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**kweyers59@aol.com****Member**- Registered: 2005-09-25
- Posts: 2

Help... a problem in combinatorics..

stadndard 52 card deck - 5 cards being dealt.. what is the probability to have at least 3 non-paired cards under 9??

Santa in Florida, USA kweyers59@aol.com

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

So can the 3, 4 or 5 non-paired cards be from Ace to 8 or from Ace to 9 or from 2 to 8 or from 2 to 9?

**igloo** **myrtilles** **fourmis**

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**Klaus****Guest**

we are interested in at least three unpaired cards from Ace to 8.. smile...

thanks for looking at this..

Klaus

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,076

I think the required probability is

(32 x 28 x 24)/(52 x 51 x 50) assuming the raminaing two cards can be either paired or unpaired, greater or lesser than nine.

Character is who you are when no one is looking.

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**santa****Guest**

Ganesh.. tks for helping but your answer must be wrong..

u come up with about 16%.. and I am darn sure it is closer to 60%...

please keep at it.. or teach m,e how to do it..

THANKS