Tossing a ring over pegs: Snaring exactly one peg?

Question

Let there be a point peg at every $\mathbb{Z}^2$ lattice point. Let a ring be a radius $r$ circle.

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Q1. Which value of $r$ maximizes the chance that a randomly placed ring will enclose exactly one peg?

Small $r$ may capture no pegs; large $r$ may capture more than one peg.

I know this is elementary, but I am not seeing an easy route to calculate $r$. My real question is this generalization:

Q2. Which value of $r$ maximizes the chance that a randomly placed $(d{-}1)$ sphere in $\mathbb{R}^d$ will enclose exactly one lattice point of $\mathbb{Z}^d$?

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Questions inspired by "ring toss":

         
          ^{(Image from gameplanent.)}

Added. Following @GussB's suggestion, I compute $r=0.541$:

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Answer 1

In order to evaluate the probability for a fixed $r$, you can draw a circle of radius $r$ around each lattice point. The proportion of $\mathbb{R}^2$ that is in precisely one circle is the probability you are looking for.

Note that it suffices to compute this proportion inside the square (0,0), (0,1), (1,1), (1,0) due to the fact that the pattern is repeated. Moreover for optimal $r$, we must have $0 < r < \frac12\sqrt{2}$, so that each point of the square is in either 0, 1 or 2 circles.

In this square, there are four quarters of a circle, with a combined area of $\pi r^2$. From that we have to subtract the overlapping area twice. Let $A_1(r)$ denote the intersection area of two circles of radius $r$ whose centers are at distance $1$ of each other, which can be computed exactly. Then the probability you are looking for is $$\pi r^2 - 4 A_1(r),$$ which should then be maximized on the interval $0 < r < \frac12\sqrt{2}$ to answer your first question.

This approach generalizes to dimension $d$, although you may have to take into account more than 2 spheres intersecting.

Answer 2

This question can be cast into the problem of finding the expectation number of lattice points say, $L_d$ that fall inside a hypersphere in $d$-dimensions. Now for large values of $d$ $$E[L_d] \approx V_d(r)$$ Where $V_d(r)$ is the volume of the hypersphere with radius $r$. In our case $E[L_d] = 1$. Therefore, solving $r$ for $V_d = 1$ should give us the solution which will become increasingly accurate as $d$ and $r$ increases. Hence, $$ V_d = \frac{\pi^{d/2}r^{d/2}}{\Gamma(\frac{d}{2}+1)}$$ $$r = \frac{\Gamma(d/2+1)^{2/d}}{\pi}$$

The error on $r$ will approach zero as $d \rightarrow \infty$ as $ L_d - V_d \approx O(r^{d-2}) $(check paper 1).

Also, notice that the value of $r$ satisfying the above conditions are much larger than $1$ for $d > 10$. Again pushing down the error. Hence, for $d \rightarrow \infty $, the obove relation holds.

The last thing is that, Is this expectation translation-invariant? And again you may verify this by noticing that the lattice points are uniformly distributed samples over $R^{n}$ and the number of lattice points in each sphere centered at a lattice point is the same. And hence this expectation value holds.

In the case of $d=2$, $r = 0.564$ (with respect to your answer to GausB), which is already quite a small error, so we can expect this error to go down fast.

NOTE: I think I am not able to make the translation invariance point clearly but it will hold and the arguments I gave were based on chapter-8 of Introduction to Geometric Probability by Klein and Rota.

paper-1 : https://www.jstor.org/stable/2003508?seq=1

paper-2 : https://en.wikipedia.org/wiki/Gauss_circle_problem

paper-3 : http://www.dtc.umn.edu/~odlyzko/doc/arch/high.dim.spheres.pdf