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.

Let $X$ denote any value between $1-5$, then the optional sequences are:


*

*$XXX66$

*$X6X66$

*$6XX66$


Calculate the probability of each sequence:


*

*$P(XXX66)=\frac56\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^3}{6^5}$

*$P(X6X66)=\frac56\cdot\frac16\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}$

*$P(6XX66)=\frac16\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}$


Add up the above probabilities:
$$\frac{5^3+5^2+5^2}{6^5}\approx2.25\%$$
