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caffeine
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.
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!
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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