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On a "bottom" disk of area $\pi$, we place "top" disks of area $\pi,\frac{\pi}{2},\frac{\pi}{3},\dots$ such that the centre of each top disk is an independent uniformly random point on the bottom disk.

Will the bottom disk be completely covered by the top disks with probability $1$?

We have a strong argument that if the bottom disk's area is less than $\pi$ then the probability that it will be covered is $1$, and if the bottom disk's area is greater than $\pi$ then the probability that it will be covered is less than $1$. But the boundary case seems to be more difficult.

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    $\begingroup$ A charming question! :) $\endgroup$
    – paul garrett
    Nov 20 at 22:27
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    $\begingroup$ There is a similar question for the unit circumference and covering by random arcs that was fully answered (see Kahane's "Some random series of functions", Chapter 11). Have you tried the techniques from there? $\endgroup$
    – fedja
    Nov 20 at 23:45
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    $\begingroup$ More generally, this is called "Dvoretzky problem". It looks like the 1D case (covering interval with smaller intervals) is completely solved, and if you believe the 2D case is the same then the answer to this question should be "yes". However, I couldn't find a reference for the 2D case. $\endgroup$
    – abacaba
    Nov 21 at 0:17
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    $\begingroup$ Let a sequence $(r_n)_{n\ge1}$ be given. Let a sequence of points $u,x_1,x_2,\dots$ be chosen independently and uniformly at random within a given open set $\Omega\subset\mathbb R^2$ of area $a$. Then, I'd say the probability that $u\notin\bigcup_{n\ge1}B(x_n,r_n)$, that is, that for all $n\ge1$ one has $|u-x_n|\ge r_n$ is $$\frac1{|\Omega|}\int_{\Omega}\prod_{n\ge1} \frac{|\Omega\setminus B(u,r_n)|}{|\Omega|} du.$$ $\endgroup$
    – Pietro Majer
    Nov 21 at 19:46
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    $\begingroup$ For any $u\in\Omega$, one has eventually $B(u,r_n)\subset \Omega$ , so the ratio of areas in the infinite product is eventually $1-\frac{B_n(u,r_n)}{|\Omega|}$, which in the present case is $1-\frac{\pi}{na}$, and this make it vanish identically. So I should get the paradoxical conclusion that $\Omega\setminus\bigcup_{n\ge1}B_n$ has zero measure with probability $1$, even for large bottom disk. Or is this argument incorrect? $\endgroup$
    – Pietro Majer
    Nov 21 at 19:46

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