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PostPosted: Wed, 30 Nov 2016 13:53:04 UTC 
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Joined: Tue, 16 Aug 2011 23:38:56 UTC
Posts: 394
REnteria wrote:
Hi guys,

can you check my work.

i would like to solve for the pair Combinations/Probability given that 8 decks are currently used.

here's my solution.

combination(13,1)*combination(32,2)*combination(416,2)

so there's 13 types of cards in a deck (A,2,3,.....,Q,K)
then each card is consist of 32 cards in 8 deck
then total number of cards is 416.

is my solution right or should i use the law of total probability. in order to get the probability of having a pair.

THANK YOU SO MUCH :)


The probability of getting a pair in two consecutive draws without replacement is:

P (pair) = \frac {31}{415}


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