Approximation for Coupon Collector Probability

Asked Apr 27 '15 at 11:39

Active Apr 28 '15 at 06:18

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I was calculating the probability to draw all items in a list of N items, by picking one randomly, replacing it in the list, etc. (Edit: It's the Coupon collector Problem)

I found this formula after n picking.

$$ \sum_{i=0}^N (-1)^{N-i}{\binom{N}{i}}\left(\frac{i}{N}\right)^n $$

But, I have to calculate this formula with N > 1000000000 in a java program so to avoid a huge loop (the sum). I'm looking for a simplification, any approximation formula will help me.

edited Apr 28 '15 at 06:18

asked Apr 27 '15 at 11:39

Crypto

http://en.wikipedia.org/wiki/Coupon_collector%27s_problem – JP McCarthy Apr 27 '15 at 12:18
You're right it is the same problem, but I don't see any formula I can use on this page. – Crypto Apr 28 '15 at 06:16
See 5.2.1 here: http://arxiv.org/pdf/1504.05387.pdf – JP McCarthy Apr 28 '15 at 09:30
For large $n$, see also http://math.stackexchange.com/questions/1755398: In this case, the probability not to draw all items is approximately the expected number of undrawn items. – joriki Jun 02 '16 at 10:28

Approximation for Coupon Collector Probability

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