Probability of getting 2 Aces, 2 Kings and 1 Queen in a five card poker hand I worked this out two ways but my answers don't match. Anyone know why they dont match?
Method 1
$P(2A \cap 2K \cap 1Q) = P(Q|2A \cap 2K)P(2A|2K)P(2K)$
$= \frac{1}{12}\frac{4 \choose 2}{50 \choose 2}\frac{4 \choose 2}{52 \choose 2}$
$= \frac{1}{12}\frac{6}{1225}\frac{6}{1326} = \frac{1}{541450}$ 
Method 2
$\frac{{4 \choose 2} {4 \choose 2}{4 \choose 1}}{52 \choose 5} = \frac{3}{54145}$

The probability that your hand has exactly two Kings is $\frac{{4\choose 2}{3\choose 52-4}}{52\choose 5}$ (two Kings and three other cards).
The probability that exactly two of the three remaining cards are Aces, given that you have exactly two kings, is $\frac{{4\choose 2}{1\choose 50-2}}{3\choose 50}$.
