On mathoverflow is the question Probability that a stick randomly broken in five places can form a tetrahedron.

For this question I'm interested in a simpler variant, looking at the triangles only.

Choose five points independently and randomly on [0,1]. Sort them to obtain six lengths. What is the probability that lengths (1, 2, 4), (1, 3, 5), (2, 3, 6), and (4, 5, 6) all form triangles? With ten million random trials, I seem to be getting a probability of about 1/54. Is that correct?

For a direct 2D problem with two points chosen independently and randomly on [0,1] the odds of getting a triangle are 1/4. For the marked variant in the opening sentence, the probability that triangles 124 135 236 456 exist and make a tetrahedron is conjectured to be 1/79.

EDIT: Permutations of the barycentric point {1, 1, 2, 2}/6 produce an octahedron with 1/54 the volume of the tetrahedron. Might be related.

octahedron with 1/54 volume of tetrahedron


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