In a simple random walk on $\mathbb Z ^2$, starting in the origin, I'm trying to find out the probability of visiting the point (1,0) before the walk returns to the origin. Means, if $O$ is the origin point, and $p$ is the point (1,0), The question is what is the probability of $\mathbb P _O(\tau_p<\tau_O^+)$, when $\tau$ is the stopping time of a point, and $\tau^+$ is the stopping time for $t>0$. I tried to apply mirroring strategies simply by counting symmetric paths etc., but it doesn't work. I also tried to project the walk on each axis and analyze it separately (playing separately with the axes made me guess that the answer should be 0.5, but I still can't see how to prove it rigorously, and I'm not sure it's indeed the right answer).
Probability of reaching a point in a simple random walk on $\mathbb Z ^2$
random walk
Best Answer
The generating function for the number of RW's that return to the origin after $N$ steps is given by $a0 = 1+4z^2+36z^4+... ={}_2F_1(1/2,1/2;1;(4z)^2)$, see A002894. The generating function of RW's that are at point (1,0) after $N$ steps is $a1=z+9z^3+100z^5+\cdots = z {}_3F_2(1,3/2,3/2;2,2;(4z)^2)$, see A060150. Plug these two into the expression $\frac{1}{2(a0-a1)}-\frac{1}{2(a0+a1)}$ by Rubin and Weiss. Insert $z=1/4$ into the generating functions to obtain probabilities, which means divide the number of walks by $4^N$ to weigh them with the $1/4$ probability of the 2D square lattice. Then $a0+a1\to \infty$, and with a little bit of Gamma-function manipulation one sees that $a0-a1\to 1$ as $z\to 1/4$. This is the conjectured result, i.e., probability of $1/2$. The full analysis is in A337869.