Find the probability of getting two sixes in $5$ throws of a die.

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:

$$\frac{5^3+5^2+5^2}{6^5}\approx2.25\%$$

The other solutions work perfectly for a modest number of rolls. If you were interested in a larger number, you might find a recursion helpful.

Let $P(n)$ be the probability that your game ends in exactly $n$ rolls. Thus your problem is asking for $P(5)$. We note that $P(1)=0,\;P(2)=\frac 1{6^2}$. For $n>2$ we remark that the first roll is either a $6$ or it isn't. If it is, then the second roll can't be a $6$. That leads to the recursion $$P(n)=\frac 16\times \frac 56\times P(n-2)+\frac 56\times P(n-1)$$

Very easy to implement this. As a consistency check, we quickly get $P(5)\sim 0.022505144$ which is in line with the direct solutions.

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 $\frac {175}{6^5}$