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


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

  • $\begingroup$ Re "we choose 4 angles at random", on the circle or on and inside? $\endgroup$ – rtybase Aug 5 '17 at 9:59
  • $\begingroup$ On the circle... $\endgroup$ – Uri Goren Aug 5 '17 at 10:07
  • $\begingroup$ The reason I was asking is, if angles are on the circle, then the problem is similar to choosing 4 points on the circle. Then 4 points on the circle must belong to the same half circle. The length of the circle is $l=2\pi R$, half a circle $\frac{l}{2}=\pi R$. Using geometric probability $\Rightarrow P= \frac{1}{2}$ $\endgroup$ – rtybase Aug 5 '17 at 10:10
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    $\begingroup$ A general answer for $n$ points/angles can be found at math.stackexchange.com/q/325141/420432 $\endgroup$ – nickgard Aug 5 '17 at 10:22
  • $\begingroup$ @rtybase , I didn't understand how did you account for the number $n=4$ in your calculation $\endgroup$ – Uri Goren Aug 5 '17 at 10:30

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.


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

  • $\begingroup$ It's not true: [0,120,240] (in degrees) $\endgroup$ – Uri Goren Aug 5 '17 at 10:06
  • $\begingroup$ Yes it is. 0 and 120 are both in [0, 180] (hence the including the diameter remark) $\endgroup$ – Erik Jurriën Aug 5 '17 at 10:07
  • $\begingroup$ I don't know why this was down-voted, it's a reasonable answer! $\endgroup$ – rtybase Aug 5 '17 at 10:12
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    $\begingroup$ It is true that out of 4 angles there will always be three in the same half-circle. However, the rest of the answer is imho not correct. First of all the "fourth angle" is not well defined, what is the "fourth angle" is all 4 angles lie in the same half-circle? Secondly, the "fourth angle" is not independent from the other angles. You introduce this dependency by deciding which angle is the "fourth angle". Lastly, in general, there will be lots of half-circles that contain the first three angles, so the probability that a random angle would lie in one of these is larger than 1/2. $\endgroup$ – Ward Beullens Aug 5 '17 at 10:16
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    $\begingroup$ look at [180,181,182], the 4th angle can be anything not between 0 and 2, a probability much larger than 1/2. $\endgroup$ – Uri Goren Aug 5 '17 at 10:20

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