So I'm pretty bad at math. At any rate, I'm trying to figure out the Coupon Collector's Problem for a set of 24. That is, how many items will I have to see if I get 1 randomly with replacement, before I've seen them all? I'm trying to see how many items I will need to see for 50% chance and for 99.9% chance. I saw something about n(log(n)), but I'm not quite sure what this gives me. Thanks!
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2An advice : as it is a very well documented problem (hundreds of web sites...), why don't you look first by yourself and come back, saying "I have seen this method/this result etc... on such and such site, but I don't understand such and such..." – Jean Marie Apr 24 '17 at 17:37
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The Erdős and Rényi limit, $\operatorname{P}(T < n\log n + cn) \to e^{-e^{-c}}, \ \ \text{as} \ n \to \infty$ used as an approximation with $n=24$, seems to give close to the correct answer for the median and about $5$ away for $99.9%$ – Henry Apr 24 '17 at 17:43