What is the probability that a unit disk centered at a random point $P$ has exactly two lattice points in its interior?

Question

A point $P$ is chosen at random in the coordinate plane. What is the probability that the unit disk with center $P$ contains exactly two lattice points in its interior?

In short I've been trying to wrap my head around this problem but I have not been able to produce anything of value. Below I provide my thinking:

In the Cartesian plane we'll use grid lines as reference points for our coordinate system (each integer is represented by the intersection of two grid lines).

As you can see in the image above (I just worked on first quadrant for sake of simplicity), we can achieve our scope if and only if the point $P$ lies exactly on any grid line but without being on any crossroads or exactly strictly below any vertical line or "strictly" to the left or right of any horizontal grid line.

(I know, I am abusing the meaning of the word "strictly".)

Answer 1

There is no uniform measure on the plane, so strictly speaking, one cannot make rigorous sense of choosing a random point in the plane. The number of points contained within the unit circle centered at a point $(x, y) \in \Bbb R^2$ is, however, invariant under adding integers to $x, y$, so it preserves the spirit of the problem to reformulate it as follows:

What is the probability $\color{#bf0000}{P_2}$ that the unit circle centered at a point in the unit square $[0, 1) \times [0, 1)$ (randomly chosen with uniform probability) contains exactly two lattice points?

(More symmetrically, one could ask this about a random point in the compact quotient $\Bbb R^2 / \Bbb Z^2$, which is just the torus.) Now, the unit circle centered at a point $P$ containing exactly two lattice points is equivalent to $P$ being with $1$ unit of exactly two lattice points. So, if we draw the unit square and the unit disks centered at the lattice points at the corners of the square, we see that the set of points in the square within $1$ unit of exactly two lattice points is precisely the red region in the following diagram:

Since the unit square has area $1$, the probability that a uniformly selected random point is in the red region is precisely the area of that region, and some elementary geometry gives that it is $$\color{#bf0000}{\boxed{P_2 = 4 - \sqrt{3} - \frac{2 \pi}{3} \approx 0.174.}}$$

Similar, the probability that the unit circle centered at a point contains $3$ lattice points is $$\color{#00bf00}{P_3 = -4 + 2 \sqrt{3} + \frac{\pi}{3} \approx 0.511,}$$ and the probability that the unit circle centered at a point contains $4$ lattice points is $$\color{#0000bf}{P_4 = 1 - \sqrt{3} + \frac{\pi}{3} \approx 0.315.} .$$

This recovers, for example, that the expected number of lattice points is just $$2 \color{#bf0000}{P_2} + 3 \color{#00bf00}{P_3} + 4 \color{#0000bf}{P_4} = \pi .$$

Answer 2

Consider the square $x_0=(0,0)$, $x_1=(1,0)$, $x_2=(0,1)$ and $x_3=(1,1)$.
If $P$ lies in/on this square, the lattice points in the circle can only be one of the four points above.
So let $P = (a,b)$, then $x_i$ lies in the interior of the circle

• iff $d(P,x_i) < 1$
• iff $P$ lies in the interior of the circle around $x_i$.

As this problem is invariant under rotation by $90^\circ$, we can consider the case $x_0$ and $x_2$ and multiply by $4$ in the end.
If $x_0$ and $x_2$ do not lie in the circle around $P$, $P$ lies above the circles around $x_0$ and $x_1$, convince yourself that $x_2$ and $x_3$ then lie in the interior.
So we are interested in the point where $y=\sqrt{1-x^2}$ and $y=\sqrt{1-(x-1)^2}$ intersect, which is at $(x,y)=(\frac{1}{2},\frac{\sqrt{3}}{2})$, then the area we are interested in is $$O = 1 - 2 \int_0^\frac{1}{2} \sqrt{1-x^2} \textrm{d}x = 1 -\frac{ \sqrt{3}} {4}- \frac{\pi}{6}.$$ Now multiply by four, convince yourself that the areas are mutually exclusive for two different pairs $x_i,x_j$ and $x_{i'},x_{j'}$ and that $x_1$ and $x_2$ cannot lie in the same circle.

Then the answer is $4-\sqrt{3} - \frac{2\pi}{3}$.

Please note that convincing yourself is easy by drawing the four unit circles all at once. Furthermore, you know have to translate this probability from one unit square to the infinite two-dimensional plane.