starting hands how many please respond

[spaminator101]

ive heard that there are 1326 starting hands but wouldnt you multiply 52 and 51 and get
2652 or 2times as many hands

[AaronBrown]

Yes, but that counts A[img]/images/graemlins/spade.gif[/img] J[img]/images/graemlins/diamond.gif[/img] as a different hand from J[img]/images/graemlins/diamond.gif[/img] A[img]/images/graemlins/spade.gif[/img].

For example, suppose you want to figure out the probability of getting AJ offsuit. You could say there are four different Aces, and three offsuit Jacks per Ace, for 12 different hands. That's 12/2,652.

Or you could divide both numbers by two, to get 6/1,326. You get the same probability either way. You just have to be consistent.

[spaminator101]

thank you

[spaminator101]

if this is so then how many hands are there for omaha
should it be 52*51*50*49/4

[jba]

[ QUOTE ]
if this is so then how many hands are there for omaha
should it be 52*51*50*49/4

[/ QUOTE ]

the denominator you're looking for is not the number of cards, it's the number of permuations possible. This can be calculated by n!.

if you have AK in holdem it is AK or KA, two ways. 2!=2

if you have AKQJ in omaho it is AKQJ or AQKJ or ....., 24 ways. 4! = 24

so I think the answer you're looking for is 52*51*50*49/24.

[LetYouDown]

Yup, 270725. C(52,4)...one of the more straightforward "X CHOOSE Y" scenarios you'll run into.

[spaminator101]

thank you very much
you dont no how much this helps

[jba]

[ QUOTE ]
thank you very much
you dont no how much this helps

[/ QUOTE ]

no problem bro

can I ask how this helps? just curious

[spaminator101]

probability of an opponent having a certain hand calculate pot odds on the end of winning to make correct calls
thats how it helps

[jba]

hope you're good at dividing by 270725!

[img]/images/graemlins/smile.gif[/img]