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.

the odds of getting two sixes, in any order, within 6 dice (or is it die??) can be easily determined by using the binomial distribution
the odds would be 1/6 * 1/6 * 5/6 ^ 4
that opening statement very simply finds the chances of getting two sixes, but there is more than one way to get two sixes, so you need to multiply that value by
2!/6!(6-2)!
goes to 15..
so 15 * 1/6 * 6 * 5/6 * 4 = about 0.2009
so basically 1/5 
