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Below, 'circle' refers to a disk.

After seeing the pepperoni pizza problem, I came up with my own question regarding random circles inside circles.

Let $G$ be a circle of unit radius. Suppose we choose $2$ points uniformly randomly from the interior of $G$ (the probability that our point lies inside a circle $C \subseteq G$ of area $A$ is $A/\pi$). Further, suppose these two points are the centers of circles $P,Q$ whose are radii chosen independently and uniformly randomly from $[0, 1]$. What is the probability that both $P$ and $Q$ lie entirely within $G$? What is the probability that $P,Q$ do not share interior points/accumulation points (in other words, their intersection is finite)?

The first question is quite easy and we need only compute $\mathbb{E}(\text{one circle lies within $G$})$ and then square it. This turns out to be $$\left (\int_0^1(1-t)2t \, \mathrm{d}t \right )^2 = \frac 19.$$

The second (bolded) question is harder. I approach this problem fixing $G$ to be centered at $(0,0)$ and $P$ centered on the left-sided $x$-axis. From here, one considers the function $R(x,y) =$ the probability that $Q$ does not intersect $P$ over the region within $G$ outside $P$.

This approach, however, is incredibly messy, causes a headache, and probably won't yield a closed-form solution (not that we should assume the answer has one, but we can hope).

So, I am posting here to ask whether anyone can find a more clever and tractable path to solving this problem.

Descartes Before the Horse
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  • By "circles", do you mean "disks"? Otherwise, the probability that two circles overlap in more than two points is zero. – John Hughes Aug 24 '20 at 17:56
  • @JohnHughes Yes,sorry! – Descartes Before the Horse Aug 24 '20 at 17:59
  • Just to be clear, you are trying to compute the probability that $x_1-x_2 \in B(0, r_1+r_2)$ where $x_k$ are uniformly distributed on the unit circle and $r_k$ are uniformly distributed on [0,1]$ (and all are independent)? – copper.hat Aug 24 '20 at 18:04
  • @copper.hat Correct. – Descartes Before the Horse Aug 24 '20 at 18:16
  • I think i have spotted a mistake in your computation. If $R$ is the distance between the centers of $G$ and $P$, then $$\mathbb{P}(R\leq r) = \frac{Area(B(0,r))}{\pi} = r^2,$$ so $R$ has the PDF $f_R(r)=2r$. If $U$ is the uniform radius of $P$, then $P$ is contained in $G$ if and only if $U\leq 1-R$, so $$\mathbb{P}(\text{P is contained in G}) = \int_0^1 (1-r)2r : dr = \frac{1}{3}$$ and the probability that both are contained is $\frac{1}{9}$. – Leander Tilsted Kristensen Aug 24 '20 at 18:30
  • @LeanderTilstedKristensen My mistake, you are correct. Will edit accordingly. – Descartes Before the Horse Aug 24 '20 at 18:47

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