Probability that a stick randomly broken in five places can form a tetrahedron Edit (June. 2015) This question has been moved to MathOverflow, where a recent write-up finds a similar approximation as leonbloy's post below; see here.

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; an example is provided below.
Furthermore, another commenter kindly points to a reference that may be of help in resolving this problem. In particular, it relates the question of when six numbers can be edges of a tetrahedron to a certain $5 \times 5$ determinant.
Finally, a third commenter points out that since one such construction is possible, there is an admissible neighborhood around this arrangement, so that the probability is in fact positive.
In any event, this problem is far harder than the classic $2D$ "form a triangle" one. 
Several numerical attacks can be found below; I will be grateful if anyone can provide an exact solution.

if stick pieces are s1 (longest) to s6 shortest.
Picture the tetrahedron with longest side s1 out of view. Then s2 is the spine and any combination of pairs from {s3,s4,s5,s6} can make the two side triangles Hence s3+s6 needs to be longer than s2 (P=0.25) And s4+s5 needs to be longer than s2. (P=0.25)
so P(can form)=0.25*0.25=0.0625
