Random plane cutting an octahedron's edges

Take the six points $(\pm1,0,0)$, $(0,\pm1,0)$ and $(0,0,\pm1)$ on the unit sphere and join every pair of them with a straight segment. A unit vector is drawn uniformly at random, and the plane through the origin perpendicular to it is used to cut the sphere in two. On average, how many of the segments does this plane cross?

9. There are $\binom62=15$ segments. A random plane through the origin separates two unit vectors with probability $\theta/\pi$, where $\theta$ is the angle between them. The 12 orthogonal pairs each contribute $\tfrac12$ and the 3 antipodal pairs (whose segments pass through the origin) contribute 1, giving $6+3=9$ by linearity.