Expected number of dice rolls before rolling "1,2,3,4,5,6" QUESTION: I roll a single six-sided die repeatedly, recording the outcomes in a string of digits. I stop as soon as the string contains "$123456$". What is the expected length of the string?
My answer so far: My initial approach is to try and find the probability mass function. If we let the random variable $X$ be the length of the string, then we can easily calculate for $x\in\{6,\ldots,11\}$,
$$\mathbb{P}(X=x) = \left(\frac{1}{6}\right)^6$$
and zero for $x<6$. 
As soon as we reach $x\ge12$, we need to consider the probability that the final six rolls are "$123456$" but that sequence isn't contained in the string before that. I believe the result for $x\in\{12,\ldots,17\}$ becomes
$$\mathbb{P}(X=x) = \left(\frac{1}{6}\right)^6 - \left(\frac{1}{6}\right)^{12}(x-11).$$
Now for $x\ge18$, we will need an extra term to discount the cases when two instances of "$123456$" are contained before the final six rolls. And indeed every time we reach another multiple of six, we need to consider the number of ways of having so many instances of the string before the final six rolls. 
I've messed around with this counting problem but I'm getting bogged down in the calculations. Any input is appreciated to help shed some light on this. Thanks!

Just to point out a simple fact for independent, identical trials with finitely many outcomes: when a string $s$ of outcomes, like "123456", has no proper initial substrings which are equal to a final substring of $s$, then the expected waiting time for $s$ is just $1$/Freq($s$) where Freq($s$) is the probability that a random string of the length of $s$ is equal to $s$ -- in this case Freq(123456)= $1/6^6$. This follows from all the various methods discussed in the solutions and is also just a slight variation of the expected value computation for a geometric random variable. Modifications are needed when proper initial and final substrings coincide.
