Talk:Coupon collector's problem

Regarding the numerical example in the first paragraph of this article:

The mathematical analysis of the problem reveals that the expected number of trials needed grows as O(nlog(n)). For example, when n = 50 it takes about 225 samples to collect all 50 coupons.

Although the O-notation is correct as it is, I think it would be better to be consistent with the following article and write n ln n + \gamma n + O(1). Especially since the numerical example is otherwise not reproducible. Unfortunately I do not know how to edit the very first paragraph of a page -- and couldn't find further information on this question. Can you help? (Netzwerkerin (talk) 22:08, 24 May 2009 (UTC))

To edit the very first paragraph of a page, click the "edit" button at the top of the page – the third of the buttons "article • talk • edit • ..." etc. (Or, go to My Preferences -> Gadgets, and in "User interface gadgets", turn on the option to "Add an [edit] link for the lead section of a page".) Shreevatsa (talk) 22:16, 24 May 2009 (UTC)

Different probabilities

There should be a section on the solution to the problem if not all coupons are equally distributed. 173.63.91.121 (talk) 00:49, 4 September 2009 (UTC)

Problem definition

I think the problem definition could be a bit clearer. It says you are collecting random coupons with replacement. Then you are asked how many coupons you are expected to draw before collecting all n of them.

It's very easy to misunderstand this (as I did): You draw one coupon, then you put it back, which means you now have 0 coupons. You can keep drawing them, but if you keep putting them back, you'll never collect more than 1. A better definition of the problem would be:

Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once.