Hope this hasn't already been covered (I tried searching through some old coupon collector threads, but didn't see this question)Suppose we have the coupon collector problem, with N coupons, each with different probability p_i of turning up in each trial. We want to collect k coupons.What is the probability for each coupon that it is one of the collected ones? Edit:Just thought that for completeness I should probably explain coupon collector problem:N different types of coupons, each availible in a large population (bag) with a different probability.After each trial the coupon is replaced (by a new one from somewhere else or something, or you just keep track of the coupons you've had) so the probabilities are the same for each trial.You keep going until you have at least one of k different types of coupon.The usual question regards the expected number of trials.I want to know, given that you have already got k different types of coupon, what is the probability that a specific type is in your collection.

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I was given this in my probability class as a freshman in undergrad (my professor was a terrible man).Nevertheless, the way to look at this is to first understand the length of time it takes to draw all 'n' different types of coupons from a bin (with replacement). This is fairly simple, as we just need to consider the expected amount of time we need to wait (as in the number of draws we need to make) after we have collected 'k' coupons to get the (k+1) coupon.On our first draw, we have a probability of pulling a novel coupon of 1 (which is obvious, since we must pull a coupon, and we haven't collected anything yet), so we have an expected waiting time of 1 draw.For the next draw, we have (n-1) novel coupons remaining, but 'n' to choose from. Our conditional probability, then, of receiving a novel coupon, given the fact that we have drawn a novel coupon once already is (n-1)/n. Our expected waiting time, then, until our next novel coupon is n/(n-1). Similarly, for the next one after this, we expect to wait n/(n-2) given that we've already drawn 2 of the 'n' coupons.The total expected waiting time, then, is just n*(1+(1/2)+(1/3)+...+(1/n)).If we're looking to draw only 'k' of the 'n' total coupon types, we're looking at an expected waiting time equal to: 1 + n/(n-1) + n/(n-2) + ... + n/(n-k+1).**Now, for your question:If we're only looking to draw 'k' different coupon types from the 'n' total types, our expected number of draws is given by the summation above (denoted by **), which is also equal to the total number of coupons that we'll be holding.In particular, you are asking for the probability of holding a particular coupon. Well, there are 'n' types available, you're holding 'k' different types, and you have the following weightings of those types:type 1 (i.e. the first one you drew): by the time you have pulled the 'k-th' coupon from the bin, you have an expected value of the number of type 1 coupons equal to the following:(probability of drawing a type 1)*[(total number of trials needed to reach the 'k-th' novel coupon) - (number of trials which certainly went to pulling a coupon of a different type)]=(1/n)*[(**) - (k-1)], since we're conditioning on the fact that we know at least (k-1) of the draws were dedicated to pulling coupons of another type.type 2 (i.e. the second one you drew):=(1/n)*[(** - 1) - (k-1)]type 3:=(1/n)*[(**- 1 - n/(n-1)) - (k-1)]...type 'k':=1 (since you stop when you get it) The probability, then, of having any particular type of coupon in your collection is just equal to the expected number of that 'type' over the total number of types (which is different for each 'type,' and given above).