A pair of fair dice is rolled repeatedly then the probability of getting doublet 4^th time in the 9^th trail

The probability of getting doublet in the random experiment of throwing two dice is $\frac{1}{6}$

To get the fourth time doublet in the 9th trail, in the first 8 trails the doublet must occur only three times, and the 9th trail must be doublet

Select three trails from the 8 trails, these three trails must be doublet and the other five trails must  be doublet. finally the 9th trail must be doublet.

Hence, the probability is $8_{C_3}{\left(\frac{5}{6}\right)}^5{\left(\frac{1}{6}\right)}^3\left(\frac{1}{6}\right)=\boxed{\frac{8_{C_3}{5}^5}{6^9}}$