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A 5-card poker hand is said to be a full house if it consists of 3 cards of the same denomination and 2 other cards of the same denomination (of course, different from the first denomination). Thus, one kind of full house is three of a kind plus a pair. What is the probability that one is dealt a full house?

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Probability of one dealt is $\frac{Total possible full house pairs}{Total possible pairs}$

Total Possible full house pair = 13*12*24=3744

For 3 of an kind we have to choose 2 different denomination of card i.e (3,4)= 3,3,3,4,4 but (4,3)=4,4,4,3,3 thus order matters so this can be done by 13P2 ways =13*12

form 4 cards of same denomination we can choose 3 in 4C3 ways =4

form 4 cards of same denomination we can choose 2 in 4C2 ways =6

Total possible full house is 13*12*4*6=3744

Finding total possible combination is easy we have 52 cards and 5 positions thus52C5 ways

P(Full House)=$\frac{3744}{^{52}C_{5}}$

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