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I wanted to calculate some poker hands, for a three of a kind I infered, 1) every card rank can form a 'three of a kind' and there are 13 card ranks, 2) there are $\binom{4}{3}$ ways to choose three cards out of the four suits of every card rank, and 3) for the remaining card I can choose two out of 49 cards, i.e. $\binom{49}{2}$. Together the formulae is $$ 13 \cdot \binom{4}{3} \cdot \binom{49}{2} = 61152 $$ But on Wikipedia I found a different formulae, namely $$ \binom{13}{1} \binom{4}{3} \binom{12}{2} \left( \binom{4}{1} \right)^2 = 54912 $$ which makes also totally sense to me (1. card rank, 2. subset of suits, 3. choose form the left card ranks, 4. assign suits). But I can't see why my first formulae is wrong, can anybody explain this to me?

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For the remaining cards, you cannot choose the same face value as that of the cards comprising the three of the kind. Nor could you choose two cards of the same suit. So, you do not have 49 cards from which to choose... –  David Mitra Nov 25 '12 at 16:15

2 Answers 2

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In counting the number of hands with three of a kind we must not include those that have four of a kind or a full house.

As @Jean-Sébastien notes in the comments, your formula counts $$\# (\textrm{three of a kind}) + 4\# (\textrm{four of a kind}) + \# (\textrm{full house})$$ or $$54,912 + 4\times 624 + 3,744.$$ The factor of four arises since $\rm AAA{\underline A} = AA{\underline A}A = A{\underline A}AA = {\underline A}AAA$.

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He counts the four of a kind $4$ times each, but I dont see why. So it suffices not to remove the $4$ of a kind and full houses. –  Jean-Sébastien Nov 25 '12 at 16:18
    
@Jean-Sébastien: Thanks for the comment. I've added something above. –  user26872 Nov 25 '12 at 16:52
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Originally, I did $\binom{13}{1}\binom{4}{3}\binom{49}{2} - |four of a kind|- |full house|$. However, it is not entirely clear to me your reasoning as to how it counts the factor of $4$. Could you try something else for me? –  CodeKingPlusPlus Dec 3 '13 at 3:51
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@CodeKingPlusPlus: Notice that $\binom{13}{1}\binom{4}{3}\binom{49}{2}$ counts hands consisting of three of a kind and any two other cards. The shufflings of the three of a kind have been accounted for by the factor $\binom{4}{3}$, but if the fourth card creates four of a kind we must correct for the fact that A1 A2 A3 A4 and A1 A2 A4 A3 are treated as distinct hands. –  user26872 Dec 3 '13 at 5:07
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Ok, I can see that now. –  CodeKingPlusPlus Dec 3 '13 at 5:19

We can count more or less like you did, using $\dbinom{13}{1}\dbinom{4}{3}\dbinom{48}{2}$ (note the small change), and then subtracting the full houses.

Or else after we have picked the kind we have $3$ of, and the actual cards, we can pick the two "useless" cards. The kinds of these can be chosen in $\dbinom{12}{2}$ ways. Once the kinds have been chosen, the actual cards can be chosen in $\dbinom{4}{1}^2$ ways, for a total of $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2.$$

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Look at my comment on the other answer. Could you explain why $\binom{49}{2}$ does not work even if you subtract the number of 4 of a kinds and full houses? I don't quite see how the factor of 4 arises? –  CodeKingPlusPlus Dec 3 '13 at 4:01
    
One can do it by subtraction. However, if we want to subtract the $4$ of a kind, starting from your expression, we have to be careful. For your $\binom{4}{3}$ gives us say King of spades, hearts, diamonds, and then your $\binom{49}{2}$ includes the possibility of King of clubs. However, your $\binom{13}{1}\binom{4}{3}\binom{49}{2}$ also allows for the possibility of King of hearts, diamonds, clubs, followed by King of spades. So your formula counts each "four of a kind" hand $4$ times. We want to count them no times, so we need to subtract $4$ tims the number of $4$ of a kind hands. –  André Nicolas Dec 3 '13 at 6:13

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