Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
My thoughts
Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find 
$$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$
By the binomial distribution formula, we get that 
$$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$  
Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls.  That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$
Am I correct in thinking this?
If so, then it follows that the required probability is 
$$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$
which, I know, can be simplified further!

I took a different approach to the question. Suppose B. There are three ways to get two 6's in twenty rolls: 


*

*$B_1$: Both 6's come in the first 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen. 

*$B_2$: One 6 comes in the first 10 rolls, and the second comes in the next 10 rolls. There are $\begin{pmatrix} 10 \\ 1\end{pmatrix} \begin{pmatrix} 10 \\ 1\end{pmatrix} = 100$ ways for this to happen.

*$B_3$: Both 6's come in the second lot of 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen. 


Now
$$P(A|B) = P(B_2|B_1 \cup B_2 \cup B_3) = \frac{100}{45+100+45} = \frac{10}{19}.$$
