Let us label the broken pieces as $x$, $y$, and $z$, where $x$ is the length of the piece closest to the left end of the stick, $y$ is the length of the middle piece, and $z$ is the length of the piece closest to the right end of the stick. Without loss of generality, we can assume that $x\leq y\leq z$.
The total sample space of all possible ways the stick can be broken is determined by choosing two points on a stick of length 6. There are a total of ${{6}\choose{2}}=15$ ways to do this.
Now we need to find the subset of this sample space in which all three pieces are shorter than 5 units. For this to happen, we must have $x < 5$, $y < 5-x$, and $z < 6-x-y$. Note that $z$ can be at most 6 minus the sum of the lengths of the other two pieces, because the pieces cannot overlap.
We can graph the region of the $(x,y)$ plane where these three conditions are all satisfied. This region is a triangle with vertices at $(0,0)$, $(5,0)$, and $(2.5,2.5)$.
[asy] size(8cm); draw((0,0)--(5,0)--(2.5,2.5)--cycle); fill((0,0)--(5,0)--(2.5,2.5)--cycle,gray(0.7)); draw((5,0)--(6,0)); draw((2.5,2.5)--(6,2.5)); draw((0,0)--(0,6)); draw((0,6)--(6,6)); draw((0,0)--(6,6),dashed); label("$x$", (6,0), S); label("$y$", (0,6), W); label("$(0,0)$", (0,0), SW); label("$(5,0)$", (5,0), S); label("$(2.5,2.5)$", (2.5,2.5), NE); [/asy]
The area of this triangle is $\frac{1}{2}(5)(2.5)=6.25$, so the probability that all three resulting pieces are shorter than 5 units is $\frac{6.25}{15}=\boxed{\frac{5}{12}}$.
