First Two Card Combinations (Hold'Em)

[monkey_love]

I know that this is a basic issue so feel free to direct me elsewhere if it is convenient.

My basic understanding is that the possible combinations for your first two cards is 13 x 13 (if you ignore suits) or 52 x 52 (if you include suits). What I want to know, however, is something in between.

How many pairs?
How many suited?
How many off-suit non-pairs?
Total

Does this make sense? I don't want to know how many "true" combinations there are (which I'm assuming is 52 x 52) but rather how many "practical" combinations there are.

Thanks for any assistance.

[Bozeman]

First, since AsAs (etc.) is not possible, there are no more than 52x51 possibilities, and since AsKs = KsAs, there are actually only 51*52/2=1326. Since AsKh will play like AdKc, there are 169 (2*13*12/2+13=13*13) possible types of hands.

Now for your question: there are 13 types of pairs, and each one can be made 6 different ways (sh,sd,sc,hd,hc,dc), which makes 6/1326=0.45% chance of getting any particular pair (220:1 against). Suited: 13*12/2=78 possibilities (4 ways s,h,d,c to make each), so probability of any one=4/1326=.3%. Also 78 unsuited types of hands (for any suited hand there is an equivalent unsuited, unpaired hand).

Does this make sense to you?
Craig

[Lost Wages]

Your first card can be one of 52, your second can be one of the remaining 51. So the total numbers of 2 card combinations is 52*51 = 2652. However, that counts A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/diamond.gif[/img] as being different from K[img]/images/graemlins/diamond.gif[/img]A[img]/images/graemlins/spade.gif[/img] so we must divide our answer by 2 to arrive at 1326 2 card combinations since we don't care what order they are in.

Of course, preflop, A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/diamond.gif[/img] has the same value as A[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/club.gif[/img] so that leads to your question; how many "true" combinations.

<font color="red">Pairs </font>
There are 13 ranks so obviously there are 13 pairs.

<font color="red">Suited </font>
Your first card can be 1 of the 13 ranks. Your second 1 of the 12 remaining ranks in that suit. Again we divide our answer by two to eliminate duplicates.
(13*12)/2 = 78 suited hands.

<font color="red">Off-suit non-pairs </font>
Your first card can be 1 of the 13 ranks. Your second 1 of the 12 remaining ranks that doesn't make a pair.
(13*12)/2 = 78 off-suit non-pairs.

Total = 13+78+78 = 169 unique starting hands.

Note that that does not mean that your chances of being dealt a suited hand are the same as an unsuited one. There are four suits and 78 suited hands so there are 4*78 = 312 suited combinations.

For each offsuit hand there are 12 ways it can be made. Take AKo for example, the A can be any one of 4 suits and the K can be any of the 3 remaining suits. So the are 12*78 = 936 offsuit combinations.

There are 6 ways to make each pair. The first card is one of 4, the second is one of 3 so (4*3)/2 = 6. So there are 13*6= 78 pair combinations.

We can check ourselves 312+936+78 = 1326 which is the same answer we got when calculating the total number of 2 card combinations.

Lost Wages

[monkey_love]

Thanks for breaking this down for me.

Mark