Expected arc covering a fixed point of the circle

Three points are chosen independently and uniformly on the unit circle, cutting it into three arcs. Let $A$ be the arc that contains the fixed point $(1,0)$. Its expected length can be written as $q\pi$ with $q$ rational. Find $q$.

$q=1$. Measure the gap from $(1,0)$ to the nearest point going clockwise: it exceeds $x$ only if all three points avoid an arc of length $x$, so its mean is $\int_0^{2\pi}(1-\tfrac{x}{2\pi})^3dx=\tfrac{\pi}{2}$. The counter-clockwise gap has the same mean, so $\mathbb{E}[\text{length}]=\pi$. (Size-biasing makes this arc longer than the average $2\pi/3$.)