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Please excuse me if I don't type this right, this is my first posted question. I hope I do this right....

I'm having problems with a question from my Intro to Math Analysis course. It is not from any book, but a worksheet that the professor made himself. I need to find the probability of getting a flush or a three of a kind in a 5-card poker hand. It is a standard deck, and I need to exclude "better hands", like straight flushes and full houses. I know the denominator is ${52\choose {5}}$. For the numerator I have calculated the following:

Number of ways to get a flush but not a straight flush: $4\cdot {13\choose {5}} -4 \cdot 10 $

Number of ways to get a three of a kind but not a full house: $13 \cdot {4\choose {3}} \cdot 4 \cdot{12\choose {1}} \cdot4 \cdot{11\choose {1}} $

I know I need to add these together, and I know I don't have to worry about subtracting the number of hands that have been counted in both because you can't have a three of a kind of the same suit anyway. So I have a numerator of $114,932$. But this is far from the answer, which has a numerator of $60,020$. Please help. Thank you.

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  • $\begingroup$ I couldn't figure out how to get the binomials in that format. Thank you, RSerrao, for fixing. I tried both the {13 \choose 5} and \binom{13}{5} formats listed in the MathJax tutorial. It's not showing up for me. $\endgroup$ – dsfsu Nov 10 '16 at 0:12
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No that looks almost okay. If Ace may be high or low, then: $$\dfrac{(\binom {13}{5}-\binom{10}1)\binom 4 1 + \binom {13} 1\binom 4 3\binom {12} 2{\binom 41}^2}{\binom{52}5} =\dfrac{ 60,020}{2,598,960}$$

In the selection of the two singleton cards in the three-of-a-kind you used $12\times 11$ rather than $^{12}C_{2}$.

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  • $\begingroup$ Why 9? I thought there were 10 possible straights for each suit, because in those cases you use the Ace as a high or low card. $\endgroup$ – dsfsu Nov 10 '16 at 0:15
  • $\begingroup$ Okay, the rule should be specified. I used ace's high/ $\endgroup$ – Graham Kemp Nov 10 '16 at 0:17
  • $\begingroup$ Okay, so I was not accounting for the other 2 cards correctly in the three of a kind. Thank you very much! $\endgroup$ – dsfsu Nov 10 '16 at 0:24

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