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
Writing it as:
$ n \cdot \sum_{i=n}^1\frac{1}{i}\! $
can help you see that the summation portion is the Harmonic number. I do not believe there is a closed form to this number.
AJ
You're right on the summation, Virgil -- silly mistake on my part. I'll fix my page.
-Brian
