Probability: 4 arbitrary points on a sphere

[tomwilliam2]

This is a well-known problem I think: take four arbitrary points on a sphere as the vertices of a tetrahedron. What is the probability that the centre of the sphere will be located within the tetrahedron?

A friend gave me this puzzle to solve, and I came up with the answer P = 1/8. This is the right answer, but my reasoning was actually completely wrong, and should have produced the result 7/8...I looked up the answer and found a different (and correct) way of solving it. I was wondering though, what was wrong with my logic:

First Step: I assumed that if the tetrahedron were not to contain the centre of the sphere, then it must be possible to slice the sphere into two hemispheres in such a way as not to incorporate any of the points. This seems to be intuitively correct to me, but maybe that's where the mistake lies.

Second Step: Calculate the probability of all four points falling within the same hemisphere:
$1 \times 1 \times 1/2 \times 1/4 = 1/8$

Unfortunately, despite giving this answer quite quickly (and getting it right), I then realized that this was the probability of the centre of the sphere NOT being located within the tetrahedron, so something has gone badly wrong. In retrospect, I'm not sure about the second step at all, because the coordinates are still to be defined for the hemispheres. I don't know if I can adapt the probability calculations in step 2 to produce the answer 7/8, or is this just rephrasing the problem and getting nowhere?

Just for clarity, I do know the standard solution: the first three points form a triangle on the surface of the sphere. The condition is met if the fourth point is located within the triangle formed by the antipodes of the first three points. The probability of this is 1/8 as the 6 points (first three plus their antipodes) form 8 triangles when interconnected, covering the surface of the sphere, and the average size will therefore be 1/8. I just wanted to know if this solution is in any way linked to my instinctive approach.

Thanks!

 

[Orodruin]

Can you provide your argument for each of the factors in your probability?

 

[tomwilliam2]

Can you provide your argument for each of the factors in your probability?

Yes, although I've just realized I wrote it down wrong.

I figured that the first point can be anywhere, so it has a probability of 1. The second would have a probability of 1/2 of falling within the hemisphere we've defined, as would the third and fourth, so that line should have read $1 \times 1/2 \times 1/2 \times 1/2 = 1/8$.
However, I've since realized that the second also has a probability of 1, as wherever it lands, you can define the hemisphere to include these first two points. I'm now realising that the third can also be made to fit within the hemisphere independently of where the first two are, so it should also be 1. Now the fourth I have no clue how to work out the probability it will be in the same hemisphere, but I have a feeling the probability is 7/8 and that I've simply redefined the original question (if my assumption is correct about hemispheres).

 

[Orodruin]

The second would have a probability of 1/2 of falling within the hemisphere we've defined,

And what hemisphere would that be that is defined by a single point? Is there any way you can have two points on a sphere that do not fall in the same hemisphere? Continue the same trail of thought also for the other points.

 

[cosmik debris]

That question is one of the Putnam questions isn't it? This video is quite good at explaining, perhaps you might get some insight from it.



Cheers

 

[tomwilliam2]

That question is one of the Putnam questions isn't it? This video is quite good at explaining, perhaps you might get some insight from it.



Cheers

Thanks, that is a really elegant explanation.

 

[tomwilliam2]

And what hemisphere would that be that is defined by a single point? Is there any way you can have two points on a sphere that do not fall in the same hemisphere? Continue the same trail of thought also for the other points.

Intuitively I would say 3 points chosen arbitrarily can always be made to fit into a correctly chosen hemisphere, but the fourth point...I can’t see an intuitive jump to explain the 1/8 probability. The video below is very clear. It must be that my method doesn’t bring any added value.

 

[Orodruin]

Intuitively I would say 3 points chosen arbitrarily can always be made to fit into a correctly chosen hemisphere, but the fourth point...I can’t see an intuitive jump to explain the 1/8 probability.

The jump is in the standard argumentation:

the first three points form a triangle on the surface of the sphere. The condition is met if the fourth point is located within the triangle formed by the antipodes of the first three points. The probability of this is 1/8 as the 6 points (first three plus their antipodes) form 8 triangles when interconnected, covering the surface of the sphere, and the average size will therefore be 1/8.