Welcome to Math Overflow

Question

1

I found the following closed form solution for the abovementioned problem:

$${1\over k^n}\cdot{k!\over (k-m)!}\cdot{\{{n\over m}\}}$$ with ${\{{n\over m}\}}$ being the Stirling Number of the second kind.

Although it seems to have some intuition and seems to work for a sample problem for which I have the solution this closed form is not from a trusted source. Unfortunately I can't find any other source.

My question: Could anyone acknowledge this closed form solution and/or give me a hint where to find a citable source.

flag	asked Feb 21 at 19:05 vonjd 1,507●4●17

1

There are $k^n$ possible dice configurations, presumably all equally likely. $\frac{k!}{(k-m)!}$ is the number of distinct combinations of $m$ (out of $k$ total) faces. The Stirling number of the second kind is the number of ways to separate n elements into m sets. – Steve Huntsman Feb 21 at 19:28

3

In other words, this is not far from the definition of the Stirling number of the second kind. See equation 11 in mathworld.wolfram.com/…. Set x=k and divide both sides by k^n. – Douglas Zare Feb 21 at 19:35

Kristal Cantwell · Accepted Answer · 2010-02-21 19:31:43Z UTC

5

Applied probability by Kenneth Lange deals with this problem on page 74. It is on Google books, here is the URL.

answered Feb 21 at 19:31

Kristal Cantwell
3,091●1●12

Probability of n k-sided dice showing exactly m different faces

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.