Greg Egan on Nostr: The solution for d=2 is given here: The general solution can be found the same way, ...

The solution for d=2 is given here:

https://mathstodon.xyz/@gregeganSF/115076414261569820

The general solution can be found the same way, by noting that the conditions on x₁ and x₂ for fixed values of y₁ and y₂ apply independently to all of the d-1 other coordinates. So the probability we used for d=2, |y₁-y₂|, just becomes |y₁-y₂|^{d-1}.

We then integrate |y₁-y₂|^{d-1} over the square 0 ≤ y₁, y₂ ≤ 1, or double the integral over the triangle where y₁ ≥ y₂. This can be done with a change of variables to:

m = y₁-y₂
p = y₁+y₂

y₁ = (p+m)/2
y₂ = (p-m)/2

The change of variables gives a factor of 1/2, and we integrate:

(1/2) m^{d-1} for p from p=m (i.e. y₂=0) to p=2-m (i.e. y₁=1).

This gives us (1-m) m^{d-1}, which we then integrate for m from 0 to 1, to get:

1/[d(d+1)]

We double to get the integral over the full square, then multiply by d to account for all pairs of opposite sides. So:

P(d) = 2/(d+1)

and:

P(17) = 1/9