Random walking and the expected value I was asked this question at an interview, and I didn't know how to solve it. Was curious if anyone could help me.
Lets say we have a square, with vertex's 1234. I can randomly walk to each neighbouring vertex with equal probability. My goal is to start at '1', and get back to '1'. How many average walks will I take before I return back to 1?

After an even number of steps, you are either at 1 or at 3; after an odd number of steps, you are either at 2 or at 4. Hence you cannot return home after an odd number of steps. To return home for the first time after $2n$ steps, you must have had "bad luck" add all odd steps (i.e. have gone to 3 instead of 1) and the choice in even steps (when you are at 4 - except at the beginning) does not matter. Therefore $P(X=2n) = 2^{-n}$ and $E(X) = \sum_{n=1}^\infty 2n P(X=n) = \sum_{n=1}^\infty n 2^{-n}$.
In case you dont recognize that series in an interview situation, observe that $E(X) = 2E(X)-E(X)= \sum_{n=0}^\infty (n+1) 2^{1-n}-\sum_{n=1}^\infty n 2^{1-n}=\sum_{n=0}^\infty 2^{1-n}=4$. Note: 2E(X)=2 $\sum_{n=1}^\infty n 2^{-n}$$=\sum_{n=1}^\infty n 2^{1-n}$
