Probability of getting an ACE of Spade in the first 4 to 13 cards of an eight deck shuffled card
October 19th, 2010 at 4:27:58 AM
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Hi, please help, how do you compute the probability, that there is an Ace of Spade in the 4th to the 13th card of an eight-deck shuffled cards, i tried to compute
the 416-card combinations , but that's too huge.
Thanks and more power
the 416-card combinations , but that's too huge.
Thanks and more power
October 19th, 2010 at 5:51:04 AM
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There are 8 aces-of-spades and 408 non-aces-of-spades in eight decks. The probability of having exactly one ace-of-spades and nine non-aces-of-spades in ten random cards is the number of ways of choosing one thing out of 8 things times the number of ways of choosing nine things out of 408 things divided by the number of choosing ten things out of 416 things. In Excel notation that's =combin(8,1)*combin(408,9)/combin(416,10), which is about 16.5%.
October 19th, 2010 at 8:16:43 AM
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His question did not say exactly one ace of spades, but, rather an ace of spades. So I would guess you would need to add the probability of 2 aces of spades, 3, 4... etc.. to get the actual percentage... probably between 17 and 18%... Right?
October 19th, 2010 at 8:33:05 AM
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(COMBIN(8,1)*COMBIN(408,9)+COMBIN(8,2)*COMBIN(408,8)+COMBIN(8,3)*COMBIN(408,7)+..etc...+COMBIN(8,8)*COMBIN(408,2)) / combin(416,10) = 17.8%
October 19th, 2010 at 10:51:56 AM
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You're forgetting that it was NOT within the first ten cards, but in a SPECIFIC ten cards. I.E. The 4th thru 13th cards.
Does that change anything?
Does that change anything?
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October 19th, 2010 at 11:08:13 AM
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Not without further constraints like "no ace in the first 3 cards".
"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
October 19th, 2010 at 7:58:22 PM
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an Ace of spade in the first three cards does not matter....
October 19th, 2010 at 7:58:26 PM
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Thanks you all
