We consider a generalisation of the classical coupon collector problem. We define a super-coupon to be any s-subset of a universe of n coupons. In each round, a random r-subset from the universe is drawn and all its s-subsets are marked as collected. We show that the time to collect all super-coupons is \((1 + o(1)) \left( {\begin{array}{c}r\\ s\end{array}}\right) ^{-1}\left( {\begin{array}{c}n\\ s\end{array}}\right) \log \left( {\begin{array}{c}n\\ s\end{array}}\right) \) on average and has a Gumbel limit after a suitable normalisation. In a similar vein, we show that for any \(\alpha \in (0, 1)\) , the expected time to collect \((1 - \alpha )\) -proportion of all super-coupons is \((1 + o(1)) \left( {\begin{array}{c}r\\ s\end{array}}\right) ^{-1}\left( {\begin{array}{c}n\\ s\end{array}}\right) \log \left( {\frac{1}{\alpha }}\right) \) . The \(r = s\) case of this model is equivalent to the classical coupon collector model. We also consider a temporally dependent model where the r-subsets are drawn according to the following Markovian dynamics: the r-subset at round \(k + 1\) is formed by replacing a random coupon from the r-subset drawn at round k with another random coupon from outside this r-subset. We link the time it takes to collect all super-coupons in the \(r = s\) case of this model to the cover time of random walk on a certain finite regular graph and conjecture that in general, it takes \((1 + o(1)) {{\frac{r}{s}}\left( {\begin{array}{c}r\\ s\end{array}}\right) ^{-1}}\left( {\begin{array}{c}n\\ s\end{array}}\right) \log \left( {\begin{array}{c}n\\ s\end{array}}\right) \) time on average to collect all super-coupons.
