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
$\frac{{4 \choose 2} {4 \choose 2}{4 \choose 1}}{52 \choose 5} = \frac{3}{54145}$
 A: 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 $\frac{{4\choose 2}{1\choose 50-2}}{3\choose 50}$.
A: More generally, one can count the number of $2$ pairs in a $5$ card poker hands with
$$
{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}.
$$
The hand you describe is on of a specific subsets hand in these hands, so we have to count them.
You can choose $2$ of $4$ kings, $2$ of $4$ aces, and $1$ of $4$ queen. This give
$$
{4\choose2}{4\choose2}{4\choose 1}=144.
$$
We can calculate your probability as follow: Define $T$ the event having two pairs and $H$ the event having the hand you specify. Then
$$
\mathbb{P}(H\cap T)=\mathbb{P}(H|T)\mathbb{P}(T)
$$
But $H\cap T$ is just $H$ so
$$
\mathbb{P}(H)=\frac{144}{{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}}\frac{{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}}{{52\choose 5}}=\frac{144}{{52\choose 5}}=\frac{3}{54145}
$$
This seems unnecessarly complicated but uses many theorems/property of probability along with combinatorial arguments and shows you that your second method was right.
