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Given a line, we can choose 3 points on it randomly and independently, what is probability that these 4 segments form a quad?

My approach: I tried numbering them a, b, c and d and tried writing equations to solve them, but unfortunately I wasn't able to reach any conclusion with 4 variables, tried to solve for a triangle too, but didn't reach any conclusion, ran a simulation found it's close to 0.5, but I don't know how. Please help!

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  • $\begingroup$ What is a quad? $\endgroup$
    – JonathanZ
    Commented Aug 10, 2022 at 3:18
  • $\begingroup$ Let we have a line. Let we choose 3 points A, B, C on it. Then there are 3 segments AB, AC, BC. What is the 4th segment? In order 4 segments to be possible to be sides of quadrilateral, longest segment must be shorter than sum of 3 others. $\endgroup$ Commented Aug 10, 2022 at 9:12
  • $\begingroup$ quadrilateral, sorry for late reply $\endgroup$
    – Charlie
    Commented Aug 10, 2022 at 12:34
  • $\begingroup$ oh @IvanKaznacheyeu the segments would be start-A,AB, BC, C-end $\endgroup$
    – Charlie
    Commented Aug 10, 2022 at 12:35
  • $\begingroup$ Let 4 segments are $AB$, $BC$, $CD$ and $DE$. Let $P$ is probability that every of these segments is less than sum of three others. $P=1-Q$, where $Q$ is probability that longest segment is longer than $AE/2$. $R$ is probability that $AB$ is longer than $AE/2$, $R=S$ where $S$ is probability that there are no points in half of $AE$ nearest to $A$, then $R=S=1/8$. Due to symmetry probabilities of each segment to be longer than $AE/2$ are equal to $R$. Then $Q=4R=1/2$, $P=1-Q=1/2$. $\endgroup$ Commented Aug 10, 2022 at 13:02

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