This is a typical geometric probability problem. The answer is $\boxed{\frac{1}{4}}$.

Let $x$ and $y$ be the lengths of the first two sticks, respectively. Then the length of the third stick is $10-x-y$. Therefore we have the possible region defined by:

$$ \begin{cases} 0 < x < 10 \\ 0 < y < 10 \\ 0 < 10 - x - y \end{cases} $$

This is shown in the following diagram:

In order to form a triangle, the following relationships must hold:

$$ \begin{cases} x + y > 10 - x - y \\ x + (10-x-y) > y \\ y + (10-x-y) > x \end{cases} $$

These define three boundary conditions:

$$ \begin{cases} x + y =5 \\ y = 5 \\ x = 5 \end{cases} $$

Therefore, the qualified region shaded below shows the final answer as $\frac{1}{4}$.