Symmetry of random walk in a random environment Consider the square grid $$\{-n, -n+1, ..., n-1, n\} \times \{-n, -n+1, ..., n-1, n\} \subset \mathbb{Z}^2$$ As usual, connect $(i,j)$ to $(i',j')$ with an edge if $|i-i'|+|j-j'|=1$. 
We randomly specify the environment as follows: At each vertex we designate one of the 4 outgoing directions to have probability 1/2, and the remaining 3 outgoing directions to have probability 1/6. The choice of which outgoing edge gets probability 1/2 is done uniformly and independently at random for each vertex. 
Then we consider the random walk starting at $(0,0)$ that moves from each vertex in each direction with the chosen probabilities $1/2$ or $1/6$. 
I want to show that for n large, if I fix a random environment and start the random walk at $(0,0)$, then the probability that the random walk hits the top edge of the grid before any of the other edges is very likely to be close to 1/4.  
 A: Your random walk can be thought of as a isotropic diffusion with a random iid drift. This is very close to the situation studied here: http://link.springer.com/article/10.1007/s00222-005-0477-5 (ok, they are replacing the discrete model with a continuum one, but this isn't really important and you can find previous work on the discrete model in their references if this makes a difference to you).
Unfortunately, that paper is for dimensions $d>2$ and is perturbative: the drift is required to be smaller than $\delta$ for some very tiny $0<\delta\ll1$. Even in the perturbative regime, the two dimensions is critical in some sense and difficult to analyze. I know some colleagues who are trying to analyze the perturbative regime in $d=2$ and claim to have results "coming soon". This would answer your question if $1/2$ was replaced by $1/4+3\delta$ and the other edges were $1/4-\delta$. 
Analyzing the situation outside the perturbative regime seems to be a hugely difficult task and no one knows what to do.
