What is the probability that the center of the sphere is within the tetrahedron formed by four points chosen uniformly on the surface of a sphere?

From Himanshu Nautiyal, my classmate from IIT Delhi (1991-1995). This problem also appeared as Problem A-6 in the 53rd Putnam Competition in 1992.

Video by 3Blue1Brown: The hardest problem on the hardest test (10 mins video by 3Blue1Brown) is wonderful! Below is a slightly different approach. The solution is also discussed here.

Let's first solve the problem in 2D:

Two dimensions:

What is the probabilty that the center of a circle lies within the triangle formed by three points chosen uniform at random on the circumference of that circle?

Fix P1. Choose P2 uniformly at random. Then probability of success = Θ / 360, as shown in the figure. 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 center, as shown in pic. What is the probability of success if P2 were replaced by P2'? It is (180 - Θ) / 360.

Now P2 and P2' are equi-probability events in the sense that if we include both, we are overcounting (counting twice).

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.

Three dimensions: How may we continue the strategy developed above into 3D?

Let's fix P1. Then P2, P3 are two more points on the sphere chosen uniformly at random. Just like we drew diameters (P1, P1') and (P2, P2') in a circle for the 2D case, we can draw diameters (P1, P1'), (P2, P2') and (P3, P3') on the sphere. For point P4 to form a tetrahedron with P1, P2 and P3, it has to fall inside the "solid angle" Θ formed by [P1', P2', P3']; the probability of this Θ / total solid angle in a sphere.

Remember, we argued for the 2D case that if we were to sum the probabilities associated with P2 and P2' while keeping P1 fixed, the sum of those two probabilities happened to be independent of Θ! Can we similarly argue that for the 3D case that if we were to keep P1 fixed but vary (P2 and P2') and (P3 and P3'), then the sum of probabilities associated with these choices has these properties:

(a) we overcount by 4, and

(b) the sum of the solid angles associated with these 4 choices is independent of Θ, and in fact equals half the total solid angle in a sphere?

For (a), we have to convince ourselves that choosing P2 and P3 is as likely as choosing P2 and P3', which is as likely as choosing P2' and P3, which is as likely as choosing P2' and Since [P2, P2'] and [P3, P3'] are diameters passing through the center of the circle, we can argue by symmetry that this is true.

(b) is more interesting.

Let's keep P1 fixed and vary (P2 and P2') and (P3 and P3'). We get 4 solid angles: Θ(P1', P2', P3'), Θ(P1', P2, P3'), Θ(P1', P2', P3') and Θ(P1', P2, P3). Do these add up to half the total solid angle in a sphere, no matter how P2 and P3 are chosen?

Well, let's look at the other 4 solid angles: Θ(P1, P2', P3'), Θ(P1, P2, P3'), Θ(P1, P2', P3') and Θ(P1, P2, P3). Basically, we replaced P1' by P1 for these 4 Θ's.

Can we argue by symmetry that the sum of the 4 solid angles associated with P1' equals the sum of the 4 solid angles associated with P1? If so, they would be exactly half the total solid angle because these 8 solid angles equal the total solid angle in a sphere.

Luckily, if lines corresponding to (x, x'), (y, y'), (z, z') pass through the center, then Θ(x, y, z) = Θ(x', y', z'). Using this property, we can see that the 4 angles in the first sum have 1-1 correspondence to the 4 angles in the second sum.

References: A generalization of this problem: Capturing the Origin with Random Points: Generalizations of a Putnam Problem by Ralph Howard and Paul Sisson, The College Mathematics Journal, Vol 27, Issue 3, 1996, pages 186-192.