With what odds do three random points on an n-sphere form an acute triangle?

Select three points a, b, and c, randomly with respect to the surface of an
n-sphere.  These three points determine a fourth, x, which is the intersection
of the sphere with the axis perpendicular to the abc plane.  (Choose the pole
nearest the plane.) I could have, just as easily, selected x, a distance d
from x, and three points d units away from x.  The distribution of d is not
uniform, but that's ok.  For every x and d, the three points abc form an acute
triangle with probability p[n-1].  By induction, p[n] = 1/4.