(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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Revision as of 08:43, 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


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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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