**Edit (June 2015):** Addressing this problem is a brief project report from the Illinois Geometry Lab (University of Illinois at Urbana-Champaign), dated May 2015, that appears **here** along with a foot-note saying: *An expanded version of this report is being prepared for possible publication*.

Below is an excerpt (though, to be clear, *this* question on MO is about finding a closed-form solution; moreover, (a) the main finding in part i agrees with an earlier **MSE response**, and (b) I have relayed the typographical error of "math.overflow.net" to the write-up's corresponding faculty mentor).

The following problem was brought to my attention by a doctoral dissertation on Mathematics Education, but - as far as I know - the solution remains unknown.

I have already asked this question on MSE, where **the post** has garnered over 90 votes, but still no canonical answer in the more-than-a-year since it has been there.

Please add or suggest different tags if it seems warranted.

Randomly break a stick in five places.

**Question:** What is the probability that the resulting six pieces can form a tetrahedron?

Clearly satisfying the triangle inequality on each face is a necessary but *not* sufficient condition.

Furthermore, the question of when six numbers can be edges of a tetrahedron is related to a certain $5 \times 5$ determinant, namely, the Cayley-Menger determinant. (See, e.g., Wirth, K., & Dreiding, A. S. (2009). Edge lengths determining tetrahedrons. Elemente der Mathematik, 64(4), 160-170. A more recent article by these authors is cited in the comments below: Wirth, K., & Dreiding, A. S. (2013). Tetrahedron classes based on edge lengths. Elemente der Mathematik, 68(2), 56-64.)

Obviously, this problem is far harder than the classic $2D$ "form a triangle" one. I would welcome any progress on finding a solution or a reference to one if it already exists in the literature.

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