How to count C(47,20) / C(49,20) on my comp?

[Supern]

What is the mathematics behind C(47,20)?
What is the formula for combinatorics?

I searched for it but couldn't find anything.
Thanks for you help!

[LetYouDown]

C(n,r) = n!/[(n-r)! * r!]

If you have excel, you can use =COMBIN(N,R)

HTH

[Supern]

Sweet!

The Excel answer was just what I was looking for.

I'm counting this problem:

Flop is QQA and I have AK for example.

I want to calculate the probability no one was dealt a Q.

10 players at the table: C(45,18)/C(47,18) = ~37%
63% chance someone has trips if everyone takes the flop

6 players at the table: C(45,10)/C(47,10) = ~62%
28% chance someone has trips if everyone takes the flop

Good to know these hard facts when playing against habitual bluffers. They can't have trips every time. /images/graemlins/grin.gif

Of course are the real threat lower because every hand that are dealt don't play.
I understand this.

[LetYouDown]

Glad to help. Looks like your calculations are right...for at LEAST trips.

[SheetWise]

It's worth noting (when you don't have a computer handy) that; C(n,r) = C(n,n-r), so C(47,45) = C(47,2)

[LetYouDown]

Amen to that. Not too long ago, I forgot that simple shortcut and cost myself a lot of manual effort.

[Supern]

You are right I don't add the AA-probability in this example.

If I want to know the risk being outkicked when I flop trips? Or if an opponent has flopped a full house (either with pocket pair or the kicker)?

Flop is JJ5, I have J7

Another player was dealt a J: 1 - (C(46, 18)/C(47, 18)) = 28%
The only jacks an opponent probably plays (if not in blinds) are probably the ones that beat me.

How do I continue from here?

[LetYouDown]

You're going to run into problems looking at it that way.

Assuming you're heads up, and you're only looking for hands that are ahead of you:

5-5, J-5, J-8, J-9, J-10, J-Q, J-K, J-A

If my head counting is anywhere near right, that's 30 combinations where you're behind. There's C(47,2) hands an individual opponent can hold. So the probability that a specific opponent is ahead of you on this flop should be:

30/C(47,2) = 30/1081 or about ~2.78%. Now, to get a close approximation, you could just factor in the number of players. To get an exact answer, you'll need BruceZ's inclusion/exclusion concept.

I vote that this post becomes a sticky:

http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&v iew=&sb=5&o=&fpart=

[Supern]

Thanks man. You are right.

I guess the trips kicker problem % is not that useful anyway.
Reads and how the betting goes are probably all I need to make a good decision.

[pzhon]

Here is another tool:

Google: 47 choose 20 / 49 choose 20 (http://www.google.com/search?hl=en&lr=&rls=GGLD%2CGGLD%3A2004-36%2CGGLD%3Aen&q=47+choose+20+%2F+49+choose+20)

[Siegmund]

If it's beyond the abilities of your computer to solve

(28 * 29) / ( 48 * 49 )

then it may be time to upgrade your hardware. From an analytical engine to a slide rule....

[Homer]

[ QUOTE ]
What is the mathematics behind C(47,20)?
What is the formula for combinatorics?

I searched for it but couldn't find anything.
Thanks for you help!

[/ QUOTE ]

As someone else alluded to, you don't actually need anything more than a 4-function calculator to solve this problem.

C(47,20) = 47!/(27!20!)
C(49,20) = 49!/(29!20!)

C(47,20)/C(29,20) = 47!/(27!20!) * (29!20!)/49!

This reduces to (47!29!)/(49!27!) since the 20! terms cancel.

47!/49! = 1/(49*48)
29!/27! = 29*28

So, it's (29*28)/(49*48).

-- Homer