TITLE: Rain droplets falling on a table
QUESTION [41 upvotes]: Suppose you have a circular table of radius $R$. This table has been left outside, and it begins to rain at a constant rate of one droplet per second. The drops, which can be considered points as they fall, can only land in such a way such that they impact the surface of the table. Once they strike the table, they form a puddle of radius $r$, centered at their point of impact. What is the expected number of droplets it takes to cover the table in water?
The answer should be left in terms of $R$ and $r$. However, if you can simulate this, while changing both $r$ and $R$ so that some regression might be applied to find the approximate relation, that would be nice as well. I'd really like some intuition as to how the two are related. 
I have tried decomposing the problem by considering only the 1-dimensional case with line segments, but even its solution has eluded me. A potential starting point could be the discrete case of marbles falling into buckets. 
The following edit was suggested by Jbeuh: 
A more formal restatement of the problem :
Consider a sequence $X_1,\ldots,X_n$ of independent random variables following a uniform distribution on a disk $D$ of radius $R$. For each $i$, let $C_i$ be the disk centered at $X_i$ with radius $r$. Let $Y$ be the first $i$ such that the disks $C_1,\ldots,C_i$ forms a cover of $D$. What is the expected value of $Y$?

REPLY [9 votes]: For any covering of the target region with shapes of diameter $\le r$, if one droplet center lies within each shape, this is sufficient to cover the entire region with droplets.  This can be used to calculate an upper bound on the expected time to cover the table.  Note that this upper bound will be tighter when the number of shapes is smaller; i.e., we want the area of each shape to be as large as possible.  Suppose the covering uses regular hexagons of diameter $r$: these have area $3r^2\sqrt{3}/8$, so there will be about
$$
N_{\rm hex} = \frac{8\pi}{3\sqrt{3}}\left(\frac{R}{r}\right)^2
$$
of them.  The coupon collector's problem then states that the expected time to hit all the hexagons is $E[T_{\rm hex}] \sim N_{\rm hex}\ln N_{\rm hex}$.
On the other hand, for any packing of the target region with disks of radius $r$, it is necessary for at least one droplet center to fall within each disk in order for the target region to be covered (in particular, the disk centers cannot be covered otherwise).  This can be used to provide a lower bound on the expected time.  Here we again use a hexagonal packing, which covers a fraction $\pi/\sqrt{12}$ of the table's surface, but now the hexagon area is $2r^2\sqrt{3}$.  The number of disks is then
$$
N_{\rm disc}=\frac{\pi}{2\sqrt{3}}\left(\frac{R}{r}\right)^2=\frac{3}{16}N_{\rm hex}.
$$
The coupon collector's problem in this case states that the expected time to hit all the disks is $E[T_{\rm disc}] \sim \frac{\sqrt{12}}{\pi}N_{\rm disc}\ln N_{\rm disc}=\left(\frac{R}{r}\right)^{2}\ln N_{\rm disc}$.  Note the extra factor of $\sqrt{12}/\pi$, which comes from the fact that each droplet only "collects a coupon" with probability $\pi/\sqrt{12}$.
Putting these bounds together, we have that the expected time to cover the table satisfies
$$
2 \mu^2 \ln \mu + O(\mu^2) = E[T_{\rm disc}] \le E[T] \le E[T_{\rm hex}] = \frac{16\pi}{3\sqrt{3}} \mu^2 \ln \mu + O(\mu^2)
$$
for $\mu \equiv (R/r) \gg 1$.  In particular, the expected time is proven to be $\Theta(\mu^2 \ln \mu)$, with a multiplicative factor bounded between $2$ and about $9.68$.