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?
Poker, number of three of a kind, multiple formulaes
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2For 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... – 2012-11-25
2 Answers
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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0@CodeKingPlusPlus: Glad to help. Cheers! – 2013-12-03
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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1One 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. – 2013-12-03