December 11th, 2010 at 6:13:16 AM
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What are the odds and expected frequency off a suited pair occurring in an 8 deck shoe when only two cards are dealt, one to player one to house.
December 11th, 2010 at 9:09:04 AM
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Do you mean identical cards? If so, assuming a randomly-shuffled, full 8-deck shoe, the probability is 7 in 415. There are 416 cards in an 8-deck shoe, and one is the first card out. Once that card is out, there are only 7 left in the deck that match it out of 415 remaining. This changes, sometimes dramatically, as cards are dealt.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
December 11th, 2010 at 8:52:17 PM
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Thanks for the reply,clarifies a couple of things.
December 14th, 2010 at 8:31:43 AM
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Am I right in assuming ttha his question has something to do witth the Perfect Pair BJ side be or am I way off?
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December 23rd, 2010 at 8:11:49 AM
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DUHHIIIIIIIII HEARD THAT!
August 2nd, 2016 at 4:38:40 AM
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Quote: MathExtremistDo you mean identical cards? If so, assuming a randomly-shuffled, full 8-deck shoe, the probability is 7 in 415. There are 416 cards in an 8-deck shoe, and one is the first card out. Once that card is out, there are only 7 left in the deck that match it out of 415 remaining. This changes, sometimes dramatically, as cards are dealt.
Is that correct, 13 cards in a deck and potential pairs of; J J, Q Q, K K.
August 2nd, 2016 at 5:56:23 AM
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There are 52 cards in a deck, if I'm doing the math properly. Not sure where the jacks queens and kings comment comes into play, though.Quote: CyrusVIs that correct, 13 cards in a deck and potential pairs of; J J, Q Q, K K.
August 2nd, 2016 at 7:28:36 AM
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There are 13 possible pairs in 8 decks, 1 ~ 10 then 3 picture cards.Quote: MathExtremistthe probability is 7 in 415
August 2nd, 2016 at 8:11:58 AM
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Quote: CyrusVThere are 13 possible pairs in 8 decks, 1 ~ 10 then 3 picture cards.
This is also true.
August 2nd, 2016 at 11:10:42 AM
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Got it after re-reading the OP's question.