Greg Egan on Nostr: SOLUTION: Take the unit square [0,1] × [0,1], and consider the two horizontal edges. ...

SOLUTION:

Take the unit square [0,1] × [0,1], and consider the two horizontal edges.

Parameterise the endpoints of a line joining those two edges as (α,0) and (β,1), then parameterise two points along that line as:

(1-λ) (α,0) + λ (β,1)
(1-μ) (α,0) + μ (β,1)

where α, β, λ, μ all range from 0 to 1.

So we have a map from the parameter space [0,1]^4 to the geometric space of pairs of points [0,1]^4:

F(α,β,λ,μ) = (β λ + α (1-λ), λ, β μ + α (1-μ), μ)

The absolute value of the Jacobian determinant of F is just:

|J(F)| = |μ-y|

The integral of this over [0,1]^4 in the parameter space gives the 4-volume in the geometric space:

∫_[0,1]^4 |μ-y| = 1/3

We multiply by two to account for the other case, where the pair of opposite edges are the two vertical edges.

So P(opposite) = 2/3.

I think this is a simpler approach than trying to specify the required subset of the geometric space in terms of a collection of inequalities in the Cartesian coordinates. My original intuition was that this subset ought to consist of a union of convex polytopes in R^4 ... but it doesn't, the boundaries are not hyperplanes!