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.
