Once again, consider rolling our 24-sided die until you get a multiple of 9...that is, either a 9 or an 18...for the third time. Once again, the probability of getting a 9 or 18 on any given roll is \(p = \frac{1}{12}\) but since we will continue rolling until we get a success for the third time, this is modeled by a negative binomial distribution and you are looking for
\begin{equation*}
f(x) = \binom{x-1}{2} \left ( \frac{11}{12} \right )^{x-3} \left ( \frac{1}{12} \right )^3 \\
= \frac{(x-1)(x-2)}{2} \left ( \frac{11}{12} \right )^{x-3} \left (\frac{1}{12} \right )^3
\end{equation*}