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If 6 numbers are chosen at random, uniformly and independently, from the interval [0,1], what is the probability that they are the lengths of the edges of a tetrahedron?

I wrote some code and simulate this probability and I suppose the answer is 1/3.

As for 2-dimensional question it is trivial to find that: If 3 numbers are chosen at random, uniformly and independently, from the interval [0,1], the probability that they are the lengths of the sides of a triangle is 1/2.

3-dimensional case seems to be very difficult to prove.

Moreover is this true (n-dimensional case):

If n(n+1)/2 numbers are chosen at random, uniformly and independently, from the interval [0,1], what is the probability that they are the lengths of the edges (1-faces) of a n-dimensional simplex?

Inspired by the 2 and 3-dimensional cases I suspect the answer is 1/n but this one seems to be a real chestnut.

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These links will probably be helpful, as they describe conditions beyond the triangle inequality that edges of a tetrahedron must satisfy eom.springer.de/t/t110020.htm math.niu.edu/~rusin/known-math/98/tetrahedral_ineq –  j.c. May 17 '10 at 16:39
    
I got the 15 arrangements worked out, but I can't figure out how to go from lots of variables to an actual numeric answer. I have "the answer is probably greater than a quarter" from trial and error, but past that, I have no idea. –  Gabriel Benamy May 17 '10 at 16:50
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1 Answer

There are four ways to interpret your question:

1) compute the probability exactly when each length $\ell_{ij}$ is chosen independently

2) compute the probability exactly when 6 numbers are chosen independently and you have a freedom to assign them to edges in any way

3) you want some (say, 0.001) lower bound in either case

4) you want a 1/3 lower bound in either case

In case 1), consider the set of possible tetrahedron 6-tuples of edge lengths as a points in $\Bbb R^6$. You basically want to compute the volume of this set. Unfortunately, this set is non-convex and is defined by rather nasty inequalities (see this Rivin's paper which I already mentioned on this MO answer). To get convexity, Rivin shows you need to consider squares of edge lengths. In other words, the desired volume is a volume of an algebraic body and is likely be non-algebraic itself. It being 1/3 is doubtful.

For 2), the problem is much harder as you have various permutations to consider. For 3), this is easy - a small enough perturbation of lengths of a regular simplex will work. For 4), this may or may not be true. I sort of doubt it if you don't allow permutations, but with permutations I have no intuition. In principle, you can simply approximate the volume of a body in $\Bbb R^6$, there are better ways for doing that than sampling random points and checking if it's in there. Either way, it is hard to imagine how you would later extend this to simplices in higher dimension.

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Definitely I was thinking to case 2) and a trial computing program suggest that the answer is in this case 1/3. I was looking indeed for a proof of case 2) ... The Rivin's paper you suggest is no longer available on the provided link arxiv1.library.cornell.edu/abs/math/0308239 (maybe you have a copy of it to share) Thanks, –  heartwork Nov 2 '11 at 12:58
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