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I understand that for 2 random chords, the probability of no intersections is 1/3 thanks to this blog post.

What happens when I have 3 random chords? Is there an intuitive explanation for calculating the number of intersections given n random chords?

I found a post that proposes the following formula for r chords, but this does not hold for 2 chords. Any ideas?

$$P(NoIntersection) = \frac{2^r}{(r+1)!}$$

Note: By random chord, I mean by randomly picking 2 points that lie on the circle.

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  • $\begingroup$ You should be careful in defining exactly what it is you mean by "random chord" (see Bertrand's Paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)). The posts you link do define it, but I'd like to just make sure you are working with the same definition as them. $\endgroup$ Jul 4 at 0:58
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    $\begingroup$ You mean $(r+1)!$? $\endgroup$ Jul 4 at 1:13
  • $\begingroup$ Quote: "Is there an intuitive explanation for calculating the number of intersections given $n$ random chords?" The number of intersections has various possible values, each with some probability; I'm not sure what you mean by calculate this number. (It's like saying, toss $10$ coins, then "calculate" the number of heads.) Anyway, is this question different from the question in the title? $\endgroup$
    – Dan
    Jul 4 at 1:42

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The first linked blog post shows that the probability that two random chords do intersect is $\tfrac 13$, which means the probability they don't intersect is $\tfrac 23$. This lines up with the formula provided in the second link.

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