Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points?What if it's a circle instead of a sphere?

QuoteOriginally posted by: outrun1/2 for both.Does that generalize to hyperspheres?

QuoteOriginally posted by: outrunQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: outrun1/2 for both.Does that generalize to hyperspheres?I think it's wrong now! I was mirroring 1 point thought the plane spanned by the other points and expecting that either one to contain the center point.I think the mistake does indeed generalize to higher dimensions :DHmmm.... Yes, it must be less than 1/2. The first two points, regardless of location define a great circle on the sphere which has a center collocated with the center of the sphere. The probability that BOTH of the next two points are on the same half of the sphere is 1/2 so the probability that the center of the sphere lies inside the tetrahedron is less than or equal to 1/2. But even if points 3 and 4 reside on opposite halves, the sphere's center may still be outside the tetrahedron (e.g., points 3 and 4 are very close to points 1 and 2 or points 1 and 2 might be very close to each other and points 3 and 4 might be on the same half of the half sphere defined by the perpendicular of the chord between 1 and 2). So the total probability is much less than 1/2.

Hi outrun, in your solution are you only using 3 points? The questions ask for 4 points. I think your solution is right for the case of 3 points though.

How's this for dubious logic:In the 0-D case: P = 1 (the 1 point always includes the "center")In the 1-D case: P = 1/2 (the 2 points are on opposite side of the center half the time)In the 2-D case: P = 1/4 (by outrun's logic)Shall we extrapolate and say the answer in the 3-D case is 1/8????

QuoteOriginally posted by: outrunFor the 4 points on a circle that might be valid!The center won't be inside an n-gon if all n points are in the same semicircle?I'd have thought the 2-D case with 4 points would have a much higher probability than the 2-D case with 3 points. If the first 3 points already contain the center, the fourth point changes nothing. If the first three points do NOT contain the center (P=0.5), there's a good chance (> 0.5) that the fourth point will create a 4-gon that contains the center.

QuoteOriginally posted by: outrunQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: outrunFor the 4 points on a circle that might be valid!The center won't be inside an n-gon if all n points are in the same semicircle?I'd have thought the 2-D case with 4 points would have a much higher probability than the 2-D case with 3 points. If the first 3 points already contain the center, the fourth point changes nothing. If the first three points do NOT contain the center (P=0.5), there's a good chance (> 0.5) that the fourth point will create a 4-gon that contains the center.Duh.. of course! My mistake.I like your semicircle model for the 2-D/3-point case and wonder if it can be extended to the 3-D/4-point case by defining a set of hemispheres on the various pair-wise points.

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 containing the center of the circle.4 points on the sphere is N=4, D=3, P = 1 - (1/2)^1 = 1/2 probability of containing the center of the sphere.Proof: the first D-1 points uniquely define a hyperplane (dimension D-1) containing the center of the hypersphere. That hyperplane splits the hypersphere in half. In order for the remaining N-D+1 points to NOT capture the center of the hypersphere, they must all land in the same half of the hypersphere (as defined by the first N-D+1 points). Probability of that is 2 * (1/2) ^(N-D+1) = (1/2)^(N-D), since there are two halves. The final answer is the complementary probability 1 - (1/2)^(N-D) since we DO want to capture the center of the hypersphere.

QuoteOriginally posted by: yegulalpGeneral 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 containing the center of the circle.4 points on the sphere is N=4, D=3, P = 1 - (1/2)^1 = 1/2 probability of containing the center of the sphere.Proof: the first D-1 points uniquely define a hyperplane (dimension D-1) containing the center of the hypersphere. That hyperplane splits the hypersphere in half. In order for the remaining N-D+1 points to NOT capture the center of the hypersphere, they must all land in the same half of the hypersphere (as defined by the first N-D+1 points). Probability of that is 2 * (1/2) ^(N-D+1) = (1/2)^(N-D), since there are two halves. The final answer is the complementary probability 1 - (1/2)^(N-D) since we DO want to capture the center of the hypersphere.Although it is true that if the remaining N-D+1 points fall on same half, then the simplex does not include the center, it does not imply falling on opposite halves guarantees the center is included.Simple counterexample: points 1 and 2 are 1 milliradian apart, point 3 is 1 milliradian above points 1 and 2, and point 4 is 1 milliradian below points 1 and 2. The result is a tiny tetrahedron located far from the center but with points 3 and 4 on opposite side of the {center, p1, p2} hyperplane.