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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 2010 at 19:05 vonjd 2,766●3●19●37

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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 2010 at 19:28

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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 2010 at 19:35

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

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

answered Feb 21 2010 at 19:31

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

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