# Probability of getting 2 Aces, 2 Kings and 1 Queen in a five card poker hand (Part II)

So I reworked my formula in method 1 after getting help with my original question - Probability of getting 2 Aces, 2 Kings and 1 Queen in a five card poker hand. But I am still getting results that differ...although they are much much closer than before, but I must still be making a mistake somewhere in method 1. Anyone know what it is?

Method 1

$P(2A \cap 2K \cap 1Q) = P(Q|2A \cap 2K)P(2A|2K)P(2K)$

$$= \frac{1}{12}\frac{{4 \choose 2}{46 \choose 1}}{50 \choose 3}\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}$$

$$= \frac{(6)(17296)(6)(46)}{(2598960)(19600)(12)}$$

$$= 4.685642 * 10^{-5}$$

Method 2

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

$$5.540678 * 10^{-5}$$

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Please make an effort to make the question self-contained and provide a link to your earlier question. –  Sasha Oct 28 '12 at 19:56
I think we would rather ahve you edit your initial question by adding your new progress. This avoids having loss of answer and keeps track of progress –  Jean-Sébastien Oct 28 '12 at 19:56
But there already answers to my original question so those answers would not make sense now that I am using a new formula for method 1. –  sonicboom Oct 28 '12 at 20:03
Conditional probability arguments can be delicate. Given that there are exactly two Kings, what's the $46$ doing? That allows the possibility of more Kings. –  André Nicolas Oct 28 '12 at 20:26
The $46$ is because have already taken two kings from the pack leaving us with 50. And now we have chosen 2 aces and we have to pick the other 1 card from the 50 remaining cards less the 4 aces? –  sonicboom Oct 28 '12 at 20:42
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$$\frac{1}{11}\frac{{4 \choose 2}{44 \choose 1}}{48 \choose 3}\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}$$
If you wrote this as $$\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}\frac{{4 \choose 2}{44 \choose 1}}{48 \choose 3}\frac{{4 \choose 1}{40 \choose 0}}{44 \choose 1}$$ it might be more obvious why they are the same.