Expected number of rolls for an unfair die to get all possibile values at least once Suppose that we have a 6-sided unfair dice, where rolling a 1 is twice as likely as rolling any other number, and the other numbers have the same likelihood. What is the expected number of rolls to get each value at least once?
Thus, $$p(1) = 2/7\qquad p(2) = p(3) = \cdots = p(6) = 1/7$$
I understand how to approach this when the probabilities are the same, as it's just the Coupon collector's problem, but throwing in a non-uniform probability distribution throws me off. I know there is a general solution for this problem, but I can't seem to get any intuition as to why it's the case.

For $k=0,1,2,3,4,5$ let $S_{k}$ denote the status where $1$ has
not been rolled yet and exactly $k$ of the other numbers have been
rolled.
For $k=0,1,2,3,4,5$ let $T_{k}$ denote the status where $1$ has
been rolled and exactly $k$ of the other numbers have been rolled.
Let $\mu_{k}$ denote the expectation of rolls still to go starting in
status $S_{k}$.
Let $\nu_{k}$ denote the expectation of rolls still to go starting in
status $T_{k}$.
To be found is $\mu_0$ and we have the following relations:
$$\nu_{5}=0\tag1$$
$$\mu_{5}=1+\frac{2}{7}\nu_{5}+\frac{5}{7}\mu_{5}=1+\frac{5}{7}\mu_{5}\tag2$$
for $k=0,1,2,3,4$:
$$\mu_{k}=1+\frac{2}{7}\nu_{k}+\frac{k}{7}\mu_{k}+\frac{5-k}{7}\mu_{k+1}\tag3$$
$$\nu_{k}=1+\frac{k+2}{7}\nu_{k}+\frac{5-k}{7}\nu_{k+1}\tag4$$
Note that (2) leads to $\mu_5=3.5$. Substituting this together with $\nu_5=0$ in (3) and (4) gives you for $k=4$ two equalities that enable you to find $\nu_4$ and $\mu_4$. This proces can be repeated till you arrive at values for $\nu_0$ and (our final goal) $\mu_0$.
