jmc

We can assume that the circle has its center at $(0,0)$ and a radius of $1$. Call the three points $A$, $B$, and $C$, and let $a$, $b$, and $c$ denote the length of the counterclockwise arc from $(1,0)$ to $A$, $B$, and $C$, respectively. Rotating the circle if necessary, we can also assume that $a=\pi /3$. Since $b$ and $c$ are chosen at random from $[0,2\pi )$, the ordered pair $(b,c)$ is chosen at random from a square with area $4{\pi }^2$ in the $bc$-plane. The condition of the problem is met if and only if $$0<b<\frac{2\pi }{3},0<c<\frac{2\pi }{3},\text{and}|b-c|<\frac{\pi }{3}.$$This last inequality is equivalent to $b-\frac{\pi }{3}<c<b+\frac{\pi }{3}$.



The graph of the common solution to these inequalities is the shaded region shown. The area of this region is $$\left(\frac{6}{8}\right){\left(\frac{2\pi }{3}\right)}^2={\pi }^2/3,$$so the requested probability is $$\frac{{\pi }^2/3}{4{\pi }^2}=\boxed{\frac{1}{12}}.$$