This is an instance of the "coupon collector's problem" in which not all the "coupons" have the same probability of appearance. Here the "coupons" are boxes of cereal, each having a prize in it. I have been struggling with it for some time.
Suppose there are 7 items (e.g., Snow White's 7 dwarfs). One per box of cereal. Except for Dopey, they are distributed evenly over the boxes of cereal. But the number of Dopeys is 1/2 the number of each of the others. The question is: What is the expected value of the number boxes of cereal that should be bought to get a complete set of dwarfs? As an example, if they were all evenly distributed, I get that you would need to buy 1+7/6+7/5+7/4+7/3+7/2+7/1 = 18.15 boxes.
Here is a solution based on exponential probability generating functions. Readers interested in learning about generating functions can find many resources here: How can I learn about generating functions?
Number the "coupon" types from $1$ to $7$, with Dopey as type number $1$, and associated probabilities $p_1, p_2, p_3, \dots , p_7$. From the problem statement we find $p_1 = 1/13$ and $p_i = 2/13$ for $2 \le i \le 7$. Let's say $T$ is the number of the draw on which we first have a complete set of coupons.
If $T \le n$ then we have a compete set of coupons on draw $n$ or earlier. The exponential generating function of $P(T \le n)$ is $$f(x) = \prod_{i=1}^7 (e^{p_i x} - 1) = (e^{p_1 x} -1) (e^{p_2 x} -1)^6$$ since $p_i = p_2$ for $i \ge 2$. Since $P(T > n) = 1 - P(t \le n)$, the EGF of $P(T > n)$ is $e^x - f(x)$. We are interested in $P(T > n)$ because by a well-known theorem, $$E(T) = \sum_{n \ge 0} P(T > n)$$ Since we know the EGF of $P(T > n)$, we can relate this infinite sum to an integral. Because $$\sum_{n=0}^{\infty} P(T>n) \frac{1}{n!} x^n = e^x - f(x)$$ and $$\int_0^{\infty} x^n e^{-x} \;dx = n!$$ we have $$\begin{align} E(T) = \sum_{n=0}^{\infty} P(T>n) &= \int_0^{\infty} e^{-x}(e^x - f(x)) \;dx \\ &= \int_0^{\infty} e^{-x}(e^x - (e^{p_1 x} -1) (e^{p_2 x} -1)^6) \;dx \\ & \approx \boxed{20.3579} \end{align}$$ on substituting $p_1 = 1/13$ and $p_2 = 2/13$. I admit to having used a computer algebra system to evaluate the integral, but a paper and pencil computation should not be too difficult.