Points on a sphere .

[jay_shark]

Suppose n points are independently chosen at random on the surface of a sphere , and we're interested in the probability that all the points lie on some hemisphere . That is , we want the probability that there is a plane passing through the center of the sphere such that all the points are on one of the halves of the hemisphere .

[bigpooch]

I've seen the case for n=3 posed as a problem before.
For n<=3, clearly the probability is one. When n=3, the
first two points (they are different with probability one)
define a great circle on the sphere. Then the third point
can be anywhere for the three points to lie in SOME
hemisphere.

I'll think about when n>=4 later.

[pzhon]

I mentioned this problem in my solution of the problem of breaking the interval to form the lengths of a triangle. That corresponded to picking 3 points on the circle (or 1-sphere) in R^2 which are not contained in any semicircle. The probability that the convex hull of points on a d-sphere contains the origin is a classic one in integral geometry.

Ignore the probability 0 degenerate configurations.

The convex hull is the intersection of the half-spaces ccontaining the points. There is at least one hemisphere containing the points if and only if the convex hull does not contain the origin.

n+k points in n space have a null space of dimension k. This is a k-dimensional space in the n+k dimensional space of coefficients. The convex hull contains the origin if some positive (weight 1 by scaling) linear combination of the points is the 0 vector, that is, if the null space intersects the positive orthant.

All orthants are equally likely to be intersected by the null space by the symmetry of replacing subsets of points by their negatives, which are also uniformly distributed on the sphere.

At this point in writing up the solution, I found the following reference, which includes a published solution from a 1962 paper:

http://www.mathematik.uni-bielefeld....yclic-polygons

J. G. Wendel; A Problem in Geometric Probability,
Mathematica Scandinavica 11 (1962) 109-111.
(Univ. Michigan, Ann Arbor, Michigan, USA and Univ. Aarhus, Denmark)

That paper uses the same argument, as does a thread from usenet:

<ul type="square">Douglas J. Zare &lt;z...@cco.caltech.edu&gt; wrote:

&gt;[...]

Let P(n,d) be the probability that the origin is contained in the convex hull of n points chosen uniformly and independently on the surface of the d-sphere.

&gt;The case of the 0-sphere is worth including; the probability that the
&gt;convex hull contains the origin is 1-(1/2^(n-1)).

Further, there are natural reasons to think of P(n,-1) as 1, for n&gt;=1. The other boundary condition may be taken to be P(1,d)=0 for d&gt;=0.

&gt;[...]
&gt;Conjecture:
&gt;
&gt;P(n,d) = (P(n-1,d-1) + P(n-1,d)) / 2.

Proof:

From previous articles, the P(n,d) equals the expected number of quadrants in R^n through which a random subspace S of dimension n-(d+1) passes, divided by 2^n. I still don't know what that random variable really is, but with probability 1 the space is in general position.

S. Smirnov had the idea that one should consider the intersection of the coordinate hyperplanes with the subspace instead of counting the quadrants. (Call the hyperplane Xi=0 the ith coordinate hyperplane.) Each quadrant of R^n is convex, hence it intersects S in at most one connected region. So, we want to count the number of components of S\{coordinate
hyperplanes}.

If we remove one of those hyperplanes, H, we would decrease the number of components by the number of regions in H\{other coordinate hyperplanes}. By induction, this number is P(n-1,d-1)2^(n-1), and the result of the
subtraction is P(n-1,d)2^(n-1), hence

P(n,d)2^n = P(n-1,d-1)2^(n-1) + P(n-1,d)2^(n-1)

P(n,d) = (P(n-1,d-1) + P(n-1,d)) / 2.

This implies that the values are partial sums of rows of Pascal's triangle, normalized.[/list]
For example, with 5 points, the probabilities that these are in the same hemisphere of the unit sphere is
0/16 in dimension 5+,
1/16 in dimension 4,
5/16 in dimension 3,
11/16 in dimension 2,
15/16 in dimension 1, and
16/16 in dimension 0.

[imfatandugly]

for n&lt;=3, 1, for n&gt;3 prob is 0

[bigpooch]

Good idea of S. Smirnov!

BTW, in the first link, the first post was made by Noam
Elkies, who happens also to be a chess composer (although he
is better known for being a mathematician and the youngest
full professor at Harvard).

Thanks for the links! (even though you gave us the answer)