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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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    Ah I see what is happening now...I see why it should be 44 instead of 46.2012-10-28

1 Answers 1

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If you change some of the values in your method one, you get

$\frac{1}{11}\frac{{4 \choose 2}{44 \choose 1}}{48 \choose 3}\frac{{4 \choose 2}{48 \choose 3}}{52 \choose 5}$

which gives the same answer as method two.

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.