Coupon Collector's problem, version with multiple coupons in a box

Question

I have been given a version of the coupon collector's problem for homework which includes a modification of the standard problem. Mainly, the modification is that there are n possible coupons, and I can get k (k<=n) coupons per box. Additionally, a coupon can be repeated in a box, so some of those k coupons can be the same coupon (for example: in a box with 10 coupons, you can get 6 different coupons while some of them can be found multiple times).

I need to find the probability distribution of this problem for the given Xn,k random variable which represents the number of boxes I need to buy to get all n coupons, and I can't use the traditional formula because of the modification. How could I implement this new rule?

Any help would be appreciated. Thank you in advance.

Answer 1

Henning's solution is correct; but since we know not only $P(X_{n,1}=x)$ for the standard problem, but also $P(X_{n,1}\le x)$ (see e.g. Probability distribution in the coupon collector's problem), we can also write this without a sum:

\begin{align} \def\stir#1#2{\left\{#1\atop#2\right\}} P(X_{n,k}\le a)&=P(X_{n,1}\le ak)\\ &=\frac{n!}{n^{ak}}\stir{ak}n\;, \end{align}

and thus

\begin{align} P(X_{n,k}=a)&=P(X_{n,k}\le a)-P(X_{n,k}\le a-1)\\ &=P(X_{n,1}\le ak)-P(X_{n,1}\le(a-1)k)\\ &=\frac{n!}{n^{ak}}\stir{ak}n-\frac{n!}{n^{(a-1)k}}\stir{(a-1)k}n\\ &=\frac{n!}{n^{ak}}\left(\stir{ak}n-n^k\stir{(a-1)k}n\right)\;. \end{align}

Answer 2

If you know how to calculate the distribution for the standard problem $X_{n,1}$, then you can simply say $$ P(X_{n,k}=a) = \sum_{j=0}^{k-1} P(X_{n,1}=ak-j) $$

Things get more involved if you're asked for the expectation instead of the distribution -- because there's a shortcut to computing the expectation for $k=1$ that doesn't seem to generalize as nicely here.