I found this question on http://www.cis.jhu.edu/~xye/papers_and_ppts/ppts/SolutionsToFourProblemsOfRollingADie.pdf. It is Q(d):
Roll x times until getting all the faces from “1” to “6”, what’s the distribution of x?
The analytical derivation starts on pg 3 of 5 towards the middle of the page.
They show that
$$ P(\text{First x-1 rolls have all faces but 6}) = P_{x-1}\left(\cap_{i=1}^5 A_i \cap A_6^c \right) $$ where $A_i$ is the event that we toss the $i$-th number.
Why is this probability not $\left(\frac{5}{6} \right)^{x-1}$? Each of the first $x-1$ rolls are independent, and there is each a probability of $\frac{5}{6}$ of not tossing a $6$.
You are computing the probability that there are no $6$s in the first $x-1$ rolls, but they want to compute the probability that faces $1,2,3,4,5$ each appear at least once in the first $x-1$ rolls.