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Medal and become Fan Rewarded Please Help me :D
Find the probability of the following fivecard poker hands from a 52card deck. In poker, aces are either high or low. Two pair (2 cards of one value, 2 of another value).
 one year ago
 one year ago
Medal and become Fan Rewarded Please Help me :D Find the probability of the following fivecard poker hands from a 52card deck. In poker, aces are either high or low. Two pair (2 cards of one value, 2 of another value).
 one year ago
 one year ago

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e.cociubaBest ResponseYou've already chosen the best response.2
The probability of fivecard poker hands for two pairs is 0.04753902. Probability describes the attitude of mind that asks the question whether a certain event will occur. It measures the confidence of a person in regards to an event occurring.
 one year ago

Needhelpp101Best ResponseYou've already chosen the best response.1
dw:1359757267374:dw
 one year ago

LizzyLove<3Best ResponseYou've already chosen the best response.0
@e.cociuba how please explain cause i need help to...
 one year ago

Needhelpp101Best ResponseYou've already chosen the best response.1
Hey Lizzy do you go to Ashworth College online?
 one year ago

LizzyLove<3Best ResponseYou've already chosen the best response.0
i dont get what ur sayin e.cociuba
 one year ago

LizzyLove<3Best ResponseYou've already chosen the best response.0
no...i just need help with something like tht
 one year ago

e.cociubaBest ResponseYou've already chosen the best response.2
hold on @LizzyLove<3
 one year ago

Needhelpp101Best ResponseYou've already chosen the best response.1
@e.cociuba please explain to me too :D thx
 one year ago

e.cociubaBest ResponseYou've already chosen the best response.2
So i did the problem again and got a different answer this time. I'm pretty sure this is the right way to do it!! First, there are 52C5 = 52! / (47! 5!) = 2,598,960 different possible fivecard poker hands. There are 13 possible ranks for the three cards of a kind, and each case leaves 12 possible ranks for the pair. There are 4 ways to select three cards of any rank (differing in which card is omitted). There are 4C2 = 6 ways to select a pair of any rank. So, the number of different full houses is 13 * 12 * 4 * 6 = 3,744 The probability is therefore 3,744/2,598,960 = 6/4,165 = about 1.44*103
 one year ago
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