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We add $n$ random lines to the unit disc. We do so by adding two points on the disc, transcribing a line between them and extending that line to the boundary of the disc, namely the unit circle. Each line is chosen independently and we do this $n$ times.

After $n$ random lines have been added, what is the probability that the largest arc on the unit circle between two points is of length $\pi/2$ or longer? To be clear, the arc itself must not have any points on it.

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  • $\begingroup$ You might find the following easier and suggestive. Pick a specific quadrant of the circle. Compute that area A of the disk with the following property: for any point P in the area, at least half the disk is comprised of points Q such that PQ intersects the quadrant. I imagine A is at least a quarter of the disk. Now the goal is to avoid the area, and when in the area, to avoid the region of points Q. This should give something like (1-area/2)^n as an upper bound for your probability. Gerhard "Tackle The Problem In Halves" Paseman, 2017.10.20. $\endgroup$ Commented Oct 20, 2017 at 20:52
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    $\begingroup$ what is here "a piece of circumference on the circle" ? does it mean, one of the 2n arcs into which the circumference is divided by the 2n points? $\endgroup$ Commented Oct 20, 2017 at 21:44
  • $\begingroup$ I will point out that the tag (discrete-mathematics) is deprecated - see the tag-info - so it would be good to choose some other suitable tag for the question. $\endgroup$ Commented Oct 21, 2017 at 0:21

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The distribution of the maximal distance between a pair of random points on the circle is known - when you scale it by $n/\log n$ you get a Gumbel distribution with scale 1, location 1., see, e.g.,

Schlemm, Eckhard, Limiting distribution of the maximal distance between random points on a circle: a moments approach, Stat. Probab. Lett. 92, 132-136 (2014). ZBL1294.60045.

so it is quite obvious the the probability of maximal distance being $O(1)$ goes to zero exponentially fast (it is obvious that it goes to zero about as fast as $(3/4)^{2n}$, in fact.

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  • $\begingroup$ So quite researchy after all then! $\endgroup$ Commented Oct 22, 2017 at 17:41
  • $\begingroup$ I guess the last bit is to prove that extrapolating a line from two end points chosen randomly inside the unit circle to the unit circle itself results in points that are themselves uniformly random. Apart from what might be a misguided belief that this is obvious, I can only suggest constructing those points from a basic geometric calculation given the end points of the line stated in polar coordinates. I'm guessing it should all just fall out. $\endgroup$ Commented Oct 22, 2017 at 17:56
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To re-ask Pietro Majer's question, what do you mean by a "piece of circumference"?


          CircChords
          $20$ random points, $n=10$ chords.
If you mean the largest section of the circumference containing no chord points, then the chords play no role: the question could be posed just in terms of $n$ (or $2n$) points, rather than $n$ "lines." On the other hand, every chord of the circle has half the circumference to one side or the other.
Added. I think James Smith's interpretation of the question makes the most sense:
          RandLines
          $10$ random points inside the circle, determining $n=5$ lines. Largest arc: $94^\circ$.


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  • $\begingroup$ OP is choosing two points in the circle, not on the circle, when drawing the lines, so it's not clear to me that the chords give the same answer as choosing points on the circle. $\endgroup$ Commented Oct 22, 2017 at 8:19
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    $\begingroup$ I agree, I think it's two points inside the circle, not on the circumference. I also think 'piece of circumference' means arc. The OP really needs to clear this confusion up! $\endgroup$ Commented Oct 22, 2017 at 10:46
  • $\begingroup$ @JamesSmith: Your interpretation makes sense. $\endgroup$ Commented Oct 22, 2017 at 12:00
  • $\begingroup$ Okay, emboldened by your comment I've edited the question, so let's see. I actually think it's quite interesting although not very researchy. $\endgroup$ Commented Oct 22, 2017 at 13:14

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