Yahoo Answers: Answers and Comments for In a poker hand consisiting of 5 cards,find the probability of holding a) 3 aces b) 4 hearts & 1 club? [Mathematics]
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From bleach
enUS
Thu, 13 Mar 2008 04:12:31 +0000
3
Yahoo Answers: Answers and Comments for In a poker hand consisiting of 5 cards,find the probability of holding a) 3 aces b) 4 hearts & 1 club? [Mathematics]
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From Leta: Can you provide more insight?
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Sun, 31 Jul 2016 03:35:42 +0000
Can you provide more insight?

From Anonymous: 94/54145
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Thu, 31 Oct 2013 10:04:51 +0000
94/54145

From Merlyn: If you have n objects and chose r of them, the...
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Sun, 16 Mar 2008 22:18:57 +0000
If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (nr)! )
this can be written as nCr
There are 52 C 5 = 2598960 total possible hands
there are ( 4 C 3 ) * (48 C 2 ) = 4512 hands with three aces
P( three aces ) = 4512 / 2598960 = 0.001736079
there are (13 C 4 ) * (13 C 1) = 9295 hands with four hearts and one club
P( 4hearts & 1club) = 9295 / 2598960 = 0.003576431

From Anonymous: 1/20, 5/24
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Thu, 13 Mar 2008 04:25:28 +0000
1/20, 5/24

From Take Our Country Back: a) Since there are 4 aces in a deck, there are...
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Thu, 13 Mar 2008 04:52:59 +0000
a) Since there are 4 aces in a deck, there are several ways to arrange 3 aces:
nCr = n!/((nr)!*r!)
4C3 = 4!((43)!*3!) = 4
Now, there are 48 cards left in the deck if you exclude the fourth ace as a possible card for the remaining 2.
The number of possibilities for those 2 cards is 48C2 = 48!/(46!*2!) = 48*47/2 = 1128
Since there are 52C5 possible poker hands, that is the denominator in the Probability equation
So P(3 aces) = 4C3 * 48C2 / 52C5 = 4*1128/2598960 = 1736 * 10^3 = 0.1736%
b) There are 13C4 ways to pick 4 hearts. There are 13C1 ways to pick 1 club
P(4 hearts + club) = 13C4 * 13C1 / 52C5 = 9295/2598960
= 3.5764 * 10^3 = 0.3576 %