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.

Here is simplified solution.
In order experiment to end at 5th trial, the last two rolls must be 6. So we have,
NNN66 $Pr=5^3/6^5$
6NN66  $Pr=5^2/6^5$
N6N66  $Pr=5^2/6^5$
$(N <6)$
Add all probabilities. 
PS. The last two roll must be (6,6), so we get $\frac {1}{6^2}$
The third roll can't be 6, so we get $\frac {5}{6}$
And first two rolls can be anything except  (6,6), so we get $\frac {35}{6^2}$
Answer is $\frac {175}{6^5}$
