Find the probability of getting two sixes in $5$ throws of a die. 
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial? 

My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.

So both the fourth and the fifth rolls need to be sixes:
$$P(4^{th},5^{th}\mbox{ rolls are sixes}) = \frac{1}{6^2}$$
There are following possible combinations of rolls for the game to not end until the fifth roll:
$$(XXXOO), (XOXOO),(OXXOO)$$
where $X,O$ represent non-six and six, respectively.
Thus, the probability is:
$$P(\mbox{end in fifth trial}) = \frac{1}{6^2}\left( \frac{5^3}{6^3} +2 \cdot \frac{5^2}{6^2} \cdot \frac{1}{6} \right) = \frac{175}{6^5}$$
