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May 24, 2012 at 21:22 comment added Douglas Zare Instead of unfolding the torus in both directions, I suggest estimating the probability by unfolding the torus in one direction. That is, estimate the probability that the loop will wind around the torus in one direction (say the $x$ direction). The events that the loop winds in the $x$ or $y$ directions are not independent, but one or the other must occur for the loop to be nontrivial, so $P(x) \le P(x~\text{or}~y) \le 2 P(x)$.
May 24, 2012 at 4:58 comment added Chris I realize I did not specify before: I am specifically interested 2 dimension random walk, so the walk will be recurrent.
May 23, 2012 at 19:30 comment added Will Sawin The probability that the walker never returns to the start is well-known to be $0$ in dimensions 1 and 2 and positive in higher dimensions.
May 23, 2012 at 19:29 comment added Will Sawin The limit as the size of the torus goes to ∞ depends heavily on the dimension, because it is the same as the probability that the walker never returns to the start. Proof: Obviously this forms a lower bound because if the walker never returns to the start then it cannot land in the trap at the origin. However, if the size of the torus is at least N, then the probability that the walker takes more than N steps to return to the start is an upper bound. As N goes to ∞ the probability gets squeezed between these two.
May 23, 2012 at 19:22 answer added Robert Israel timeline score: 2
May 23, 2012 at 19:00 history asked Chris CC BY-SA 3.0