# Probability that centre of circle is contained in triangle formed by three points on the circumference

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?

• I don't follow how you get $\pi<\phi<\pi+\theta$ What if $\theta=\pi-\varepsilon, \phi = \pi+\varepsilon?$ – saulspatz Apr 17 '18 at 20:41