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Problem:

Consider $3$ points randomly chosen on a circle, find the probability that all of them lie in some semi circle

I tried to solve this problem but I am getting an incorrect answer. I have seen correct solutions to this problem but I am unable to see what is wrong with my reasoning which is worrying me because I may make similar mistakes in future.

My solution:

Number the points $1,2,3$ (the order in which I choose). After choosing first point, the event that the points $2$ and $3$ lie on the same semicircle is equivalent to the event that points $2,3$ lie either on the semicircle measured counter clockwise from point $1$ or clockwise from point $1$. Both of them have probabilities $1/4$. Since the events the disjoint, the total probability is $1/2$.

What is wrong with the above solution?

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    $\begingroup$ Your reasoning is flawed. If 2 and 3 are equidistant from the point antipodal to 1, but on opposite sides, then they don't both lie in either semicircle ending at 1. (In terms of angles on the unit circle, suppose $\theta_1=0$, $\theta_2=\pi-\epsilon$, and $\theta_3=\pi+\epsilon$.) $\endgroup$ – MPW Jul 6 '17 at 17:51
  • $\begingroup$ Thanks for pointing it out. If you post it as an answer, I shall accept it. $\endgroup$ – Shadow Jul 6 '17 at 18:01
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[Comment converted to answer]

Hint: Your reasoning is flawed. If 2 and 3 are equidistant from the point antipodal to 1, but on opposite sides, then they don't both lie in either semicircle ending at 1.

(In terms of angles on the unit circle, suppose $\theta_1=0$, $\theta_2=\pi-\epsilon$, and $\theta_3=\pi+\epsilon$.)

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