Maximum of all distances between any pairs of vertices of a random triangle

Question

Given a set $S_n$ of $n$ points selected uniformly at random in a $d$-ball, I want to estimate the average length of the longest side of a triangle having as vertices any three distinct points of $S_n$, when $n\gg d\gg 1$. Below you can find my approach, which unfortunately does not scale well with the number of points $n$ nor the number of dimensions $d$. Is it possible to improve it to make it more scalable?

n:=300; d:=30;
p := RandomPoint[Sphere[d], {n}];
x:=EuclideanDistance @@@ Subsets[p, {3}][[1 ;; , 1 ;; 2]]; 
y:=EuclideanDistance @@@ Subsets[p, {3}][[1 ;; , 2 ;; 3]]; 
z:=EuclideanDistance @@@ Drop[Subsets[p, {3}], None, {2}];
Mean[Map[Max, Transpose[{x,y,z}]]]

Answer 1

Maybe

SeedRandom[1]
n = 300; d = 30; 
p = RandomPoint[Sphere[d], {n, 3}]; 
xyz =  Apply[EuclideanDistance, Subsets[#, {2}] & /@ p, {2}];

Mean[Map[Max, xyz]]

1.5164

If we use Ball[d] instead of Sphere[d] we get 1.47179.

And a modification of your code (eliminating repeated invocations of Subsets[#,3] on p):

SeedRandom[1]
n = 300; d = 30;
p = RandomPoint[Sphere[d], {n}];
xyz = With[{s3 = Subsets[p, {3}]}, 
   Apply[EuclideanDistance, Subsets[#, {2}] & /@ s3, {2}]];
Mean[Map[Max, xyz]]

1.51594

If we replace Sphere[d] with Ball[d] we get 1.46985.

Answer 2

Seems no so effective.

Here we use RandomPoint to select uniform points in Ball[] and use RandomSample to select three points to construct a random triangle.

SeedRandom[1];
n = 300;
d = 30;
pts = RandomPoint[Ball[d], n];
Table[Max @@ 
   EuclideanDistance @@@ Subsets [RandomSample[pts, 3], {2}], {i, 
   200000}] // Mean

1.46926