After 6n roll of dice, what is the probability each face was rolled exactly n times?

Question

This is closely related to the question "If you toss an even number of coins, what is the probability of 50% head and 50% tail?", but for dice with 6 possible results instead of coins (with 2 possible results). Actually I would like a more general approximation formula, for dice with m faces.

Answer 1

This is solved with multinomial coefficients. Out of $6n$ dice, we need to select $n$ of them to be $1$, $n$ of them to be $2$, $\ldots$, $n$ of them to be $6$. There are ${6n\choose n,n,n,n,n,n}$ ways to do this. This counts the desired rolls, and there are $6^{6n}$ total rolls, including both desired and undesired. Combining, we get $$\frac{{6n\choose n,n,n,n,n,n}}{6^{6n}}=\frac{(6n)!}{(n!)^66^{6n}}$$

If you have $m$ faces instead of $6$, replace the three $6$'s with three $m$'s in the above.

Answer 2

Think of creating a string of length $6n$, where each letter is one of the numbers on the face of the die. For instance, for $n=1$, we could have the string $351264$. There are $6^{6n}$ such strings, as can be seen by elementary counting methods. The strings you are interested in are ones in which each letter appears $n$ times. Again, from combinatorial results, there are $$ \frac{(6n)!}{(n!)^6} $$ such strings. To get the probability, now you just have to divide the two, as each string is equiprobable.

This result can be easily generalized to dice with $m$ faces.

Answer 3

To add to the other two answers, you can use Stirling's approximation of the factorial to get a sense of the asymptotic behavior of this expression: $$ \frac{(mn)!}{(n!)^mm^{mn}} \approx \frac{\sqrt{2\pi mn}(mn/e)^{mn}}{(2\pi n)^{m/2}(n/e)^{mn}m^{mn}}=\frac{\sqrt{m}}{(2\pi n)^{(m-1)/2}} $$

So, for fixed $m$, the probability decays like $n^{-(m-1)/2}$; in particular, when $m=6$, it decays like $n^{-5/2}$.