Finding a red ace before black ace in a deck We draw cards one by one from a deck.If for the first time we see a red ace at position 6, what is the probability after that to see a red ace before black ace?.
What I have tried is the following - Let $A$ be the event of finding a red ace at position 6 and let B be the event : there is a red ace at position >6 and less than the position of the last black ace .
so we have to find : $P(B|A) = P(BA)P(A) = P(B \cap A)P(A) $.
P(A) is $\frac{46}{{52}\choose{2}}$ , because for the denominator the 2 aces can be anywhere in the 52 cards , and for the numerator if the first is at position 6 then we have the second red ace to be from positions [7;52] which is ${46}\choose{1}$ which is 46.
But how can I compute $P(BA)$ ?
 A: You're making it too complicated with the $46$ remaining cards. Just consider this:  If in the first $5$ cards there were no black aces, then there are three aces (two blacvk and one red) remaining in the $46$ cards, and since each is equally likely to show up next, the probability of the next ace being red is $\frac{1}{3}$
Likewise, if one of the first five cards was a black ace, then the probability of the first case after the first red ace being red is $\frac{1}{2}$
And if both black aces were among the first $5$ cards, then the probabilitt of the next ace being red is $1$.
Now, the probability of not getting any black aces in the first $5$ cards, given that the first red ace is in position $6$ is the probability of not getting aces at all in the first $5$ cards, and that is 
$$\frac{{48 \choose 5}}{{52 \choose 5}}$$
The probability of getting $1$ black ace in the first $5$ cards and not getting any red aces is 
$$\frac{{2 \choose 1}\cdot {48 \choose 4}}{{52 \choose 5}}$$
And the probability of getting both black aces but no red aces is:
$$\frac{{2 \choose 2}\cdot {48 \choose 3}}{{52 \choose 5}}$$
So, the probability is:
$$\frac{{48 \choose 5}}{{52 \choose 5}}\cdot \frac{1}{3}+\frac{{2 \choose 1}\cdot {48 \choose 4}}{{52 \choose 5}}\cdot \frac{1}{2} + \frac{{2 \choose 2}\cdot {48 \choose 3}}{{52 \choose 5}} \cdot 1$$
