I have come across variation of the coupon collector's problem that goes like this. The coupons are of $n$ different types and in infinite number (or sampled with replacement after each draw, where "draw" is defined next). At each draw, you get a variable number of coupons $X$, where $X$ is the number of consecutive successes before the first failure in Bernoulli trials with success $\Pr=q$. Whenever you get multiple coupons, the type of the first one is chosen uniformly at random, but the types of any additional ones are consecutive to the first. That is, if the first coupon is $i$, the next will be $i+1$, and so on. If you reach coupon $n$, the current draw stops and a new one starts. What is the expected number of draws until I have at least one coupon of each type?
I have tried to keep track of the expected number of "holes" left by the drawing process among the $n$ coupon types, but the variable number of coupons and the fact that there is a "wall" at $n$ seem to imply that you have to take into account which coupons you have and not only how many...
Any suggestions on how to make progress are welcome! I would be happy with an iterative solution or a solution involving computable quantities like the harmonic numbers in the original problem, even if an explicit expression cannot be found.
EDIT: I should have realized earlier that one can get an easy upper bound through the standard coupon collector's problem. In fact, if I assume that each draw includes only one Bernoulli trial with success $\Pr=q$, the number of draws needed to get at least one coupon of each type is $1/q$ times the number in the standard problem (which can be viewed as having Bernoulli trials with $\Pr=1$). Now I would be happy with a decent lower bound!