# Probability that a random tetrahedron over a sphere contains its center

I got interested in this problem watching the YouTube channel 3Blue1Brown, by Grant Sanderson, where he explains a way to tackle the problem that is just … elegant!

I can't emphasize enough how much I like this channel. For example, his approach to linear algebra in Essence of linear algebra is really good. I mention it, just in case you don't know it.

## The problem

Let's talk business now. The problem was originally part of the 53rd Putnam competition on 1992 and was stated as

Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is in- dependently chosen relative to a uniform distribution on the sphere.)

As shown in the mentioned video, the probability is $1/8$. Let's come with an algorithm to obtain this result —approximately, at least.

## The proposed approach

The approach that we are going to use is pretty straightforward. We are going to obtain a sample of (independent) random sets, with four points each, and check how many of them satisfy the condition of being inside the tetrahedron with the points as vertices.

For this approach to work, we need two things:

1. A way to generate random numbers uniformly distributed. This is already in numpy.random.uniform, so we don't need to do much about it.

2. A way to check if a point is inside a tetrahedron.

### Checking that a point is inside a tetrahedron

To find if a point is inside a tetrahedron, we could compute the barycentric coordinates for that point and check that all of them are have the same sign. Equivalently, as proposed here, we can check that the determinants of the matrices

\begin{equation*} M_0 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_1 = \begin{bmatrix} x &y &z &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_2 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x &y &z &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_3 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x &y &z &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_4 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x &y &z &1 \end{bmatrix}\, , \end{equation*}

have the same sign. In this case, $(x, y, z)$ is the point of interest and $(x_i, y_i, z_i)$ are the coordinates of each vertex.

## The algorithm

Below is a Python implementation of the approach discussed before

from __future__ import division, print_function
from numpy import (random, pi, cos, sin, sign, hstack,
column_stack, logspace)
from numpy.linalg import det
import matplotlib.pyplot as plt

def in_tet(x, y, z, pt):
"""
Determine if the point pt is inside the
tetrahedron with vertices coordinates x, y, z
"""
mat0 = column_stack((x, y, z, [1, 1, 1, 1]))
det0 = det(mat0)
for cont in range(4):
mat = mat0.copy()
mat[cont] = hstack((pt, 1))
if sign(det(mat)*det0) < 0:
inside = False
break
else:
inside = True
return inside

#%% Computation
prob = []
random.seed(seed=2)
N_min = 1
N_max = 5
N_vals = logspace(N_min, N_max, 100, dtype=int)
for N in N_vals:
inside_cont = 0
for cont_pts in range(N):
phi = random.uniform(low=0.0, high=2*pi, size=4)
theta = random.uniform(low=0.0, high=pi, size=4)
x = sin(theta)*cos(phi)
y = sin(theta)*sin(phi)
z = cos(theta)
if in_tet(x, y, z, [0, 0, 0]):
inside_cont += 1

prob.append(inside_cont/N)

#%% Plotting
plt.figure(figsize=(4, 3))
plt.hlines(0.125, 10**N_min, 10**N_max, color="#3f3f3f")
plt.semilogx(N_vals, prob, "o", alpha=0.5)
plt.xlabel("Number of trials")
plt.ylabel("Computed probability")
plt.tight_layout()
plt.show()


As expected, when the number of samples is sufficiently large, the estimated probability is close to the theoretical value: 0.125. This can be seen in the following figure.