## Sphear of Probabilities

Posted by Karl Sharman on June 20, 2011 – 1:07 am

Choose four points on the surface of a sphere. Each point is independently chosen relative to a uniform distribution on the sphere.

What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points?

The reason this is titled Sphear is that it is a mix of sphere and fear – regular visitors will know of my inability to grasp probabilities, and that Chris will be waiting in the wings to pounce on my weakness!

June 20th, 2011 at 11:34 am

I could first try with 3 points on a circle in the plane…

let S be a circle, O its center.

First point A can be anywhere.

Second poind B is located with the angle t = (AOB) (in ]-Pi,Pi]).

Then C must be in an area which makes an angle of measure t exactly (draw a picture and see for yourself where C should be…)

That is a probability of abs(t/(2*Pi)) that the center 0 is inside the triangle ABC.

That’s where things become strange…

Then there’s kind of a dt/(2*Pi) probability for B to be where it is, and you sum the probabilities, that is you integrate :

Int(abs(t)*dt/(4*Pi²), t=-Pi..Pi) = 1/4.

Then I think you have counted each position twice as ABC and ACB, so the final answer in the case of a circle is 1/8.

I’m not quite sure about the “dt/(2*Pi)” though… It’s something our physics (whatever you call this discipline) teacher uses, something to evade any mathematically formal explanation, you just say “dt” is something small, like B has to be in a small area around where it is to be in the situation we are in…

Now I’m kind of in trouble with the 3D-sphere, because if I use this method again, I will have to deal with complex surfaces on a sphere…

June 20th, 2011 at 12:18 pm

Following a month’s sojourn in another hemisphere my mental faculties are well short of what they should be, so let me put in the first wrong answer to this puzzle . . .

The first three points of the tetrahedron define a plane which would lie on one side of the sphere’s centre. Let’s say that the sphere is our own planet. Then we can take these three points to define a latitude on the southern side of the equator. The fourth point then clearly has to lie north of the equator. It seems obvious that the further south the latitude of the plane is in the south, then the further north the fourth point has to be to to keep the centre inside the tetrahedron and therefore the smaller the chance of this happening.

Here’s my leap into the dark: The average southerly latitude of the plane (with a 50% probability) has to be matched by the fourth point not only being in the northern hemisphere (with 50% probability) but also being at least as far north as the plane is to the south, with a further 50% probability.

So my answer is 25%.

June 20th, 2011 at 2:47 pm

Good evening everybody! I am alive but still busy and ashamed that I still don’t have answer for the fourth mathematical challenge, so this is why I am not active. But we have a nice question here, and although I just have a sketch in my head and nothing on paper, I claim the answer should be 20%. To give some kind of a hint – long time ago I asked Chris to post a problem on my behalf regarding the probability of obtaining an acute triangle if three points are chosen at random on a circle. There were several correct solutions (including post #1 here by Karys) but I wanted a particular geometrical construction that leads to the answer, which so far nobody had. If I am not mistaken (many beers and nothing written down) this construction can be straightforwardly extended to the problem above. Cheers

June 20th, 2011 at 4:22 pm

Hi all. It’s re-union night. It seems we’ve all been busy together.

Hi slavy, me too (maths challenge 4)! I’m completely gutted that despite thinking about it a fair bit, I have made no progress either.

This problem looks hard – I hope there’s a really nice solution to it.

I’m not even sure what the probability of 4 points being in the same hemisphere is, yet alone forming a tetrahedron that contains the centre of the sphere.

June 21st, 2011 at 4:06 am

OK, my first retraction . . .

I suggestd in post 2 above that three points making a plane in the southern hemisphere together with the remaining point being being at least at a higher latitude in the northern hemisphere would be sufficient.

This obviously wouldn’t work if all four points were in (say)the western hemisphere.

This leads me to believe that no matter how you orient the hemispheres (NS, EW or anywhere in between) you must have at least one point in each half of the sphere.

So my next guess is that the points must be distributed at one to each of the four “demi-hemispheres”, i.e. NW (western half of northern hemisphere), NE, SW and SE.

There are 256 possible distributions of four points to the four demihemis. 24 of these allocate one to each. So, my new answer is 24/256 or 9.375%.

I’m sure I’ll have another answer in a day or two.

June 21st, 2011 at 8:10 am

I stink at probabilities enough to say I have 0 probability of getting this right.

It seems to me that the first 3 points can reside anywhere on the sphere and the fourth must reside on the opposite side of the sphere within an area the exact same shape and size as the triangle formed by the first 3 points. Furthermore the location of the 4th point must reside in a triangle which is a mirror image of the first.

