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Given 4 points ( not all on the same plane ), what is the probability that a hemisphere exists that passes through all four of them ?

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Related thread: mathoverflow.net/questions/2014/… –  Yoo Jan 7 '10 at 18:30
I may be being slow here (and the fact that no one else is asking suggests that I am), but... in order for the question to make sense, don't you need to specify what probability distribution you're using? I.e. when you say "at random", what do you mean? I guess you mean the points to be in R^3, so you need a probability distribution on (R^3)^4 = R^12. Even if the points are chosen independently and according to the same distribution, don't you need to give us at least a distribution on R^3? –  Tom Leinster Jan 7 '10 at 23:03
Hmm, the answers of Michael and Dmitri (and the paper of Wendell that Michael links to) seem to assume that the points all lie on S^2, presumably uniformly distributed. They seem to have magically guessed what you meant. –  Tom Leinster Jan 7 '10 at 23:06
Tom, if you decide indeed that all the points are on the fixed sphere (as we have done), then it is clear from my answer, that you can take any measure on S^n, invariant under the central symmetry. You just need to ask that the measure has no atoms. This need not be a measure induced by the standard metric. –  Dmitri Jan 8 '10 at 0:47
a bit of my thought process: "hemisphere" implied that there was a sphere, and I generally default to a uniform distribution when possible. –  Michael Lugo Jan 8 '10 at 2:01
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3 Answers

up vote 8 down vote accepted

See J. G. Wendell, "A problem in geometric probability", Math. Scand. 11 (1962) 109-111. Available online here. 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.

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nice.. am stil tryin to understand the solution.. i expected that the answer would be simpler than that !! To think this was asked as a question at the interview for an app-developer post. –  sanz Jan 7 '10 at 17:03
I seem to recall seeing a simpler solution somewhere in the specific case you asked about. –  Michael Lugo Jan 7 '10 at 17:05
If my mental picture is right, the fourth point will be inside triangle A'B'C' one eighth of the time because we may replace any subset of {A, B, C} with their antipodes, and the eight triangles formed in this way cover the sphere. –  Reid Barton Jan 7 '10 at 17:46
Kind of nice to combine the two solutions for the following puzzle: Choose $n$ points randomly. Suppose they belong to the same hemisphere. What is the (conditional) expected area of their spherical convex hull? –  t3suji Jan 7 '10 at 18:05
Computing expected areas of convex hulls is tricky enough that I don't think I'd call that a "puzzle". –  Michael Lugo Jan 7 '10 at 18:06
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We chose n points on $S^{n-2}$ and want to show that the probability for them to be in one half-sphere is $1-2^{1-n}$. A simple way to solve this question is to notice that up to a linear transformation there exsits a unique collection of generic $n$ lines in $R^{n-1}$ through 0. This reduces whe problem to a combinatorial one. End of solution.

Here are details. Namely, Instead of chosing points on the whole shpere, it is sufficient to chose these points among $2n$ points of intersection of the sphere with generic lines $L1,...,L_n$. We just need to chose one point on one line. We call these $2n$ points $P_1, -P_1,...,P_n, -P_n$

Lemma. For generic $L_i$ there will be only two choices of n points $\pm P_i$, such that the obtained simplex is not contained in the demi-sphere.

Proof for n=4. It is sufficient to check this statement for the verticies of the regular cube. Indeed, for generic 4 lines in $R^3$ there is a linear transformation that takes these lines to the axes of the cube.

"Proof" for any n. For n generic lines in $R^{n-1}$ it is alway possible to send them to the lines generated by vectors $1,0,...,0$,... $0,0,...,1$ and $1,1,...,1$. It is sufficient to check the lemma for 2n points repersenting intersections of these lines with $S^{n-2}$.

From this lemma we get the answer. Number of all choices of $n$ points is $2^n$, two choices are bad, so we get $(2^n-2)2^{-n}$.

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The probability is 7/8ths.

Consider throwing 3 darts at a sphere on average the darts will land with one on each of the ends of the Cartesian coordinates, i.e. (0,0,1), (0,1,0), (1,0,0). Or on average the SA of the spherical triangle made using the three darts as vertices will be 1/8th of the sphere volume.

This is easy to verify. The position the dart lands can be described using 2 coordinates and the equation for a sphere in Cartesian. Randomly choose an x value between -1 and 1, randomly choose a y value between -1 and 1 and the equation for a sphere will give you the z value.

On average x will be 0, y will be 0 and z will be 1.

Do this using the others and you can see why the triangle will on average have a SA of 1/8th of sphere volume.

Now consider placing a great circle around any 2 points. There are 3 ways this can be done. On average These 3 great circles will make up the x, y and z planes. Or better to describe as, on average these three great circles will be on orthogonal planes.

So there are 8 octant to choose from, Sa of 7 of these octants can be included in the same hemisphere as the 3 points by choosing different great circles. So only if the 4th dart lands in the 8th octant do we not have a great circle that can be used to split the sphere into 2 hemispheres encompassing all 4 darts.

The 8th octant will be the dipodal spherical triangle of the "average position of the darts landing. that is draw lines through the darts, through the centre of the sphere and make another spherical triangle using the intersection of the before mentioned lines on the opposite side of the sphere.

Think about it, that is the only octant that can not be encompassed using any of the three great circles.

and again

throw 3 darts, 3 great circle can be drawn, on average the planes these three great circles lie on will be orthogonal. The intersection of these three planes creates 8 octants. There can be a spherical triangle drawn around the three darts. If and only if the fourth dart lands on the antipodal spherical triangle of the first three darts will it not be in the same hemisphere.

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