<t>QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: yegulalpMonte 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 ...

<t>" 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). "I don't think this statement is true. In the case of a circle when D=2, the first point and the center cuts the c...

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.

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?

Here is a python 2 code for some of the algorithms mentioned here.