Suppose there are two different coupons and we want to find the average number of coupons drawn until we get the two coupons.
The probabilities of getting the two coupon types are not equal. Denote the probabilities by $p$ and $1-p$.
I thought about the next solution:
$$1 + p\cdot \frac{1}{1-p} + (1-p) \cdot \frac{1}{p}$$
Explanation: $1$ is for the first coupon. with probability p the fisrt one was of the first type, and now (by geometric distribution) we will have to try $\frac{1}{1-p}$ times in average. the third component $(1-p) \cdot\frac{1}{p}$ is the complementary situation.
Is this correct?
Community wiki answer so the question can be closed as answered: As noted in a comment, yes, that's correct.