# Billiards in a holey square

Suppose you start a point-billiard (or light ray) in a square at a random location, shooting off at a random angle, reflecting with angle-of-incidence equals angle-of-reflection. In general, because the point coordinates and direction vector are irrational with probability $$1$$, the path will fill the square.

Left: Starting from blue point, $$100$$ bounces; Right: $$500$$ bounces.
Now suppose you remove all rational points from the boundary. If the square is $$[-1,1]^2$$, remove points $$(\pm 1, r)$$ and $$(r, \pm 1)$$ for every rational $$r \in [-1,1]$$. So now the boundary has an infinite (but countable) number of holes: there is a hole at $$(\frac{1}{2},1),(\frac{1}{32},1),(\frac{171}{541},1)$$, etc.

Q1. Is it the case that the probability that the billiard / light ray escapes through a boundary hole is zero?

I believe the answer is Yes, but it certainly strains intuition, so I want to be certain.

Q2. Under what conditions on the starting position and initial ray direction will the escape probability be positive, presumably $$1$$?

In that situation, we have removed countably many points from the plane (countably many on each of countably many line segments). Thus the set of slopes connecting the starting point to one of the missing points is countable so the probability that a randomly selected departure angle will have one of those slopes is $$0$$.