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 firstname.lastname@example.org
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
we are interested in at least three unpaired cards from Ace to 8.. smile...
thanks for looking at this..
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.
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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..