Simple question I think - Gambling and Probability

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Simple question I think

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03-10-2017 , 08:00 PM

cicakman

banned

Join Date: Jul 2008 Posts: 9,229

You draw two cards and neither is an ace. Your friend is feeling generous and let's you draw 10 more. Still no ace. You know there are 40 cards left, 4 of which are aces.

If your friend now draws 2 cards, what's the probability she gets exactly 1 ace?

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03-10-2017 , 10:23 PM

snark

old hand

Join Date: Jun 2007 Posts: 1,234

P(ace) x P(not ace) = (4/40) x (36/39) = 9.23%

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03-11-2017 , 02:38 AM

nickthegeek

journeyman

Join Date: Sep 2011 Posts: 273

Quote:

Originally Posted by snark

P(ace) x P(not ace) = (4/40) x (36/39) = 9.23%

You have to double that. The above is the probability to draw an Ace and a non-Ace in that specific order. You have to consider the opposite order also.

- Two non-Ace cards: C(36,2)/C(40,2) = 0.807692308
- One Ace and one non Ace: C(4,1)*C(35,2)/C(40,2) = 0.184615385
- Two Aces: C(4,2)/C(40,2) = 0.007692308

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