Probability of rolling product of 200 
Alec rolls a fair $6$-sided die repeatedly until the product of the
  rolls is at least $100$, at which point he stops rolling. Compute the
  probability that he stops rolling with a product of $200$.

My approach was to make a map: something like $200$ can come from $40$ and $50$, $40$ can come from (rolls here), etc.
This method is very bashy, and I'm curious if there is a better method, or if this method even works.
 A: First, you can consider that the die has only $5$ faces, numbered from $2$ to $6$, and is fair, since rolling a $1$ doesn't change the product.
Then, factorising $200$ gives you $200=2^3\times5^2$. It can be done either in 4 rolls (two 5's, one 2, one 4) or 5 rolls (three 2's instead of one 2 and one 4).
The probability of reaching 200 in 4 rolls is $(\frac15)^4\times(\frac{4!}{2!})$. The only problem here is that 2 can't be the last number rolled because we would stop at 100. Hence, the actual probability of this happening is $(\frac15)^4\times(\frac{4!}{2!}-\frac{3!}{2!})=(\frac15)^4\times9$.
When we want to count in how many ways we can reach 200 in 5 rolls without having 2 as the last roll, we are just left to find in how many ways we can order one 5 and three 2's, since the last roll will be a 5 for sure. This can be done in 4 different ways. Hence, the probability of reaching 200 in 5 rolls is: $(\frac15)^5\times 4$.
So the total probability is $(\frac15)^4\times9+(\frac15)^5\times4$.
