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I have been looking at this question.

I thought I knew a solution, but it appears to be at odds with the answers there. Can anyone tell me what is wrong with my solution?

My Attempted Solution.

Without loss of generality, our circle is the unit circle and we can take one of the three points to be $(1,0)$. Let the other points be $(\cos\theta,\sin\theta), (\cos\phi,\sin\phi)$, $\theta,\phi\in(0,2\pi), \theta<\phi$.

In order for the centre to lie inside the triangle, we require $\pi<\phi<\pi+\theta$.

Hence, for fixed $\theta$, the probability that the centre lies inside the triangle is $\dfrac{\theta}{2\pi}$.

Now, we can average over all possible values of $\theta$: $$\frac{1}{2\pi}\int_0^{2\pi}\frac{\theta}{2\pi}\,d\theta=\frac{1}{4\pi^2}\cdot\frac{4\pi^2}{2}=\frac{1}{2}.$$

But according to the answers to the existing question, the correct probability is $\dfrac{1}{4}$.

Where did I go wrong?

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  • $\begingroup$ I don't follow how you get $\pi<\phi<\pi+\theta$ What if $\theta=\pi-\varepsilon, \phi = \pi+\varepsilon? $ $\endgroup$ – saulspatz Apr 17 '18 at 20:41
  • $\begingroup$ @saulspatz Then still \pi<\phi<\pi+\theta$ $\endgroup$ – Hagen von Eitzen Apr 17 '18 at 20:45
  • $\begingroup$ youtu.be/OkmNXy7er84 $\endgroup$ – user29418 Apr 17 '18 at 20:48
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    $\begingroup$ you do not need to integrate to 2π $\endgroup$ – user29418 Apr 17 '18 at 20:54
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    $\begingroup$ Yes, you are right. You have exactly what I have written down. I must be cross-eyed today. $\endgroup$ – saulspatz Apr 17 '18 at 20:56
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Without loss of generality, we can assume $0 < \theta < \pi$ (i.e., the second point is in the top half).

If the third point is in the top half, the probability of success is $0$, and if $\phi \ge \theta+\pi$, the probability of success is also $0$, so success requires $\pi < \phi < \theta + \pi$.

Hence as you argued, for fixed $\theta$, the probability of success is ${\large{\frac{\theta}{2\pi}}}$.

Then the integral becomes $$\frac{1}{\pi}\int_0^{\pi}\frac{\theta}{2\pi}\,d\theta=\frac{1}{4}$$

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As already remarked, the issue lies in the integration bounds. Anyway such question and its generalization to the $n$-sphere can be solved by symmetry in a much slicker way. Assume that three points $A',B',C'$ have been chosen on the circumference. Let $A'',B'',C''$ be their antipodes. In any case, among the eight triangles $A^? B^? C^?$ two of them contain the center of the circle and six of them do not. The wanted probability is so $\frac{2}{8}=\frac{1}{4}$ since a point and its antipode are equally likely to be picked.

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