What is the average pathlength and probability to randomly traverse any given graph? 
To get specific first off, it's about this graph:


I want to get from $A$ to $B$. Every edge has the same length (e. g. 1 m). The shortest walk from $A$ to $B$ is easily found ($A-2-5-B$ and $A-3-6-B$). But how can I calculate the average length of a walk and its probability, when I am navigating randomly through this graph (what is the length/probability of $n$ steps)? At every node, there is a chance of $p= \frac{1}{degree}$ to choose any next edge (also back, where I just came from).
For example, there would be a chance of walking in circles for a long time, but probability decreases - also to get straight through.
I wrote a computer simulation to find a statistic answer: After a megaattempt (1000000 attempts), it averaged at about 20.329 edges for this specific graph here. 
I'd also like to know how to calculate such things in general, of course ;)
Path vs. Walk: Some authors (e.g. Bondy and Murty 1976) use the term "walk" for a path in which vertices or edges may be repeated, and reserve the term "path" for what is here called a simple path. - https://en.wikipedia.org/wiki/Path_(graph_theory)

It's difficult to claim an average because there exist many possible infinite length paths that never reach $B$.  You may want instead to calculate the probability that you have reached $B$ after $n$ steps.  
@draks ... is on to something, but if you're going to determine the probably that you have "seen" node $B$ after $n$ steps, then you definitely need to include the probability as weights in the adjacency matrix.  
Question: For the given graph $G$, what is the probability of having seen node $B$ at the $n^{th}$ step?
Answer: Let us write the adjacency matrix of $G$ weighted with edge probabilities as
$$M = \begin{bmatrix} 
    0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 & 0 & 0   \\
    \frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3} & 0 & 0 & 0  \\
    \frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0  \\
    \frac{1}{3} & 0 & \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3} & 0  \\
    0 & \frac{1}{4} & \frac{1}{4} & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{4}  \\
    0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4}  \\
    0 & 0 & 0 & \frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3}  \\
    0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0   \\
\end{bmatrix}$$
Let $N$ be the matrix $M$ with the last row and column removed.  Then the probability of having seen $B$ at step $n$ is then equal to 1 minus the sum of the first row of $N^{n}$ (because that is the row corresponding to the node we started from $A$).  
Here's a pretty picture of how the probability changes as $n$ increases:

