I'm trying to figure out the probabilities for various 5 card poker hands, and I'm not having a very good time of it.(adsbygoogle = window.adsbygoogle || []).push({});

Consider "two pairs".

For the 1st pair, there are [itex]\binom{13}{1}[/itex] choices for denomination. Out of that, there are [itex]\binom{4}{2}[/itex] choices for suit.

For the second pair, there are [itex]\binom{12}{1}[/itex] choices for denomination (not 13, otherwise it'd be "4 of a kind"). And again, there are [itex]\binom{4}{2}[/itex] choices for suit.

For the last card, there are [itex]\binom{11}{1}[/itex] choices for denomination and [itex]\binom{4}{1}[/itex] choices for suit.

There's a grand total of [itex]\binom{52}{5}[/itex] things that can happen for 5 draws from 52 cards. So the probability for 2 pair should be:

[tex]

\frac{\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{2}\binom{11}{1}\binom{4}{1}}

{\binom{52}{5}}

[/tex]

which works out to be .095. A book I have says that two pair should be .048.

As another example, consider 3 of a kind.

There are [itex]\binom{13}{1}[/itex] choices for denomination of the "3 of a kind", and [itex]\binom{4}{3}[/itex] choices for suit.

Now there are 2 cards left, and 12 denominations. So there are [itex]\binom{12}{2}[/itex] ways of choosing a denomination for the two single cards, and each card can have any of [itex]\binom{4}{1}[/itex] suits. So the probability for 3 of a kind should be:

[tex]

\frac

{\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}}

{\binom{52}{5}}

[/tex]

which works out to be .0079. The book says .021.

At this point, I'm ready to admit there's a huge conceptual flaw in my head. There's something fundamental about counting that I'm just not "getting". Can someone please help clear my conceptual misunderstanding about counting?

Many thanks!

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# Homework Help: Probability of poker hands

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