A point is performing Brownian motion (a "drunkard's walk") in the plane. It is now at (0,1). It will eventually reach the x axis. When it next does, what is the probability that it will be at a point where x > 1?
We draw the green line x=1. By symmetry, the point is equally likely to reach the green line or the x-axis first. If it reaches the x-axis first, it does so with x<1. If it reaches the green line first, it is then equally likely to reach the x axis with x<1 or x>1. So the answer is 1/4.
To generalise the solution, we restate the problem slightly:
A point is performing Brownian motion in the plane. It is now at x, y. It will eventually reach the x axis. When it next does, what is the probability that it will be at a point where x > 0, as a function of x and y?
We start with Figure 3. If the point is on the y axis, the probability that when it next reaches the x axis it will have a positive x value is 1/2. If it is on the x axis, the probability is either 0 or 1, as shown the Figure.
If the point is midway between the "1/2" line and the "1" line of Figure 4, i.e. it is anywhere on the bisector of the angle between them, then it is equally likely to reach either of them next, so the probability of it reaching the x axis with positive x is the mean of 1/2 and 1, which is 3/4. This is shown in Figure 4.
We can repeat this angle-bisection indefinitely. If the point is on the bisector of the "1/4" line and the "1/2" line, the probability is 3/8, and so on.
Thus we see that the probability that when the point next reaches the x axis, it is on some specified section of the x axis, is proportional to the angle subtended at the point by that section. So it is given by 1/2 + arctan( x/y ) / π.
This is one of several pages on using symmetry in mathematics.