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Observing cutting patterns in my wooden plate I came to the question: What is the expected (most likely) shape* of a triangle formed by three randomly chosen, but exactly in three points intersecting, lines?

Does the answer change if we limit the whole plane to a unit circle? Related but different questions are here and here.

*)The shape of the triangle shall be given by the ordered tupel of the two largest angles.

An intuitive geometrically derived solution would be $90°$ for the largest angle (the average value of the intervall $(0°,180°)$) and $45°$ for averaging all possible line positions given two other lines intersecting at $90°$.

Raphael J.F. Berger
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    How do you randomly choose the lines? – Michael Lugo Jun 15 '20 at 12:54
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    Defining a 'random triangle' is a tricky business. There is whole wikipedia page dedicated to the issue: https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) – Vincent Jun 15 '20 at 12:55
  • @MichaelLugo: Well since its only three lines it should be straightforward: The first line is arbitrary, the second line is defined by only one angle from $(0,\pi)$. Then we have two lines now we can fix the angle bisector of the first two lines then the third line is defined by the angle how it cuts the bisector, those maybe from $[\pi/2;\pi)$. Alternatively one could use 6 randomly chosen points.... – Raphael J.F. Berger Jun 15 '20 at 13:54
  • @MichaelLugo. The 6 points would be obviously only working for the second problem, the one on the unit disc. – Raphael J.F. Berger Jun 15 '20 at 14:08

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Hint:

Random lines over the whole plane is an undefined concept. But as you are only interested in angles, we can answer for random directions.

So we should draw three angles uniformly and look for the distribution of the largest and second largest cyclic difference. The computation is a little technical. It is a double integral where you keep the largest absolute difference fixed (or the second largest), and all cases are equiprobable. The expression of the domain seems a little complicated.