You are not logged in.
Post a reply
Topic review (newest first)
- santa
- 2005-10-07 16:41:23
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
- 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