Example:

Point 1 resides on the globe at 45o S and 0o W.

Point 2 resides on the globe at 45o S and 90o W.

Point 3 resides on the globe at 45o S and 180o W.

In order for the center of the sphere to be included in the tetrahedron, Point 4 must reside on the globe in a triangle formed with mirrored coordinates from Points 1, 2, and 3.

Mirrored triangle coordinates:

Point 1 resides on the globe at 45o N and 0o W.

Point 2 resides on the globe at 45o N and 90o E.

Point 3 resides on the globe at 45o N and 180o W.

Given that a globe has 180o in the southern hemisphere and 360o there are 64800 points where our degree lines intersect with infinite positions in between. Assuming all of our points must reside on an intersection of degree lines should get us in the ballpark for an answer.

This is where my brain hurts too much to go on.

Good problem!

I am guessing in the 1.4% neighborhood

June 21st, 2011 at 9:23 am

Oh dear, it’s just dawned on me that a sphere has three dimensions, not two. This is beyond embarrassing.

If we stand in front of a globe of the world, facing longitude 0 degrees, we see that the four points defined in my previous post (NE, NW, SE and SW) could all be clustered around the intersection of longitude 0, latitude 0. Such a tetrahedron would go nowhere near the centre of the sphere. That was my latest mistake in post 5.

Now let’s see what we can do to overcome this. My latest thought (almost certainly wrong) is that we need to work with what could be called semi-demi-hemispheres (SDHs). That is, again facing our globe at longitude 0, we have NE front, NE back, NW front, etc. There are thus eight of these SDHs.

If we now colour our SDHs chequerboard style so that no two adjacent SDHs have the same colour, and if we stipulate that one point of our tetrahedron must reside in each of the like coloured SDHs, then we hopefully should have mapped out all the possible tetrahedra that enclose the sphere centre.

So, out of a total of 8^4 = 4096 possible distributions of points among our eight SDHs we have 24 possible distributions among the black SDHs and 24 among the whites. My latest guess is therefore 48/4096 = 1.171875%. This is not too far from John24’s estimate in post 6 above, which gives me some hope that I’m getting warm.

June 21st, 2011 at 10:40 am

Now is obviously not a good time to let you know that I don’t have the answer yet. But my take on this is as follows, and please note that I have certainly helped myself from the contributed work so far!!

Pick three points A, B, C, giving you a spherical triangle defined by those points – T (for triangle-ish). The centre of the sphere lies in the convex hull of A, B, C, and another point (P) if and only if it lies in the convex hull of T and P. This happens if and only if P is on the opposite side to T. So the desired probability is the expected fraction of the sphere’s surface area which is covered by T……

To rip off Wizard of Oz’s hard work:

Now let’s see what we can do to overcome this. My latest thought (almost certainly wrong) is that we need to work with what could be called semi-demi-hemispheres (SDHs). That is, again facing our globe at longitude 0, we have NE front, NE back, NW front, etc. There are thus eight of these SDHs.

The eight spherical triangles NE front, NE Back, NW Front, NW Back, SE Front, SE Back, SW Front, SW Back. I’m calling these SDH’s – T1, T2…T8. The fourth point for each tetrahedron T1-T8 will therefore be P1-P8

The total fractional area of T1-T8 is exactly 1: the eight triangles cover the sphere exactly once. Hence each T has an expected fractional area 1/8. So go back to the end of para. 1… the desired probability is the expected fraction of the sphere’s surface area which is covered by T……

I have most likely oversimplified / under-thunk this, but with my track record of probabilities, like John24, I am most probably wrong.

June 21st, 2011 at 4:04 pm

Probability is a very thin ice and needs to be approached gently and carefully. It is very dangerous to fix something a priori and let the rest vary afterwards. In 99.9% of the cases this leads to a fatal end. So far I don’t see the other 0.01% in your answers/solutions/arguments.

Let us start with something much simpler – the 1 dimensional case. What is the probability that if you pick 2 points at random on the interval [0,1] (uniformly distributed of course), the line segment they cut out contains the point 1/2?

Then, as I already mentioned but maybe nobody read, go the the 2D case with three points on a circle and the probability of an acute triangle (actually today I halfened the amount of beers in my organism and I realized that Karys was wrong in the first post). This should be enough to convince you that my answer is correct. However, I would like to see more efforts from your side so I don’t intend to publish any solutions in the near future