E as opposed to P
I am not entirely certain, but since $ \frac{n - i + 1}{n}\! $ is the probability of getting a different coupon for each one, souldn't the expected value be:
$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $
The sum of the individual expected values should be the expected value of the sum, right? Just a thought. I am not sure, if that is right.
Perhaps, someone else could expand on this idea with more word-for-word, verbatim note-copying.
Virgil, you are right. The question is, can this be simplified to a closed form (non-summation) equation? Still trying to determine this.
Ken
the $ \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}\right) $ portion is the harmonic number. I do not believe their is a closed form to this number
AJ
