What are the chances of 4 random angles being on the same half-circle

Question

Given a circle, we choose 4 angles at random (uniformly distributed).

What are the chances that the four of them would be at the same half-circle ?

The way I thought about solving it, is to define to look at the segment $$[0,2\pi]$$ And define the first angle as $\theta_1=0$,

And then I reduced the question, to

"what is the probability of difference between the max and min being less than pi"

However, my approach is problematic when dealing the cyclic nature of the angle

$$\theta_1=0,\theta_2=\epsilon,\theta_2=2\pi-2\epsilon,\theta_3=2\pi-\epsilon$$

I'd love to hear your ideas on how to approach this problem

Answer 1

In stead of putting $\theta_1 = 0$ like you suggest, you can put $\theta_1 = \pi$. We then have that any semicircle that contains $\theta_1$ is of the form $[a,a+\pi]$ with $a \in [0,\pi]$. So you don't have to worry about the interval being a circle anymore.

Now the question is reduced to: what is the probability that $$ \max(\pi , \theta_2 , \theta_3 , \theta_4) - \min(\pi , \theta_2 , \theta_3 ,\theta_4) $$ is less than $\pi$ , where the $\theta_i$ are uniformly distributed in $[0,2\pi]$? You can convince yourself that this probability is the same as the probability that $$ \max(\theta_2 , \theta_3 , \theta_4) - \min(\theta_2 , \theta_3 ,\theta_4) $$ is less than $\pi$, and you can try to calculate this probability.

Answer 2

Out of the 4 angles there are always 3 that are on the same half-circle (including the diameter). Take the first angle to be 0 then the other angles are in [0,$\pi$] or in [$\pi$,$2\pi$] and at least 2 of the 3 angles we have left have to be in the same set.

Quick and dirty for the rest: The 4th angle now has a 0.5 chance of being in the same half-circle. So the chances that the four angles would be at the same half-circle is 0.5