(New page: '''E as opposed to P''' I am not entirely certain, but since <math>\frac{n - i + 1}{n}\!</math> is the probability of getting a different coupon for each one, souldn't the expected value...) |
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Perhaps, someone else could expand on this idea with more word-for-word, verbatim note-copying. | Perhaps, someone else could expand on this idea with more word-for-word, verbatim note-copying. | ||
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| + | Virgil, you are right. The question is, can this be simplified to a closed form (non-summation) equation? Still trying to determine this. | ||
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| + | Ken | ||
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| + | Writing it as: <br> | ||
| + | <math> n \cdot \sum_{i=n}^1\frac{1}{i}\!</math> | ||
| + | <br> can help you see that the summation portion is the Harmonic number. I do not believe there is a closed form to this number. | ||
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| + | AJ | ||
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| + | You're right on the summation, Virgil -- silly mistake on my part. I'll fix my page. | ||
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| + | -Brian | ||
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Latest revision as of 19:25, 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
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
