# 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

$\frac{{4 \choose 2} {4 \choose 2}{4 \choose 1}}{52 \choose 5} = \frac{3}{54145}$

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The first calculation makes no sense to me unless we very carefully define the events. Probability of two Kings is wrong. Indeed only the $\frac{1}{12}$ in the calculation is right. –  André Nicolas Oct 28 '12 at 19:00

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 $\frac{{4\choose 2}{1\choose 50-2}}{3\choose 50}$.

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If exactly two of the remaining cards are aces dont we have to select the remaining one card from (52 - 2 kings - 4 aces) 46 cards? –  sonicboom Oct 28 '12 at 19:54

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.

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