If you pick a random $3$-digit number, what is the probability that its hundreds digit is triple the ones digit?
1. Let the 3-digit number be represented as $\overline{abc}$, where $a$, $b$, and $c$ are the hundreds, tens, and ones digits respectively.
2. According to the problem, the hundreds digit $a$ must be triple the ones digit $c$. Therefore, we have the relationship $a = 3c$.
3. Since $a$ is a digit in the hundreds place of a 3-digit number, it must be between 1 and 9. Similarly, $c$ must be a digit between 0 and 9.
4. We need to find the valid values of $c$ such that $a = 3c$ is also a valid digit (i.e., between 1 and 9). 
5. The possible values for $c$ are:
   - If $c = 1$, then $a = 3 \times 1 = 3$.
   - If $c = 2$, then $a = 3 \times 2 = 6$.
   - If $c = 3$, then $a = 3 \times 3 = 9$.
6. Therefore, the valid pairs $(a, c)$ are $(3, 1)$, $(6, 2)$, and $(9, 3)$. This gives us 3 possible values for $a$.
7. The tens digit $b$ can be any digit from 0 to 9, giving us 10 possible values for $b$.
8. Thus, the total number of 3-digit numbers where the hundreds digit is triple the ones digit is $3 \times 10 = 30$.
9. The total number of 3-digit numbers is $900$ (from 100 to 999 inclusive).
10. Therefore, the probability is given by:
   \[
   \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{30}{900} = \frac{1}{30}
   \]

The final answer is $\boxed{\frac{1}{30}}$