Yahoo Answers: Answers and Comments for Find the probability of getting 2 Aces, 3 kings, and one 4, and one other card when being dealt 7 cards from a standard 52card deck.? [Mathematics]
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From Maya
enUS
Mon, 27 Apr 2015 20:45:34 +0000
3
Yahoo Answers: Answers and Comments for Find the probability of getting 2 Aces, 3 kings, and one 4, and one other card when being dealt 7 cards from a standard 52card deck.? [Mathematics]
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From Puzzling: First count the number of ways to form the des...
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Mon, 27 Apr 2015 20:53:52 +0000
First count the number of ways to form the desired hand. For the "other" card, I assume it can't be another ace, king or four.
From the 4 aces, pick 2 > C(4,2) = (4 x 3) / 2! = 6 ways
From the 4 kings, pick 3 > C(4,3) = 4 ways
From the 4 fours, pick 1 > C(4,1) = 4 ways
From the 40 other cards (4 suits, 10 ranks), pick 1 > C(40,1) = 40 ways
Now multiply those together to get the total number of hands:
6 x 4 x 4 x 40
= 3,840 ways
To get the probability, we need to divide by the total ways to pick *any* 7 cards.
From the 52 cards, pick *any* 7 cards:
C(52,7) = (52 x 51 x 50 x 49 x 48 x 47 x 46) / 7!
= 133,784,560 ways
P(2 aces, 3 kings, 1 four and 1 other card (not ace/king/four))
= 3,840 / 133,784,560
That reduces to:
= 48 / 1,672,307 (or 0.0000287 or about in 1 in 34,800)

From Anonymous: 4C2 for the aces
times
4C3 for the kings
times...
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https://answers.yahoo.com/question/index?qid=20150427204534AAItr7X
Mon, 27 Apr 2015 20:51:57 +0000
4C2 for the aces
times
4C3 for the kings
times
4C1 for the 4
times
40 for the other card not an ace, king or 4.