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santa
2005-10-07 16:41:23

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

ganesh
2005-10-05 21:08:43

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.

Klaus
2005-10-05 08:48:41

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

thanks for looking at this..

Klaus

John E. Franklin
2005-10-05 08:00:54

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?

kweyers59@aol.com
2005-09-26 17:36:58

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