Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**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

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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**

Offline

**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: 22,097

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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**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