Expected arc covering a fixed point

HardProbability Expected Value~9m

Three points are chosen independently and uniformly on the circumference of a unit circle (total circumference $2\pi$ ). They split the circle into three arcs. What is the expected length of the arc that contains the fixed point $(1,0)$ ? The answer has the form $q\pi$ for a rational $q$ ; find $q$ .

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The fixed point lands in a given arc with probability proportional to that arc's length - this is length-biasing, so the answer beats the mean arc $2\pi/3$ .
Expected containing length $= \tfrac{1}{2\pi}\mathbb{E}[\sum G_i^2]$ . For uniform spacings (Dirichlet), $\mathbb{E}[G_i^2]=(2\pi)^2\tfrac{2}{n(n+1)}$ with $n=3$ .

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Asked at: Jane Street, Two Sigma

Expected arc covering a fixed point

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