# Probability that the center of a circle lies within three chosen points on a circle

The following question has been answered on MSE before and is indeed quite famous. I have one way of solving it and it seems quite wrong. Could anyone please help me correct it?

Consider points $$A$$ and $$B$$ on the circle such that they subtend an angle $$\theta$$. For the centre to lie inside the triangle, $$C$$ should belong to the arc that is antipodal to arc $${AB}$$. The probability will be $$\frac{\theta}{2\pi}$$. Integrating $$\theta$$ doesn't yield the probability. It seems like the fixing of the point $$A$$ is causing the issue but I don't know what to do otherwise. I would appreciate any help regarding this.

• Pls. give a link to the MSE solution. – zoli Jun 15 '19 at 6:53
• – Mathejunior Jun 15 '19 at 6:57
• Your way of solving this quite famous problem is not quite wrong but quite nice. – drhab Jun 15 '19 at 7:53

WLOG you may fix point $$A$$.
If $$\Theta$$ denotes the length of (shortest) arc $$AB$$ then $$\Theta$$ has uniform distribution on $$[0,\pi]$$.
If $$E$$ denotes the event that triangle $$ABC$$ will contain the center of the circle then:$$P(E)=\frac1{\pi}\int_0^\pi P(E\mid\Theta=\theta)d\theta=\frac1{\pi}\int_0^\pi\frac{\theta}{2\pi}d\theta=\frac1{\pi}\left[\frac{\theta^2}{4\pi}\right]^\pi_0=\frac14$$ which is correct.
• Why the $\frac{1}{\pi}$ at the beginning? – Mathejunior Jun 15 '19 at 7:42
• Because the PDF of uniform distribution on $[0,\pi]$ is $\frac1{\pi}$ on interval $[0,\pi]$. – drhab Jun 15 '19 at 7:46