One in Eight Random Tetrahedra Hold the Centre

Three Points on a Circle

Pick three points on a circle, every place equally likely, and join them into a triangle. Sometimes the centre is inside, sometimes not. The model picks new points again and again, and the counter keeps the share that hold the centre.

Fix P₁ and P₂ first. Draw a line from each through the centre, and the circle is cut into four arcs. The triangle holds the centre only when P₃ lands on the arc opposite P₁ and P₂.

The angle θ between P₁ and P₂ is spread evenly from 0 to 180°, so the opposite arc, θ/360° of the circle, is spread evenly from 0 to a half. On average it is a quarter. So the probability is 1/4.

Lines First

There is another way to pick a point, with exactly the same result: pick a random line through the centre, then flip a coin to choose one of its two ends. Every place on the circle is still equally likely.

Pick two lines, then P₃, and flip the two coins last. Once the lines and P₃ are set, the coins have four outcomes, all equally likely. Exactly one of them puts both P₁ and P₂ opposite P₃, and the triangle holds the centre.

Wherever the two lines and P₃ fall, it is one outcome in four. So the probability is 1/4.

Four Points on a Sphere

Up one dimension: pick four random points on a sphere and join them into a tetrahedron. Problem A6 of the 53rd Putnam competition, in 1992, asked how likely the centre is to lie inside. The counter settles near 1/8.

The first way carries over: fix P₁, P₂ and P₃, and P₄ has to land in the spherical triangle with corners −P₁, −P₂ and −P₃. That triangle is sometimes large, sometimes small, and on average exactly an eighth of the sphere, but finding that average directly takes a hard integral.

Eight Outcomes

Switch to lines: pick three lines through the centre, then P₄, and flip three coins last to choose which end of each line to use. Three coins have eight outcomes, all equally likely.

Write −P₄ as a mix of the three line directions, −P₄ = a₁u₁ + a₂u₂ + a₃u₃. When every line takes the end on the side of its aᵢ’s sign, the centre is inside the tetrahedron; flip any one and it is outside. Exactly one outcome in eight, so the probability is 1/8.

A Different Question

The two ways ask different questions. The first asks how large the target is, and in three dimensions it runs into an integral. The second takes the way the points are picked apart and puts it back together in a new order, lines first and coins after, and the question becomes counting: one in four, one in eight.

The lines through the centre were first drawn only to see which arc counts. Made into the first step of the random choice, they take away the need for the hard average. Here the model does only that: three lines and P₄, and of the eight outcomes it lights the one that works.

The same count works in any dimension and for any number of points. In 1962 J. G. Wendel proved that n random points on the unit sphere in d dimensions (the circle is d = 2) all lie in some half-sphere with probability 2⁻⁽ⁿ⁻¹⁾·Σ C(n−1, k), for k from 0 to d−1. Their hull holds the centre with 1 minus that: one half for four points on a circle, 5/16 for five on a sphere.