Figure 2.

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.

Figure 3.

Figure 4.

Figure 5.

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**.

The
probability
density function of where the point next reaches the x axis is thus
Cauchy.