<s>[INST] 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.
 [/INST] 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 $\frac {175}{6^5}$
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