Expectation to collect complete sets of multiple types of coupons? Say I am collecting coupons, where the coupons are of some $N$ types, each type having $M_N$ distinct coupons. The coupons "arrive" as a group of $N$, one each of the $N$ types. What is the expectation of the number of groups I must collect to see complete sets for all $N$ types?
E.g., say there are 4 types, with 10, 5, 2, and 2 of types 1 through 4 respectively.
I can easily calculate the probability of achieving a complete set of all types on a given group draw and use that to calculate the expected number of groups (~29.5 in this example), but I'm wondering if there's a more compact form to do this, similar to using the number of distinct coupons $X$ and the harmonic number ($X H_X$) for the usual canonical example.
 A: We can extend the standard inclusion-exclusion treatment to this case to get a finite multiple sum for the expected number of groups. The probability for all sets to be complete after $t$ groups is
$$
\textsf{Pr}(T\le t)=\prod_{i=1}^N\sum_{k_i=0}^{M_i}(-1)^{M_i-k_i}\binom{M_i}{k_i}\left(\frac{k_i}{M_i}\right)^t
$$
(where $0^0=1$). Thus the expected number of groups is
\begin{align}
\sum_{t=0}^\infty\textsf{Pr}(T\gt t)
&=\sum_{t=0}^\infty\left(1-\prod_{i=1}^N\sum_{k_i=0}^{M_i}(-1)^{M_i-k_i}\binom{M_i}{k_i}\left(\frac{k_i}{M_i}\right)^t\right)
\\
&=
\sum_{\vec k\ne\vec M}\sum_{t=0}^\infty\prod_{i=1}^N(-1)^{M_i-k_i-1}\binom{M_i}{k_i}\left(\frac{k_i}{M_i}\right)^t
\\
&=
\sum_{\vec k\ne\vec M}\frac1{1-\prod_{i=1}^N\frac{k_i}{M_i}}\prod_{i=1}^N(-1)^{M_i-k_i-1}\binom{M_i}{k_i}\;.
\end{align}
In your example, the outer sum has $11\cdot6\cdot3\cdot3-1=593$ terms. Here's code that calculates the sum. The result is
$$
E[T]=\frac{692692199243669826206869938184224933187}{23509569241912552284566542690183560696}\approx29.464\;,
$$
only about $0.174$ groups more than the expected $10H_{10}\approx29.290$ coupons required for the set of $10$.
