You throw N pebbles into a round pond, uniformly distributed.

Well, for starters, three points will form a triangle like this if and
only if they are not all contained in a semicircle.  So the probability
to look for is the probability that N points will all be in a
semicircle.  (Distance from the center doesn't matter.)

The probability I got was 2N(.5)^N, which can be arrived at either by
highly suspicious hand-waving or involved math.  So the probability of
finding the triangle in question is 1-2N(.5)^N.

The hand-waving argument goes like this: pick a semicircle.  The p of
all N points being in that particular semicircle is (.5)^N.  But how
many possible semicircles are there?  Well, a bearing determines a
semicircle, and any one of the N pebbles can provide that bearing,
leaving the other N-1 to flip coins:
        p = N(.5)^(N-1) = 2N(.5)^N

The next dimension up is, of course, points scattered through a
spherical space.  F'rinstance, find four stars within 100ly of Earth
that form a tetrahedron enclosing the Earth, which means that they can
never all be seen simultaneously from any point on the (perfectly
round) Earth.  My straight math method bogs down when it runs into the
average perimeter length of convex polygons of a given area on the
surface of a sphere.  The hand-waving method suggests that the
probability of N stars all being in a hemisphere together is
4N(N-1)(.5)^N.  Can anybody suggest a rigorous but simple proof?

And for 4-points in a demihypersphere, I'd guess 8N(N-1)(N-2)(.5)^N,
and in general:
                 N!       (N-d+1)
               ------ (.5)
              (N-d+1)!

-- Benjamin Thompson <thompson@spot.Colorado.EDU>

[ Editor: the guesses for higher dimensions cannot be right: for N = 4
stars, 4N(N-1)(.5)^N = 48/16 = 3, which exceeds 1.  See also
geometry/points.on.sphere. ]