Forgetful Coupon Collectors: Urn Selection with Null Option The famous problem of the Coupon Collector caught my attention recently.  Suppose we change this problem so that there is some element of probability involved.  For example, suppose that there is a chance, upon any selection step, that no coupon will be selected at all.  That is, they forget to draw a coupon for some reason or another.  Is there a similar way of addressing the problem, stated in the same terms, with this additional element?  I imagine it will just be the expected value of the current solution, but I was curious if there is more to it that I am missing.  
Moreover, I am curious if there are well-known examples of similar problems, possibly addressing what happens as more probabilistic elements are introduced, or maybe as the number of coupons is distributed into an increasing number of urns.  Essentially, I am interested in what extensions to this problem exist, with a particular focus on probabilistic extensions.
 A: One way of modelling your null option would be to introduce another coupon into the collection. You can assign this null coupon whatever probability you like, and then scale the probabilities of the original coupons accordingly. This results in a collection of coupons with non-uniform probabilities, and as it turns out, this was one of the very first generalizations of the coupon collector's problem. Here is a link to the English translation of the original 1934 paper by von Schelling on this problem. Of course, you'll probably want to remove the influence of your null option on the expectation from von Schelling's computation, but I think this would be not be too difficult.
As far as adding more probabilistic elements goes, you might be interested is this paper from 2013, which analyzes the coupon collector's problem when not only are coupons obtained at random, but the numbers of coupons themselves to make a collection are random.
A: By linearity of expectation, the expected number of draws required if each draw yields a coupon with probability $p$ is $\frac{nH_n}p$ (where $nH_n$ is the expected number of draws in the standard coupon collector's problem), since all completion probabilities in the standard solution of the standard problem are multiplied by $p$.
