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What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different kinds and a fifth card of a third kind)?
in Set Theory & Algebra by Loyal (7.2k points)
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Select 2 out of the 13 kinds of cards first then select 2 from each kind.

$\large \binom{13}{2} \times \binom{4}{2} \times \binom{4}{2}$

now the 5th card can be chosen from remaining 11 kinds.

Total cards of remaining 11 kinds = $52-8=44$

choosing 1 card out of 44 = $\large \binom{44}{1}$

$\text{Probability} =\Large \frac{ \binom{13}{2} \times \binom{4}{2} \times \binom{4}{2} \times \binom{44}{1}}{\binom{52}{5}} $ $\approx 0.0475$
by Boss (35.4k points)
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Sorry for the silly question. Weak in card terminology :(

Kind means Spades, Hearts, Diamonds, and Clubs?

13C2 means all possible pairs of cards suppose we got (Ace,4).

what is next 4C2 * 4C2?

 No kinds mean Ace, 2 ,3, 4 ,5,....

after selecting any 2 kinds for pairs we selected total 8 cards because there are 4 cards of each kind(1 from every suit). 

From those 8 cards we need to select 1 pair of each kind. So for selecting 2 pairs $\binom{4}{2} \times \binom{4}{2}$

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