<t>QuoteCase in point - several members of the research group at IBM working on the problems I mentioned above have recently been lured away to work for a CT based billionaire's hedge fund. Most of them are smart, seasoned AI and text analysis researchers who were working on cancer genomics, among o...

<t>My explanation of flipping the signs on the points was a bit sloppy, try this:Given a random set of points [$]\vec{p_i}[$] on the sphere, we can find a set of weights [$]w_i[$] such that [$]\sum w_i = 1[$] and [$]\sum_i w_i \vec{p_i} = 0[$]. This is true because we have N equations and N unknown ...

<t>Yes, I think Vanubis is on to the right idea. For the case of N=D+1 points [$]\vec{p_i}[$] on a hypersphere of dimension D, the convex hull of the points is the set [$]H(p) = \{\sum_{i=0}^{N-1} w_i \vec{p_i} | \sum_i w_i = 1, w_i \geq 0\}[$].We want to know if the origin is in [$]H(p)[$]. To do t...

There is an amazingly complicated formula for the distribution of the area of a random triangle on the sphere in this paper.Sure enough it gives an average area of pi/2 by numerical integration. Looks like it would be fiendish to do that average analytically.

For D=2 I agree with tlin123's solution. For D>=3 and N>D+1, I think it only makes sense to define the problem in terms of the convex hull of the N points - otherwise we get ambiguities about inside/outside as noted above. No clue about the answer though.

<t>If we draw three random great circles on the sphere, they partition it into exactly 8 spherical triangles. So the average area of a random spherical triangle (generated in this way) must be 1/8 of the sphere = pi/2. So how do we show that random great circles generate the same distribution as thr...

<t>Monte Carlo is giving me P = 0.125 = 1/8 for 4 points on the sphere.An observation:For 4 points on the sphere, think of the first three points as defining a spherical triangle T. For the fourth point to generate a tetrahedron capturing the center of the sphere, the 4th point must lie in the regio...

Monte Carlo for D=2 is easier since we can just generate random angles, sort them, and find the longest arc. I get the following results for the circle:N=3, P = 0.24998977N=4, P =0.50010848N=5, P=0.68752146 (probably 11/16 in exact form)N=6, P=0.81246003 (13/16??)

good point, so my result is only a bound

<t>General case: we put N points at random on a hypersphere in dimension D. The probability that the center of the hypersphere is contained in the convex hull of the N points is P = 1 - (1/2)^(N-D) for N>=D. For N < D, P=0.So 4 points on circle is N=4, D=2, P = 1-(1/2)^2 = 3/4 probability of contain...

Doh! Took too long to type it in.

<t>It follows from Fermat's little theorem: a^(p-1) = 1 mod p for p prime, a not divisible by p.(p-3)^(p-1) = 1 mod p (by setting a = p-3 in Fermat)(p-1)^(p-1) = 1 mod p (by setting a = p-1 in Fermat)Now expand:(p-1)^(p-1) = ((p-1)^(p-3)) * (p-1)^2 = ((p-1)^(p-3)) * (p^2 - 2p + 1) = (p-1)^(p-3) mod ...

<t>For the last 3 remaining people, we require that the total life span is 27 people-years and each must have life span of at least 5 years (since they all survive the first four years). So we can have life spans of (9,9,9), (8,9,10), or many other cases. The most extreme would be (5,5,17). So we kn...

<t>I solved it algebraically, one year at a time. Let n_k be the number of people with life span of k years (measured as of 2012). Let a_k be the average life span of all people still alive after k years (measured from 2012).From the problem definition, we havea_0 = 1.6 (average life span of all peo...