Random solving of a Rubik cube . After playing a little with a Rubik cube I thought of the following problem : 

Suppose we start with a solved Rubik cube (a general one , with $n^3$ cubes) . Then we choose one of the moves , each having a probability of $\frac{1}{6n}$ of being chosen , and apply it on our cube . We continue doing so ( choosing a new move randomly and then applying it and so on...) , until we reach the solved position again . What is the expected number of moves needed to solve the cube in this way ?

I thought about it and I think the answer should be $\infty$ (at least for $n \geq 3$) because there are a lot of random sequences of moves that will take a long time to solve the cube but I don't have a rigorous way to prove it.
For $n=2$ I'm not that sure if the answer should be finite or infinite (but I still tend to think it's infinite).
Thank you for your time to help me!

Consider a simpler scenario: suppose you have a biased coin, which shows Heads with probability $p > 0$. What is the expected number of attempts required to throw a Head? The probability that exactly $n$ throws are required is $q^{n-1}p$, where $q = 1-p$. So the expected number of throws is $\sum_{n=1}^\infty nq^{n-1}p = \dfrac{p}{(1-q)^2} = \dfrac{1}{p}$. Note in particular that this is finite!
Now, let $N$ be the maximum number of moves required to solve an $n \times n \times n$ Rubik's cube. Thanks to the efforts of countless dedicated investigators, we know that for $n=3$, we have $N=20$ according to the half-turn metric, and $N=26$ according to the quarter-turn metric (see for instance this link). But the exact value of $N$ doesn't matter; the number of positions is finite, and therefore the maximum number of moves required to solve a solvable position is also finite.
Suppose the number of moves available at each turn is $m$. You say that $m=6n$, and I won't argue with that; again, the important thing is that $m$ is finite. Then all you have to do to solve a position in $N$ moves or less is to pick the optimal move each time. The probability of doing this by chance is at least $p=m^{-N}$.
So if you consider a "coin flip" to be $N$ random moves in your Rubik's cube, then the expected number of coin flips is at most $\dfrac{1}{p} = m^N$. So the expected number of moves is at most $\dfrac{N}{p} = Nm^N$.
