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Let's say I have $n$ fair (six-sided) dice ($n>=6$) and I want to figure out the probability of getting each side at least once. I figured out a formula through elementary combinatorial reasoning: $$\frac{6!\cdot\binom{n}{6}\cdot6^{n-6}}{6^n}$$

First, of all, is this formula correct?

Second, somewhere I read that I could do this with the sieve formula. How do I use the sieve formula for this one?

ghthorpe
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  • I think you are looking at an instance of the coupon collector's problem. (I had to look twice to be sure, but I think this is right.) – Brian Tung Oct 24 '17 at 17:08
  • That page may not have the expression you want. You may be able to find other pages, however, using that name. Let me think about whether that expression is right. Note that you can simplify $6^{n-6}/6^n$ to just $1/6^6$. – Brian Tung Oct 24 '17 at 17:09
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    No, your reasoning is not correct. Consider the case where the die has only two sides (like a coin). Your formula and reasoning would imply that in three rolls of the die the probability we see each at least once would be $\frac{2!\cdot\binom{3}{2}\cdot 2^{3-2}}{2^3}=\frac{12}{8}>1$ which is clearly incorrect as probabilities cannot be larger than one. – JMoravitz Oct 24 '17 at 17:11
  • @JMoravitz Could it be that my $6!$ term in the numerator is wrong?Remving it would make the formula yield the correct answer for your example. – ghthorpe Oct 24 '17 at 17:26
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    Possible duplicate: https://math.stackexchange.com/a/123125/265452 – Steve D Oct 24 '17 at 17:26
  • @JMoravitz hmm, I just realized that choosing $n = 5$ would disprove my adjusted formula... – ghthorpe Oct 24 '17 at 17:31
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    @ghthorpe the error is that when you selected the positions of a set of six rolls where all occur at least once you didn't account for the rest of the positions having some repeats as well. For three sides to the die for example, you counted the sequence $\color{red}{123}1$ differently than $1\color{red}{231}$ (where the red numbers are those selected as the set guaranteed to have at least one of each). To fix your mistake, one of the easiest approaches is to use inclusion-exclusion instead (as is suggested in the linked duplicate) – JMoravitz Oct 24 '17 at 17:31
  • okay, I accept your answer. I guess this post is a duplicate then. – ghthorpe Oct 24 '17 at 17:33

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