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Old 09-12-2005, 06:08 PM
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Default very simple math questions

hoping someone can help me...

1) wanted to know the odds of each pocket pair, each suited combo and each unsuited. i don't know how to do the math easily. i could do it brute force but i assume someone will know.. basically want to know odds of getting AA, AKs, AK on flop.

and also, what are the odds that someone else at the table has a pocket pair pre-flop??? i guess the odds are 1/16 taken to 10 players, but i know that's not 10/16.

thanks in advance.
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  #2  
Old 09-12-2005, 06:49 PM
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Default Re: very simple math questions

Any pocket pair 1 in 17
Individual pocket pairs 1 in 221
Any particular pair of values, suited 2 in 663 (1 in 331)
Any particular non-pair values (not necessarily suited) 8 in 663 (1 in 83)
Chance that your opponent has a pair when you also do, heads up 1 in 72. Chance the opponent has a pocket pair when you do not, 1 in 73.
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Old 09-12-2005, 06:56 PM
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Default Re: very simple math questions

thanks rufus, i'll have a look at those #'s...

you did the odds for one opponent having pair, but was wondering odds of nine opponents having at least one pair. i can probably do it as a giant string in excel, but was wondering if you knew the #.

thanks again.
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Old 09-13-2005, 05:17 AM
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Default Re: very simple math questions

[ QUOTE ]
Any pocket pair 1 in 17
Individual pocket pairs 1 in 221
Any particular pair of values, suited 2 in 663 (1 in 331)
Any particular non-pair values (not necessarily suited) 8 in 663 (1 in 83)
Chance that your opponent has a pair when you also do, heads up 1 in 72. Chance the opponent has a pocket pair when you do not, 1 in 73.

[/ QUOTE ]

Rufus -- How did you arrive at odds opponent has a pair when you do / do not have one? I believe the chance your opponent has a pocket pair is roughly 1/17 regardless of your hole cards (although it is actually very slightly more likely if you hold a pair). Whereas the odds two players playing heads up both hold pocket pairs in a given hand is roughly 1/17 * 1/17, or 1/289.
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Old 09-13-2005, 06:04 AM
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Default Re: very simple math questions

[ QUOTE ]
[ QUOTE ]
Any pocket pair 1 in 17
Individual pocket pairs 1 in 221
Any particular pair of values, suited 2 in 663 (1 in 331)
Any particular non-pair values (not necessarily suited) 8 in 663 (1 in 83)
Chance that your opponent has a pair when you also do, heads up 1 in 72. Chance the opponent has a pocket pair when you do not, 1 in 73.

[/ QUOTE ]

Rufus -- How did you arrive at odds opponent has a pair when you do / do not have one? I believe the chance your opponent has a pocket pair is roughly 1/17 regardless of your hole cards (although it is actually very slightly more likely if you hold a pair). Whereas the odds two players playing heads up both hold pocket pairs in a given hand is roughly 1/17 * 1/17, or 1/289.

[/ QUOTE ]
Ah... to answer my own question, I think you meant there were 73 possible pair combinations left if you had a pair and 72 possible pair combinations left if you did not... out of 1225 total hole card combinations, this comes to approximately 1/17... [img]/images/graemlins/smile.gif[/img]
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Old 09-13-2005, 08:47 AM
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Default Re: very simple math questions

[ QUOTE ]

Ah... to answer my own question, I think you meant there were 73 possible pair combinations left if you had a pair and 72 possible pair combinations left if you did not... out of 1225 total hole card combinations, this comes to approximately 1/17... [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Argh! Brain fart.
You are, of course, correct.
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  #7  
Old 09-13-2005, 01:19 PM
AaronBrown AaronBrown is offline
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Default Re: very simple math questions

I would state this a little differently.

There are 52*51/2 = 1,326 possible two-card combinations. A specific pair can be formed 4*3/2 = 6 ways. A specific suited combination can be formed 4 ways. A specific unsuited combination can be formed 4*3 = 12 ways. Just to check, there are 13 possible pairs and 13*12/2 = 78 possible non-pair combinations. So there are 13*6 + 78*4 + 78*12 = 1,326 possible hands.

There is one chance in 17 of getting a pair, so out of 10 players the expected number of pairs is 10/17. However, that is not the probability that at least one player has a pair, because there will be some deals with more than one pair among the 10 players. The exact computation is not simple, but a good approximation is to say the probability of no one getting a pair is exp(-10/17) = 56%. Under that assumption, the probability of getting exactly one pair is 33%, two pairs is 10% and three pairs is 2%.
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  #8  
Old 09-13-2005, 01:38 PM
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Default Re: very simple math questions

thanks aaronbrown. i love watching you on channel 4 in seattle.

is it worthwhile to learn all these odds?? or does it just confuse???

for me to sit there and say 33% chance someone has pocket pair, but then someone acts like he has pocket pair, i'm not really sure.

i do remember harrington getting into some basic math (nothing to do with my question). i.e. you have pocket 10's and one overcard hits. he says opponent could equally have J,Q,K,A or perhaps a couple of those, so your odds aren't that bad i.e. good chance your opponent only has one of the other 3 overcards. this used to be a big leak in my game. i would always assume the worst i.e. overcard hits and i'm dead, but i've learned that's not the case.
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