^ke^{-k}}{\sqrt{2\pi k}~k^ke^{-k}}\right)^6}}=\frac{n!}{\sqrt{2\pi n}~n^ne^{-n}} \frac{\sqrt{2\pi n}~n^ne^{-n}}{(\sqrt{2\pi k}~k^ke^{-k})^6}\frac{(\sqrt{2\pi k}~k^ke^{-k})^6}{(k!)^6}\frac{1}{6^n}\\ &=\frac{n!}{\sqrt{2\pi n}~n^ne^{-n}} \frac{(\sqrt{2\pi k}~k^ke^{-k})^6}{(k!)^6}\frac{\sqrt{2\pi 6k}}{8\pi^3 k^3}\frac{6^n}{6^n}. \end{align*} If we apply the Stirling approximation and the above limit we see that for all $n$ with $n\mod 6=0$ we have: $$ \lim\limits_{n\to\infty}\frac{n!}{\sqrt{2\pi n}~n^ne^{-n}} \frac{(\sqrt{2\pi k}~k^ke^{-k})^6}{(k!)^6}\frac{\sqrt{2\pi 6k}}{8\pi^3 k^3}=1\cdot1\cdot 0=0. $$ (Note that $n\to\infty\implies k\to\infty$).
Is this correct? It seems a bit counterintuitive because I thought if one extends the stochastic experiment to infinite die throws each number must appear with the same frequency as the die is assumed to be fair. So my first guess was that the probability should be $1$?!