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The original coupon collector problem was answered here:

Probability distribution in the coupon collector's problem

The subset version of the coupon collector's problem is:

There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes $(N)$ one has to buy to collect $q$ types of coupons, where $q<m$?

My question is, what is the "closed form" (as a sum perhaps) of $P(N\leqslant n)$?

Note that the collector is seeking a specific given subset of the $m$ coupons, rather than any $q$ of them.

Jack Tiger Lam
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  • This doesn't seem materially different from the original problem. Can't we just assume that he starts out with all the unwanted coupon, and ask how long it takes to complete the set? – saulspatz Dec 18 '20 at 15:23
  • I do not see how that simplifies the problem. After all, the unwanted coupons will still show up even if you have already collected them. – Jack Tiger Lam Dec 18 '20 at 15:30
  • But then they're just duplicates, as in the original problem. – saulspatz Dec 18 '20 at 15:34

1 Answers1

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You can solve this exactly like the original coupon collector's problem. At the beginning, the probability that a randomly-selected coupon is one of the desired ones is $\frac qn$, so the expected waiting time is $\frac nq$. The expected waiting time for the second coupon is $\frac n{q-1}$ and so on, down to the last coupon, which has an expected waiting time of $n$. The overall expected waiting time is $$\sum_{k=1}^q\frac nk = nH_q\approx n\log q$$

saulspatz
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