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Suppose an $n$-sided die is rolled $t$ times. What is the probability that each side comes up at least $k$ times ($nk\leq t$)?

I've tried several combinatoric approaches, but every time I get stuck counting outcomes multiple times. For example

$$\binom{t}{nk}\frac{(nk)!}{k!^n}$$

yields the number of ways of placing the $nk$ constraining rolls, but the remaining outcomes cannot be placed freely (a factor of $n^{t-nk}$) without duplicating configurations from other placements of the constraining ones.

Is there a different approach I should be looking at, or a well-known solution?

R.. GitHub STOP HELPING ICE
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  • If $t=nk$, then the value is $$\frac{\binom{nk}{k,k,\dots,k}}{n^{nk}}$$ where we are using the multinomial. I would be surprised if there was a simple formula for the general result. – Thomas Andrews Jul 09 '17 at 04:46
  • If $n=2$, then the value is $$\sum_{i=k}^{t-k} \binom{t}{k}$$ which I think as a rule does not have a simpler closed form. – Thomas Andrews Jul 09 '17 at 04:55
  • @ThomasAndrews: If $t=nk$, the approach in the question body works because there is no overcounting (no pseudo-free choices) and should be equivalent to the multinomial form. But I'm interested in all values of $k$ including $k=1$ and in general $nk$ much less than $t$. – R.. GitHub STOP HELPING ICE Jul 09 '17 at 05:06
  • I realize that, I was just starting with an example $t$ that we can explicitly answer. $k=1$ is a case where you might use inclusion-exclusion. It gets harder to use inclusion-exclusion when $k>1$. – Thomas Andrews Jul 09 '17 at 05:10
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    The case $k=1$ is the Coupon Collector's Problem.https://en.wikipedia.org/wiki/Coupon_collector%27s_problem The Wikipedia article points to a paper by Donald J. Newman and Lawrence Shepp where $m$ copies of each coupon are needed: http://www.jstor.org/stable/2308930?origin=crossref&seq=1#page_scan_tab_contents – awkward Jul 09 '17 at 13:55
  • @awkward: Thanks. Most of the material there seems to be stated in terms of expected time to get each result (once or $k$ times) rather than, for a given number of rolls, the probability of seeing each at least $k$ times. Is there a good approach for adapting it to my form of the problem? – R.. GitHub STOP HELPING ICE Jul 10 '17 at 01:45

1 Answers1

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Nothing like a complete answer, just one specific case.

When $k=1$, we see that we are counting the set of "onto" functions $\{1,2,\dots,t\}\to\{1,2,\dots,n\}$. This is related to the Stirling numbers of the second kind, namely it is:

$$n!S(t,n) =\sum_{i=0}^n (-1)^{n-i}{n \choose i} i^t$$

The formula on the right can be proved using inclusion-exclusion.

The probability is thus:

$$\frac{n!S(t,n)}{n^t} = \sum_{i=0}^{n}(-1)^{n-i}\binom{n}i\left(\frac in\right)^t$$

The inclusion-exclusion formula cannot be easily adjusted to deal with other values of $k$.

Thomas Andrews
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