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SansSuit

Probability question

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I don't have the math skills to figure this out. I don't have the google skills to phase this in order to get an answer. I'm sure one of you all will have the knowledge.

Nine playing cards, three of which are aces. They are shuffled and spread out face down. You have one attempt to pick out the 3 aces. What is the probability (or is it odds?) of this happening?

Thanks!
Peace,
-Dawson.
http://www.SansSuit.com
The Society for the Advancement of Naked Skydiving

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Quick guess, off the top of my head:

1 in 84

First Card: 3 Aces in Nine Cards (1 in 3)

Second Card: 2 Aces in Eight Cards (1 in 4)

Third Card: 1 Ace in Seven Cards (1 in 7)

3*4*7=84

1 in 84



I agree.
"There are only three things of value: younger women, faster airplanes, and bigger crocodiles" - Arthur Jones.

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1 in 121.5

1 in 729 chance of picking 3 cards in a particular order.
6 possible combinations of aces.
so 6 in 729 chances of picking 3 aces in no particular order.

OOPS! right 1 in 84

same reasoning but each card can be used only once (obviously) so 7*8*9=504 and 504/6=84

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1 in 729 chance of picking 3 cards in a particular order.
6 possible combinations of aces.
so 6 in 729 chances of picking 3 aces in no particular order.


Actually, 1 in 504 chance of picking 3 cards in a particular order (9 cards for first choice, 8 cards for second choice, 7 cards for third choice. 9 x 8 x 7 = 504. Or, to use the formula: 9! / (9-3)! = 9! / 6! = 9 x 8 x 7 = 504).

So, 6 possible combination of aces = 6 in 504 chances of picking 3 aces in no particular order = 1 in 84 chances.

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