Maybe it's a classical one!You have n points on a sphere, what is the probability that all of them are on the same hemisphere?Sorry if it has already been posted

QuoteOriginally posted by: quantorYou have n points on a sphere, what is the probability that all of them are on the same hemisphere?Maybe it's really simple. 1 = 100%. Two equals 100% * 50%. Three equals 100% * 50% * 50% * 50% or something like that...

(1/2)^(n-3) for n>=3 and 1 if n<=3. The first two determine a great circle dividing the sphere into two halves. The third point determines which half we want the points in, and then assuming uniform distribution,a 1/2 factor for each additional point. EDIT: I was assuming you are picking points one after another, not sure that was implied...

Nope, that's only a lower bound.I would be tempted to conjecture that it would be asympotically the best upper bound if instead of considering the prob of being in the same hemisphere, you consider the probability of being in a piece a little smaller than a hemisphere (with the extra hypothesis that the piece remain convex (and with the sphere geometry notion of convex))

Farmer : for one two or three points, they always are on the same hemisphere.adgyboy is right because aym is not considering all the cases. It's just a lower bound. The first two don't determine a great circle in all cases.

QuoteOriginally posted by: quantor...The first two don't determine a great circle in all cases.You are right about that. I ignored the case when the points are not distinct as a point matches the previous one with 0 probability(My assumptions are: a) P(pt in A) proportional the to the area of A, 0 on measure 0 sets, and rotation-invariant. b)Draws are consecutive).When 2 points are distinct they determine a great circle. Perhaps the non-generic case has a non-zero contribution...I'll wait and see what you come up with.

Maybe i am wreong : Let's consider a sphere of radius one. For me a great circle is a circle of radius one. Two points A and B can't define one great circle unless the distance AB = 2.

quantor, on the sphere, there is a unique great circle that passes through any two (distinct) points P & Q.It can be visualized as the intersection of the plane determined by the triple O (the center of the sphere),P & Q, and the sphere.Of course it would then have the required diameter...

How stupid am i?you're right!I think that the sphere thing is a little hard. Try the same exercise with a circle instead of a sphere and a semicircle instead of an hemisphere.start with 3 points

I think the answer in this case is n*2^{1-n}The events that the points all lie in the semicircle starting at position i are all disjoint, and the probability of each of these is 2^{1-n}.But this argument doesn't seem to generalize well to higher dimensions, as the events of all points lying within a hemisphere determined (in some way) by a subset of the points are no longer disjoint.

QuoteOriginally posted by: adgyboyNope, that's only a lower bound.I would be tempted to conjecture that it would be asympotically the best upper bound if instead of considering the prob of being in the same hemisphere, you consider the probability of being in a piece a little smaller than a hemisphere (with the extra hypothesis that the piece remain convex (and with the sphere geometry notion of convex))I agree, it is a lower bound. One can convince oneself by looking forward after 3 points are drawn. There are at this point 3 possible great circles one can choose by looking at 2 out of the 3 points and see the 3rd as being in the hemisphere it determines on one side of the great circle determined by the two points looked at. So for the 4th point (and so on) there is an area greater than 1/2 of the sphere that can be selected etc...

QuoteOriginally posted by: quantorFarmer : for one two or three points, they always are on the same hemisphere.Not in my universe.

QuoteOriginally posted by: farmerQuoteOriginally posted by: quantorFarmer : for one two or three points, they always are on the same hemisphere.Not in my universe.That's one strange universe you got yourself, then.

You got it right zarnywhoop in the circle case.An other way of doing it is dividing the perimeter into 2p equal arcs, then only considering the semi-circles composed of the arcs. One can apply theinclusion-exclusion principle and only the let k go to infinity.This reasonning, as well as a discussion on the sphere case can be found on :http://www.mathpages.com/home/kmath327.htmWhat do you think? do you have a better demonstration?

Hmm. I think a quick summary of that web page would be 'I don't know how to compute this, but here are some approximations'. I was hoping for more. Have I misunderstood?