starting hands how many please respond

spaminator101 · #1 08-01-2005, 03:25 PM

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

AaronBrown · #2 08-01-2005, 04:06 PM

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 · #3 08-01-2005, 04:13 PM

thank you

spaminator101 · #4 08-02-2005, 12:33 PM

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

jba · #5 08-02-2005, 02:02 PM

[ 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 · #6 08-02-2005, 02:09 PM

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

spaminator101 · #7 08-02-2005, 02:12 PM

thank you very much
you dont no how much this helps

Paxosmotic · #8 08-02-2005, 11:27 PM

[ QUOTE ]
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.

[/ QUOTE ]
I'm too tired to figure out where the error in this is, but I am 100% certain that there are 12 ways to get AJo in the 1,326 hands, not 6.

spaminator101 · #9 08-02-2005, 11:54 PM

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]
I'm too tired to figure out where the error in this is, but I am 100% certain that there are 12 ways to get AJo in the 1,326 hands, not 6.

[/ QUOTE ]
ya know that might be why he said that

Paxosmotic · #10 08-03-2005, 12:25 AM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]
I'm too tired to figure out where the error in this is, but I am 100% certain that there are 12 ways to get AJo in the 1,326 hands, not 6.

[/ QUOTE ]
ya know that might be why he said that

[/ QUOTE ]
Yeah but he's saying 12 of 2,652, it's actually 12 of 1,326. The problem is that the method he's using looks pretty good, I'm not really seeing where he missed it.

The important thing to know is that 2,652 is how many ways you can receive your starting cards. That number is meaningless except that it is exactly double of the next number.

1,326 starting hand combinations.