Suppose you are only allowed to move along the edges of a square. At each vertex, you have an equal probability of picking any of the available routes (including doubling back on yourself). Is the probability greater that you arrive back where you started or at the diagonally opposite corner?
Well this is quite straightforward: you start walking and hit the first corner. There is a 1/2 chance of proceeding to the next corner and a 1/2 half chance of going back to the start. They are equally probable and the expected number of moves for both will be 2 (although I'm not sure why - what kind of distribution is this?)
OK. Now the harder case. Go to a cube. Each vertex you have 3 options. Is the probability higher for returning to starting point or ending up at diagonally opposite corner. I'm not really sure how to start this - should it be something to do with Markov chains?
Can we generalise it to an $n$-dimensional cube?
Thanks.
end is the diagonally opposite vertex
p -> Probability to return at starting point before reaching at end from vertex at a distance 1
q -> Probability to return at starting point before reaching at end from vertex at a distance 2
A -> Probability to return at starting point before reaching at end from starting point
A = p and
p = 1/3 + 2/3 *q and
q = 2/3* p
So p = 1/3 +4/9p
5/9 p = 1/3
p = 3/5