First the problem should be clarified to explicitly make clear what three points at random on a circle mean - otherwise we start to get into trouble with Bertrand's paradox - https://en.wikipedia.org/wiki/Bertrand_paradox_(probability). But assuming the simple definition that the points are distributed uniformly according to equal arc length (or equal angles) in a range from zero to 2*pi, I will give an analytic solution, that is probably way harder than necessary for this problem but very general and useful if the probability distribution is different from uniform. So starting from the point (1,0) let x,y,z be the value of the cut points arc length measured counterclockwise and for simplicity for now use a range of 0-1 instead of 0-2*Pi (afterwards we can scale up the answer by multiplying by 2pi). So essentially we have chosen three points in the 0-1 interval uniformly, this can be visualized as choosing three points from a cube of side 1 uniformly. There are six possible orderings for the values of the three points, so the cube is divided into six regions, but by simple symmetry arguments one can show that the value of integrating the arc length containing the point (1,0) on the circle is the same for all regions of the cube. So assume x<y<z this corresponds to the region of the cube 0<x<1, x<y<1, y<z<1. The arc length of the point containing (1,0) on the circle is given by 1+x-z (easy to see this as total length of circle 1 minus middle interval z-x. So integrating 1+ x - z over the region of the cube given by the prior intervals gives the average arc length corresponding to the x<y<z ordering. The evaluation of the multiple integral is routine, and the result is 1/12. Multiplying by 6 to account for all the orderings, and by 2*pi to rescale as indicated above leads to an answer of pi.