Breaking a stick to form a triangle A stick is randomly broken into $n$ pieces. What is the minimum value of $n$ such that there always exists three pieces that can form a non-degenerate triangle? Preferably without calculus.
I know how to compute the probability of forming a triangle when $n=3$ but I don't think I can apply that to this problem.
 A: Community wiki answer so the question can be marked as answered:
As Ed pointed out in a comment, the counterexample of a stick breaking into pieces with lengths given by the first $n$ powers of $2$ shows that the existence of three pieces that can form a non-degenerate triangle is not guaranteed for any $n$.
A: Partial answer for Ed Pegg's follow-on question: For $n = 3$, and the two breaking points chosen independently and uniformly from the unit interval, the probability is $1/4$.  Let the breaking points be $x$ and $y$.  Then the three pieces form a triangle if and only if


*

*$|x-y| < \frac{1}{2}$; and

*Either $x < \frac{1}{2} < y$, or $y < \frac{1}{2} < x$.
Plotting this area reveals the covered area to equal $1/4$.
I haven't given much thought to how straightforward it is to generalize this to larger $n$, though.  Of interest is the observation that the addition of a third breaking point can turn a qualifying set of three pieces into a non-qualifying set of four pieces.  For instance, suppose we have three pieces, of lengths $9/20, 7/20, 1/5$.  These clearly form a triangle.  But suppose the last break snaps the $1/5$ piece into two $1/10$ pieces.  Now, no three pieces can form a triangle, because the smallest of these will always be a $1/10$ piece, and that piece plus the middle piece can never exceed the largest piece.
