Triangular Cuts

The key idea here is that if one of the three pieces is longer than $\frac{1}{2}$ of the length of the stick, a triangle cannot be formed.

As the stick is uniform and straight, it is symmetrical about its midpoint. Let us define $1$ as the length of the stick, and $x$ as the distance of the first cut from the midpoint.

There are two cases:

1) The second cut is in the same half as the first. In this case there is one piece which is longer than half of the original stick. The remaining two pieces together are shorter than this, so a triangle cannot be formed.

2) The second cut is in the other half of the stick. In this case a triangle might be formed, but only if the second cut is within $\frac{1}{2}$ of the first cut.

If the second cut is in the blue area, a triangle can be formed.

Therefore, the probability that a triangle can be formed, given a first cut at a distance $x$ from the midpoint is:

$P(x)= \frac{\frac{1}{2}-|x|}{1}=\frac{1}{2}-|x|$

Graphing this, from $x=-\frac{1}{2}$ to $x=\frac{1}{2}$ , we get:

The area under this graph gives us our probability, so our final probability is:

$\frac{1}{2} \times \frac{1}{2} \times 1=\boxed{\frac{1}{4}}$