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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

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.

we are interested in at least three unpaired cards from Ace to 8.. smile...thanks for looking at this..Klaus

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?

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