Probability that the convex hull of random points contains sphere's center What is the probability that the convex hull of $n+2$ random points on $n$-dimensional sphere contains sphere's center?
 A: This problem is discussed in J. G. Wendel; A Problem in Geometric Probability, Mathematica Scandinavica 11 (1962) 109-111. Wendel showed that the probability of $N$ random points lying on the surface of the unit sphere in dimension $n$ all lie on one hemisphere is
$2^{-N+1}\sum_{k=0}^{n-1} {{N-1}\choose k}$
I've found this here.
A: This is one of those old chestnuts that come up again and again.
To be precise, the probability that the convex hull of $n+2$ points in $S^n$
(the unit sphere in $\mathbb{R}^{n+1}$) contains the origin is $2^{-n-1}$.
There's a brief argument at Wolfram's mathworld which I don't find
entirely convincing but which certainly can be patched to form
a convincing argument. In brief, show that for random points $P_1,\ldots,P_{n+2}$
on the sphere, then with probability one, exactly one choice of signs
will put the centre in the convex hull of $\pm P_1,\pm P_2,\ldots,\pm P_{n+1}$
and $P_{n+2}$.
Added (3/8/2010)
Thanks to Grigory for his comment. Changing the notation slightly,
one can show that under some fairy weak hypotheses, if we choose $m+1$
points randomly and indepedently in $\mathbb{R}^m$ the probability their convex
hull contains the origin is $2^{-m}$.
Take a probability distribution on $\mathbb{R}^m$ and choose a sequence
of points (which we identify with vectors) independently from that distribution.
Our first condition on this distribution is that $m$ vectors $v_1,\ldots,v_m$
chosen independently from it are linearly independent with probability one.
This can fail if say some point occurs with nonzero probability or the
distribution lies in a hyperplane through the origin. Assume this condition.
Now a sequence $v_0,v_1,\ldots,v_m$ of random points chosen according to
our distribution are linearly dependent: there are reals $a_i$ not all zero with
$\sum_i a_i v_i=0$. By our condition, with probability one, the sequence
$(a_0,\ldots,a_m)$ is unique up to constant multiple, and moreover all the
$a_i$ are nonzero. So we may assume $a_0=1$ and $a_1,\ldots,a_m$
are nonzero and uniquely determined. Then the convex hull of the $v_i$
contains the origin if an only if all the $a_i$ are positive.
Now we introduce another condition: that the distribution is centrally symmetric;
in detail the probability that a random vector $v$ lies in a set $A$ equals
the probability that $-v$ lies in $A$. A condition like this is clearly necessary;
it stops the distribution being supported on a small region far from the origin.
This condition shows that all the $2^m$ possibilities of signs
for $a_1,\ldots,a_m$ are equiprobable; since changing the sign of some
$v_i$ changes the sign of $a_i$.
To conclude, if our probability distribution on $\mathbb{R}^m$ satisfies
these two condition, the probability that the convex hull of $m+1$
indepdendently chosen points contains the origin is $2^{-m}$.
These conditions are satisfied by the uniform distribution on a sphere
with centre at the origin, but also by many others.
