Expected number of rolls for an unfair die to get all possibile values at least once
Consider the die as a $7$-sided die with two $1$s and wait for all sides to appear. With probability $\frac27$, you unnecessarily waited for the second $1$ for an expected number of $7$ rolls at the end. Correcting for that yields
$$7H_7-\frac27\cdot7=\frac{323}{20}\;.$$
We describe the scenario using an eleven state markov chain. (tedious, I know...if all six die results have different probabilities though, could go upwards of $2^6$ states though, so at least we could trim it down to only eleven...)
Here is a transition diagram.
To give names to states, I will refer to them as $(yes,k)$ for if $1$ has been seen and $k$ not-1 numbers have been rolled, and $(no,k)$ for if $1$ has not been seen and $k$ not-1 numbers have been rolled.
The transition matrix with order of rows/columns as $(yes,5),(yes,0),(yes,1),\dots,(yes,4),(no,1),(no,2),\dots,(no,5)$
$$\begin{bmatrix} 1&0&0&0&0&1/7&0&0&0&0&2/7\\ 0&2/7&0&0&0&0&0&0&0&0&0\\ 0&5/7&3/7&0&0&0&2/7&0&0&0&0\\ 0&0&4/7&4/7&0&0&0&2/7&0&0&0\\ 0&0&0&3/7&5/7&0&0&0&2/7&0&0\\ 0&0&0&0&2/7&6/7&0&0&0&2/7&0\\ 0&0&0&0&0&0&1/7&0&0&0&0\\ 0&0&0&0&0&0&4/7&2/7&0&0&0\\ 0&0&0&0&0&0&0&3/7&3/7&0&0\\ 0&0&0&0&0&0&0&0&2/7&4/7&0\\ 0&0&0&0&0&0&0&0&0&1/7&5/7\end{bmatrix}$$
Following properties of absorbing markov chains, we look to the fundamental matrix $(I-R)^{-1}$
Using a calculator,
Now, since we start in state $(yes,0)$ with probability $2/7$ and start in state $(no,1)$ with probability $5/7$, multiplying our fundamental matrix by the corresponding vector, we get:
$(7/5+7/4+7/3+7/2+7)2/7 + (7/12+7/5+14/5+98/15+7/6+14/15+7/10+7/15+7/30)5/7 = \frac{303}{20}$
Since this calculation assumed however that we started in one of those two states however, the initial roll must also be considered, so our final expected time is found by adding one to it, giving $\frac{323}{20} = 16.15$ rolls on average.
As a side note, the method employed is the same as that of dr.hab, however with the use of technology and the language of matrices, we can conveniently write and compute the result.