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Ken | Ken | ||
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| − | + | this is also <math>n \cdot \left(\frac{1}{n} + \frac{1}{n-2} + \cdots + \frac{1}{1}\right) </math>. The <math>\left(\frac{1}{n} + \frac{1}{n-2} + \cdots + \frac{1}{1}\right) </math> portion is the harmonic number. I do not believe there is a closed form to this number | |
AJ | AJ | ||
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Revision as of 17:20, 6 October 2008
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
this is also $ n \cdot \left(\frac{1}{n} + \frac{1}{n-2} + \cdots + \frac{1}{1}\right) $. The $ \left(\frac{1}{n} + \frac{1}{n-2} + \cdots + \frac{1}{1}\right) $ portion is the harmonic number. I do not believe there is a closed form to this number
AJ
