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Covering a circle randomly with arcs has been well studied in the past (Geometric Probability - Solomon).

But the problem when the circle is changed to a line segment doesn't seem to have been studied before.

I'd like to know if there's any work out there who already obtained the probability distribution of the number and the length of the connected line segments that you get when randomly covering a line segment with another set of shorter segments, which may all be of equal length or have some kind of distribution.

Thanks!

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    $\begingroup$ Erm... What's the difference between the circle and the line segment? You just have a small endpoint effect, which can shift the counts by $1$ here and there but I would have pretty hard time designing a meaningful problem in which either the answers would be essentially different, or ,at least, the reduction of one case to another would be non-trivial. $\endgroup$
    – fedja
    Commented Jan 6, 2012 at 12:52
  • $\begingroup$ Fedja, if you pick pairs of points (and if that carries little meaning, adjust the distribution so that nearby points are more likely to be chosen), and define the interval selected by order (or in the case of the circle, by the shorter arc), then I find it challenging to relate the line problem to a circle problem because the line problem "seems to avoid" a certain point on the circle. Perhaps you can find an easy reduction that makes the circular form of the line problem give a nice solution to the line problem? I don't see one. Gerhard "Don't Ask Me About Reductions" Paseman, 2012.01.06 $\endgroup$
    – Gerhard Paseman
    Commented Jan 6, 2012 at 19:13
  • $\begingroup$ I don't understand the question. Is the idea that you take a line segment; randomly produce smaller line segments; keep going until you've covered the original segment? You're then asking how many you need? Do you include line segments that are entirely contained in the union of segments that you've already put down? Please be more precise... $\endgroup$
    – Anthony Quas
    Commented Jan 7, 2012 at 4:29
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    $\begingroup$ Also posted to MSE. In future, if you are going to post to both sites, PLEASE indicate this in your question. That way we avoid duplicated effort $\endgroup$
    – Yemon Choi
    Commented Feb 28, 2012 at 6:22
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    $\begingroup$ math.stackexchange.com/questions/96297/… $\endgroup$
    – Yemon Choi
    Commented Feb 28, 2012 at 6:22

1 Answer 1

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This problem can actually be solved using the exact same method as Chapter 4 of Solomon's geometric probability using the inclusion-exclusion principle in a similar fashion. A brief outline is available here (although it may contain small errors).

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