Number of attempts to see all faces $1, 2, 3, ..., 6$ when rolling a dice? Recently I have been thinking about the following random experiment: we repeatedly roll a dice until we see all the faces $1, 2, 3, 4, 5, 6$ of the dice at least once. 
Let $X$ = number of attempts necessary to see all the faces.
Obviously $X(\Omega) = \{6, 7, 8, ...\}$
Can we describe precisely the law (and maybe also the expected value) of $X$?
(It did not look as simple as it seems, thus this question).
Note: linked to Expected time to roll all 1 through 6 on a die but in my question here, the law of $X$ is also discussed ($P(X=k)$ for $k \geq 6$).

Set up the problem as a Markov chain with states being the number of faces that have shown up to present, thus from 1 to 6. The transition matrix (with columns adding to 1) is $$P=\frac{1}{6}\begin{pmatrix}1&0&0&0&0&0\\5&2&0&0&0&0\\0&4&3&0&0&0\\0&0&3&4&0&0\\0&0&0&2&5&0\\0&0&0&0&1&6\end{pmatrix}$$ The 'absorbing' state is 6, so let $P=\begin{pmatrix}Q&0\\x&1\end{pmatrix}$ where $Q$ is a $5\times5$ sub-matrix of $P$ that represents the chain for the first five states. The required expectation is given by $$(I+Q+Q^2+\cdots)\mathbf{i}=(I-Q)^{-1}\mathbf{i}=6\begin{pmatrix} \frac{1}{5} & 0 & 0 & 0 & 0 \\ \frac{1}{4} & \frac{1}{4} & 0 & 0 & 0 \\
 \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
 1 & 1 & 1 & 1 &1 \end{pmatrix}\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix}=\begin{pmatrix}\frac{6}{5}\\ \frac{3}{2}\\ 2\\ 3\\ 6\end{pmatrix}$$ This vector signifies the expected number of steps each state has been visited before reaching the sixth state. Hence their sum is what is needed: $137/10$. Since this model assumes it starts with state 1 already, you need to add 1 for the first throw. Hence the expected number of times before all six faces appear is $$147/10.$$
