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Suppose there is a line of length $L$ cm. And it begins to rain at a constant rate of one droplet per second. Once a drop strike the line and it wets 1 cm of the line. What is the expected number of droplets it takes to wet the whole line?

The following condition was suggested by Henry:
To avoid the boundary problem, each point on the line within a distance of $\frac{1}{2}$ the point struck is wetted. The centre of the drop can be anywhere on the line (so if near the end would cover less than 1 but at least $\frac{1}{2}$ cm of the line.

The discrete version of this problem is a classical coupon collector problem，I am curious what would happen in the continuous case. The related post of this problem is rain droplets falling in a table.

Any help will be appreciated. Thanks in advance.

sundaycat
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  • As stated, the extreme ends of the line might have an expected infinite time to get wetted as it is highly unlikely that any drop manges to cover the whole first or last centimetre. Ways round this could include (a) the line is actually a circle so has no ends, (b) the centre of the drop can be anywhere on the line (so if near the end would cover less than $1$ but at least $\frac12$ cm of the line), or (c) the centre of the drop might be up to $\frac12$ cm beyond the ends of the line (so could cover less than $1$ but more than $0$ cm at the end of the line ) – Henry Jul 02 '17 at 21:49
  • @Henry I don't see anything in the question that prevents a raindrop from falling less than 1 cm from the edge, or raindrops to overlap. The circle idea is elegant though. – orlp Jul 02 '17 at 21:57
  • @orlp When the questions says "Once a drop strike the line and it wets 1 cm of the line", that suggests to me that each end of the $1$cm is on the line, and therefore the centre is at least $\frac12$cm from each end – Henry Jul 02 '17 at 22:05
  • @Henry I think what the OP meant is that each point on the line within a distance of $\frac{1}{2}$ of the point struck is wetted. That at least physically makes sense of a rain drop hitting a point. – orlp Jul 02 '17 at 22:09
  • @orlp Perhaps that is what was intended: that would be my suggestion (b), though I suspect (c) would be easiest to solve – Henry Jul 02 '17 at 22:09
  • @Henry Thanks for the advice. The boundary needs to be taken care of. I have updated the problem. – sundaycat Jul 02 '17 at 22:46
  • @orlp Thanks for the advice. The boundary needs to be taken care of. I have updated the problem. – sundaycat Jul 02 '17 at 22:46

1 Answers1

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Your problem is equivalent to breaking a stick of length $1$ into $n$ pieces1, and then asking what the expected number of $n$ is such that every piece has length $\leq \frac{1}{L}$. This problem is studied in "On the lengths of the pieces of a stick broken at random" by Lars Holst.

The exact chance of that happening after breaking the stick into $n$ pieces is:

$$\sum_{v=0}^n (-1)^v\binom{n}{v}\max\left(0, 1 - \frac{v}{L}\right)^{n-1}$$

They also proved that if it takes $N$ pieces such that each piece $\leq \frac{1}{L}$, then

$$P\left(\frac{N}{L} - \ln(L) - \ln(\ln(L)) \leq x\right) \rightarrow \exp(-e^{-x})$$

So $N \approx L(\ln(L) + \ln(\ln(L))$.

1To handle the edges you can add $\frac{1}{2}$ to both sides of the line. So, effectively, my whole answer above should use $L' = L + 1$.

orlp
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  • Thanks for the answer. Could you explain a little bit more why those two problems are equivalent? – sundaycat Jul 03 '17 at 00:00
  • @SundayCat In order for the whole line to be covered two adjacent raindrops may not be further apart than distance $1$. Now replace raindrops with knives. So it's equivalent to breaking a stick of length $L$ and checking if every part has a length $\leq 1$. This is equivalent to breaking a stick of length $1$ and checking if every part has length $\leq 1/L$. – orlp Jul 03 '17 at 00:03
  • Thanks for the further explanation. When cutting the stick, are the cutting points randomly selected? That is, even thought the length of a piece of stick is less than 1/L, does it still have a chance to be cutted? – sundaycat Jul 03 '17 at 01:41
  • @SundayCat Correct, you can see it as if $n$ random uniform cutting locations are selected in advance, and only then the stick is cut at those locations. – orlp Jul 03 '17 at 02:02
  • Thanks again :). If I understood correctly, the previous cutting locations wouldn't be able to be selected in furture cutting. It is equivalent to say that the landing postiton of drops are not allowed to be the same. In our case, two drops can land on the same position. There is a simlation programe(https://jsfiddle.net/dongogogo776/d3x5oqn2/), it shows that L = 50 takes around 2500 drop to wet the whole line. But according to the formula given here, it only takes 960 drops. – sundaycat Jul 03 '17 at 02:14
  • @SundayCat Considering that a cut on a stick can be arbitrarily close to a previous cut, I don't see the distinction. But I also see no issue with a zero length cut. – orlp Jul 03 '17 at 02:24
  • Let us continue this discussion in chat. – sundaycat Jul 03 '17 at 09:08