Probability that a random jump decreases a distance to a previously specified point

Question

I am thinking about the following problem that probably is already solved somewhere. Perhaps you have an idea.. Lets assume that a particle jumps constantly in a random direction (isotropic) but with a constant jump length. The question is, as the title already says, what is the probability that a jump takes the particle closer to a specific point. This should depend of course on the distance to the point and the jump length. I would like to have a solution for 2 dimensions. Could you help me?

Answer 1

Below is a drawing of the point and the particle, then added a black circle centered at the point with radius equal to the distance (say $d$), and a green circle centered at the particle, with radius equal to the jump length (say $r$).

You want the length of the arc between the two green lines, as a proportion of $2\pi r$. Alternatively, the angle $\angle BAC$ as a proportion of $360^\circ$.

We have $$ \angle BAC = 180^\circ - \frac12\angle BPC = 180^\circ - \angle APC $$ Now note that $\triangle APC$ is an isosceles triangle with sides $d, d, r$. Thus we can use trigonometry to find the angle at $P$. The law of cosines yields $$ r^2 = d^2 + d^2 - 2d^2\cos\angle APC\\ \cos\angle APC = \frac{2d^2 - r^2}{2d^2} =1- \frac{r^2}{2d^2}\\ \angle APC = \arccos\left(\frac{2d^2 - r^2}{2d^2} =1- \frac{r^2}{2d^2}\right) $$ Thus the probability of the particle getting closer after a jump is $$ \frac{\angle BAC}{360^\circ} = \frac{180^\circ - \arccos\left(1-\frac{r^2}{2d^2}\right)}{360^\circ} $$ (If you do this on a calculator, make sure to either convert this answer to radians, or set the calculator to degrees.)