What is the expected number of tickets in the coupon collector's problem where duplicates are allowed, but some tickets have a different, arbitrary number of duplicates required until the collector's job is finished?
Example: Pool of five different tickets where each ticket has their own number of duplicates: $1, 2, 4, 6, 7$ tickets each. So having only one of each ticket is not enough, but specific tickets need multiple. Of course, each ticket still has the same probability to draw throughout.
One approach is via exponential generating functions, EGFs for short. We will use the example problem for illustration, but the method is easily generalized. Readers unfamiliar with generating functions can find a number of resources in the answers to this question: How can I learn about generating functions?.
Let's say $p_n$ is the probability that we have a complete set of tickets (1 of type 1, 2 of type 2, 4 of type 3, 6 of type 4, and 7 of type 5) on or before the $n$th draw, for $n \ge 0$. In general, the exponential generating function of $p_n$ is defined as $$f(x) = \sum_{n=0}^{\infty} \frac{p_n}{n!} x^n$$ For convenience, we also define functions $e_b(x)$ for $b=0,1,2,\dots$ by $$e_b(x) = \sum_{n=0}^b \frac{1}{n!} x^n$$ Using these definitions and the fact that the probability of drawing any single type of ticket is $1/5$ (and some experience with generating functions), $$f(x) = g_1(x) \cdot g_2(x) \cdot g_4(x) \cdot g_6(x) \cdot g_7(x)$$ where $$g_i(x) = e^{x/5} - e_{i-1} (x/5)$$ Let's say $X$ is the number of the first draw on which we have a complete set of tickets, and let $q_n = P(X>n)$, so $q_n$ is the probability that we do not have a complete set of tickets on or before the $n$th draw. Then $q_n = 1 - p_n$, and since $f(x)$ is the EGF of $p_n$, the EGF of $q_n$ is $$h(x) = e^x - f(x)$$ By a well-known theorem, $$E(X) = \sum_{n=0}^{\infty} P(X > n) = \sum_{n=0}^{\infty} q_n$$ There is an easy trick by which we can extract this infinite sum from $h(x)$. Since $$n! = \int_0^{\infty} x^n e^{-x} \; dx$$ we have $$E(X) = \int_0^{\infty} e^{-x} h(x) \; dx$$ I suppose it is possible to evaluate this integral in our example by pencil and paper, but I used Mathematica for the evaluation, with the result that the expected number of draws to get a complete set of tickets is $$E(X) = \boxed{40.6861}$$
