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View Full Version : starting hands how many please respond

spaminator101
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
08-01-2005, 04:06 PM
Yes, but that counts A/images/graemlins/spade.gif J/images/graemlins/diamond.gif as a different hand from J/images/graemlins/diamond.gif A/images/graemlins/spade.gif.

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
08-01-2005, 04:13 PM
thank you
spaminator101
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
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
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
08-02-2005, 02:12 PM
thank you very much
you dont no how much this helps
jba
08-02-2005, 02:47 PM
[ 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
08-02-2005, 03:03 PM
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
08-02-2005, 03:06 PM
hope you're good at dividing by 270725!

/images/graemlins/smile.gif
spaminator101
08-02-2005, 03:11 PM
i think ill do it on a calculater while playing online till i do it naturally
MegumiAmano
08-02-2005, 11:21 PM
[ QUOTE ]
i think ill do it on a calculater while playing online till i do it naturally

[/ QUOTE ]

Your calculation assumes that your opponent will play any two cards dealt to him with equal regularity. I guess that true with some people though.

Sidenote to this. There are 169 different hand "types". AA, AKo, AKs, etc. Not that I can calculate it, but that's what pokertracker shows.
Paxosmotic
08-02-2005, 11:27 PM
[ QUOTE ]
Yes, but that counts A/images/graemlins/spade.gif J/images/graemlins/diamond.gif as a different hand from J/images/graemlins/diamond.gif A/images/graemlins/spade.gif.

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
08-02-2005, 11:54 PM
[ QUOTE ]
[ QUOTE ]
Yes, but that counts A/images/graemlins/spade.gif J/images/graemlins/diamond.gif as a different hand from J/images/graemlins/diamond.gif A/images/graemlins/spade.gif.

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
08-03-2005, 12:25 AM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Yes, but that counts A/images/graemlins/spade.gif J/images/graemlins/diamond.gif as a different hand from J/images/graemlins/diamond.gif A/images/graemlins/spade.gif.

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.
Paxosmotic
08-03-2005, 12:27 AM
Oh I see why. Aaron, when you're doing the AJo calculation, you'd doing the math on how many aces times the offsuit jacks but you're not factoring in that you can receive them in either of two orders.

4 aces * 3 jacks * 2 ways to get it = 24 of 2,652 hands

Thus, 12 of the 1,326 starting hands are AJo. Knew we both missed somethin.