Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > General Gambling > Probability

Reply
 
Thread Tools Display Modes
  #1  
Old 09-11-2003, 02:08 AM
t_rex t_rex is offline
Junior Member
 
Join Date: Sep 2003
Posts: 6
Default Basic holdem combination question.

OK. For your hole cards in holdem, there are 169 distinct combinations. I don't understand how this number is mathematically calculated though.

I get that there are 13 pairs, and 78 suited cards (2 card combinations out of 13 cards - suit is irrelevant, so no need to multipy by 4). What I don't really understand is how the 78 unsuited combinations are calculated. Could someone explain this to me?

Thanks
Reply With Quote
  #2  
Old 09-11-2003, 02:28 AM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: Basic holdem combination question.

I get that there are 13 pairs, and 78 suited cards (2 card combinations out of 13 cards - suit is irrelevant, so no need to multipy by 4). What I don't really understand is how the 78 unsuited combinations are calculated.

Exactly the same way as the suited cards. For every suited hand there is a corresponding offsuit hand, and visa versa. Of course there are 3 times more offsuit hands (12 vs. 4) but we are counting all 12 of them as the same hand.
Reply With Quote
  #3  
Old 09-11-2003, 09:19 AM
t_rex t_rex is offline
Junior Member
 
Join Date: Sep 2003
Posts: 6
Default Re: Basic holdem combination question.

i have seen the formula "13*(4 2)" used to calculate the number of offsuit cards (78) [the numbers in paren indicate a combination of 2 cards out of 4]. this doesn't make sense to me. Can you explain?

Thanks
Reply With Quote
  #4  
Old 09-11-2003, 09:46 AM
StevieG StevieG is offline
Senior Member
 
Join Date: Jan 2003
Location: Baltimore, MD, USA
Posts: 157
Default Re: Basic holdem combination question.

(4 2) or C(4 2) is a specific case of the function C(n k) (often read as "n choose k"). If you have n distinct objects, and choose k from that set, and the order does not matter, the function tells you how many different combinations of k items you get:

C(n k) = n!/(k! * (n-k)!)

here ! indicates factorial
n! = n * (n-1) * (n -2) * .. * 1

in other words, multily all thenumbers from 1 to n together

so C(4 2) = 4!/(2! * (4-2)!) = 4*3*2*1/(2*2) or 6.
Reply With Quote
  #5  
Old 09-11-2003, 09:49 AM
StevieG StevieG is offline
Senior Member
 
Join Date: Jan 2003
Location: Baltimore, MD, USA
Posts: 157
Default Re: Basic holdem combination question.

So now we try to calculate the possible pairs. There are 13 different ranks that can pair. A pair is 2 suits out of four, so for each rank we have (4 choose 2) 6 pairs avaailable. Hence, 78.
Reply With Quote
  #6  
Old 09-11-2003, 09:58 AM
StevieG StevieG is offline
Senior Member
 
Join Date: Jan 2003
Location: Baltimore, MD, USA
Posts: 157
Default Re: Basic holdem combination question.

[ QUOTE ]
So now we try to calculate the possible pairs. There are 13 different ranks that can pair. A pair is 2 suits out of four, so for each rank we have (4 choose 2) 6 pairs avaailable. Hence, 78.

[/ QUOTE ]

D'oh! Lousy brain

You want distinct combinations, not total numbers of pairs.

The number turns out to be 78 but thta's a quirk of teh number. the calcualtion is different.

The number of non-pair combos is
13 (start with any rank)
* 12 (next rank must be different(
/2! (the number of ways to arrange two cards)

which also equals 13*6 or 78.

So total number of starting hands =
13 pairs
+ 78 combos (suited)
+ 78 combos (offsuit)
----
=169
Reply With Quote
  #7  
Old 09-11-2003, 10:46 AM
t_rex t_rex is offline
Junior Member
 
Join Date: Sep 2003
Posts: 6
Default Re: Basic holdem combination question.

i follow that, i think, but i've seen the calculation for unsuited combinations (not pairs) given as 13*(4 2) - choose 2 from 4. The result is 78. I see how you've arrived at the answer, but the logic seems slightly different than 13*(4 2). Could you explain?
Reply With Quote
  #8  
Old 09-13-2003, 09:30 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: Basic holdem combination question.

It is just (13,2) = 13*12/2 = 78. That is, there are 13 ways to pick the first rank, and 12 ways to pick the second rank, and then you divide by 2 because otherwise you would count every hand twice since you can get the cards in either order (like J,T or T,J are the same hand). You ignore suits entirely and just deal with 13 numbers. This is the same as 13*(4,2) = 13*4*3/2 = 13*12/2 = 78. I can't think of why anyone would come up with that right off.
Reply With Quote
  #9  
Old 09-14-2003, 12:56 AM
Foo King Foo King is offline
Junior Member
 
Join Date: Jul 2003
Posts: 6
Default Re: Basic holdem combination question.

Disregarding individual suits. 13 possible ranks for the first card X 13 possible ranks for the second card = 169 combinations

13 x 13 = 169 [img]/images/graemlins/grin.gif[/img]
Reply With Quote
  #10  
Old 09-14-2003, 01:11 AM
Foo King Foo King is offline
Junior Member
 
Join Date: Jul 2003
Posts: 6
Default If you want to take it a step further

To calculate the number of possible hands and taking the suit into account, the calculation would be 52 possible cards for the first card X 51 possible cards for the second card = 2652 individual two card hands are all that there is possible from a standard 52 card deck.

52 X 51 = 2652 [img]/images/graemlins/grin.gif[/img]

Hope these two posts help.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 01:45 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.