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I have a large circle of radius $B$. I choose a random point inside the large circle, and draw a small circle of radius $A$ around it ($A<B$). What are the chances that, after drawing $n$ such small circles, the union of their area covers the large circle (i.e., includes every point in the large circle)? What is the expected value of the number of small circles I must draw to cover the large circle?

My progress so far: If I pick any point in the large circle, the chance that it has not been covered by $n$ small circles is:

1 minus the chance that it has been covered

= 1 minus the chance that the origin of a small circle is within $A$ of this point, after $n$ tries

= 1 minus (the small circle of radius $A$ around this point divided by the total area of the large circle), after $n$ tries

= $\left(1 - \frac{\pi A^2}{\pi B^2}\right)^n$

= $\left(\frac{B^2 - A^2}{B^2}\right)^n$

So, the chance that any point has been covered is $1 - \left(\frac{B^2 - A^2}{B^2}\right)^n$. How do I use this info to know whether all point have been covered after $n$ small circles? And find the expected value of the number of small circles it takes?

Also, even this oversimplifies it: If the point I pick in the large circle is within $A$ away from the edge, a circle of radius $A$ around it isn't completely contained in $B$, so the ratio of "hits" is actually less than $\frac{A^2}{B^2}$. I assume we can calculate the amount of overlap for the "average" small circle, which is 100% for most of them but less than that (is it $\frac{3}{4}$?) for $\frac{A}{B}$ of them.

jamaicanworm
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  • Whether or not a given point is covered is not independent from whether another point has been covered. For instance, if the two points are very near its more likely that they're both covered or both uncovered. This seems like it would make it hard to use the expression you have to find the probability that all points are covered. – overfull hbox Oct 13 '19 at 21:17
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    Even the $1$-dimensional version of this problem (cover an interval of length $B$ using smaller intervals of length $A$ each) does not seem easy to me... – antkam Oct 16 '19 at 04:52
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    I've only done some skimming, but this paper may contain some pertinent results, or at least interesting generalizations. – Kajelad Oct 16 '19 at 18:51
  • @antkam nice comment. in the 1-D setting, induction is very tempting, but I'm not sure how to use it [e.g. if $(0,1)$ is the big interval and we remove $(\frac{1}{3},\frac{2}{3})$, we can't pretend the problem is now two copies of the original problem but with $(0,\frac{1}{3})$ and $(\frac{2}{3},1)$ instead of $(0,1)$, since the center for a randomly chosen small circle can still be chosen from $(\frac{1}{3},\frac{2}{3})$] – mathworker21 Oct 17 '19 at 01:24
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    See https://math.stackexchange.com/questions/176383/rain-droplets-falling-on-a-table -- in particular, see my answer to it :). – mjqxxxx Oct 17 '19 at 02:50
  • @mjqxxxx Wow, that looks like the exact same question, right? Want to post an abbreviated version of your answer here as well (with the probability of coverage after n trials, not just the expected value N)? – jamaicanworm Oct 22 '19 at 14:41

1 Answers1

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Here is a lower bound and an upper bound, for large values of $B/A$. It is based on the Coupon Collector's problem.
For the lower bound, take as many non-overlapping circles of radius $A$ as possible. In the hexagonal pattern, they will cover $k:=\pi\sqrt3/6$ of the area, so there will be about $M:=kB^2/A^2$ of them. There must be a point in each of them so that all their centres are covered. By the Coupon Collector, that will take about $M\ln M$ points. Since some points lie in the gaps between circles, I expect $M(\ln M)/k$ points are needed.
For the upper bound, consider circles of radius $A/2$, again in a hexagonal pattern, but separated by $A\sqrt3/2$ instead of $A$ so they just cover the big circle. The area of a hexagon is $q:=3A^2\sqrt3/8$ so there are $N:=\pi B^2/q$ of them. If there is a point in each small circle then each small circle is covered, so the big circle is covered. This will take about $N\ln N$ points. There is a chance that a point is in two of these circles, so I would reduce the expected number of points by a factor $p=(\pi A^2/4)/q$.
To conclude, I expect the answer, $n$ tries, to be in between
$$\frac{B^2}{A^2}\ln\left(\frac{\pi\sqrt3B^2}{6A^2}\right) \le n\le \frac{4B^2}{A^2}\ln\left(\frac{8\pi B^2}{3\sqrt3A^2}\right)$$

Empy2
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  • You're right. I closed this as a duplicate of https://math.stackexchange.com/questions/176383, where it's shown that the factor before the logarithm (that you have between $1$ and $4$) is $2$ and the factor inside the logarithm (that you have between $0.9$ and $4.8$) is $4$. – joriki Feb 28 '22 at 09:24