Puzzle – Probability of 4 points in a sphere

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?

Reference: https://www.physicsforums.com/threads/probability-4-arbitrary-points-on-a-sphere.934671/

Consider 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?

Puzzle – 4 points in a sphere

Solution:

Let’s first look into the problem in 2D:

Points selection for 2D plane

Step 1:

So, what will be the probability that the centre of a circle lies within the triangle formed by three points chosen uniform at random on the circumference of that circle?

• First, of let’s fix P1, then the probability of success = θ/360, as shown – this is the probability that the third point P3 lies within the arc between P1′ and P2′.
• Let P2′ be the “reflection” of P2 via centre, as shown in the pic. What is the probability of success if P2 were replaced by P2′? It is (180 – Θ) / 360.
• Now P2 and P2′ are equiprobability events in the sense that if we include both, we are over-counting (counting twice).
• The combined probability associated with P2 and P2′ put together is [Θ + (180 – Θ)] / 360 = 1/2, which is independent of Θ. Since we have been counting twice.

we get total probability = half of 1/2 = 1/4.

Step 2:

Now, let’s look into the problem in 3D:

Here we have a circle to start with, draw a line to the centre, and draw lines to 45 degrees to this line, and then draw points where they intersect the circle. Then connect these points – this gives a diameter bisecting the circle through the centre. If the two points are the other side of this diameter the triangle includes the centre, if they are on the side closer to the first point they do not.

The probability that the triangle resulting from three points chosen randomly on a circle contains the centre of the circle is therefore 1/2.

The sphere has three planes, for the centre of the sphere to be inside the tetrahedron it must satisfy the requirement for each plane. Hence probability is 1/2^3 = 1 / 8.