Given 4 points ( not all on the same plane ), what is the probability that a hemisphere exists that passes through all four of them ?
Probability – What Is the Probability That 4 Points Determine a Hemisphere?
pr.probabilitypuzzle
Related Solutions
A good way to think about a uniformely random point on the sphere is to think of it as the normalization of a vector of normally distributed i.i.d random variables. In fact, you don't even need to normalize. Consider hemisphere $v_i$ and $v_j$, form the $2 \times N$ matrix $V$, $v_i$ and $v_j$ being the rows of $V$. Let $X$ be a random Gaussian vector with covariance matrix $I$ (identity). $V.X$ is a vector with two components, whose sign represent membership in the hemisphere $i$ and $j$... $V.X$ follows a multivariate normal distribution with mean $\left(\begin{array}{cc}0\\\0\end{array}\right)$ and covariance matrix $n \left[\begin{array}{cc}1 & 1-2h/n\\\1-2h/n & 1\end{array}\right]$ where $h$ is the hamming distance. Notice that any covariance matrix can be approximated, up to a multiplicative factor, for $n$ large enough. The measure of the intersection of the two hemispheres is the integral of the the probability density over an orthant.
For the intersection of two hemispheres, the orthant is a quartant and the intersection is given as
$$\frac{1}{4} + \frac{\sin^{-1}(1-2h/n)}{2\pi}$$
(note that this is not the area but the fraction of the $(n-1)$ sphere area that is in the intersection)
There is also a formula for the trivariate case, i.e. if you're considering the intersection of three hemispheres
$$\frac{1}{8} + \frac{1}{4\pi}( \sin^{-1}(1-2h_{12}/n) + \sin^{-1}(1-2h_{23}/n) + \sin^{-1}(1-2h_{13}/n) )$$
Unfortunately, it is known that there is no closed-form expression of the probability of the orthants of a general multivariate normal distribution above the trivariate case. In your case, the covariance matrix have a certain structure, so it is not definite evidence that there is no formula for the area of the intersection of more than three hemispheres, but it is pretty good evidence.
The difference between the positions is another random walk in the same dimension. You can either view the steps as different, or sample a random walk at even times. So, the probability is $0$ if meeting is ruled out by parity, and $1$ in the plane if meeting is possible.
Best Answer
See J. G. Wendell, "A problem in geometric probability", Math. Scand. 11 (1962) 109-111. The probability that $N$ random points lie in some hemisphere of the unit sphere in $n$-space is
$$p_{n,N} = 2^{-N+1} \sum_{k=0}^{n-1} {N-1 \choose k}$$
and in particular you want
$$p_{3,4} = 2^{-3} \sum_{k=0}^2 {3 \choose k} = {7 \over 8}$$.
A second solution: A solution from The Annals of Mathematics, 2 (1886) 133-143 (available from jstor), specific to the (3,4) case, is as follows. First take three points at random, A, B, C; they are all in the same hemisphere and form a spherical triangle. Find the antipodal points to those three, A', B', C'. Now either the fourth point is in the same hemisphere as the first three or it is in the triangle A'B'C'. The average area of this triangle is one-eighth the surface of the sphere.
This gets the right answer, but I'm not sure how I feel about it; why is the average area one-eighth the surface of the sphere? One can guess this from the fact that three great circles divide a sphere into eight spherical triangles, but that's hardly a proof. Generally this solution seems to assume more facility with spherical geometry than is common nowadays.