"X combinations of being dealt...XX"

Hey everyone... quick question... often when I'm putting opponents on a hand and trying to calculate the number of hands I beat versus the number of hands that I beat, I often find myself tripping over the numbers...

For example I hold KK and I put the opponent on another High pocket pair... well there are 6 ways of being dealt AA and there are 6 ways of being dealt KK (of which my hand prevents 3 of them) and 6 ways of being dealt QQ, JJ, 10-10

So...

Part 1) I am ahead of 18 possibilities, tied with 3 possibilities and behind 3 possibilities.

Part 2 (Where I fall down) Of the hands that I am ahead, there are 6 possible cards that can come to beat me (a card to make trips) ... so that is 6:48 or I'm an 8-1 favorite?

Of the hands I am behind, there are two cards that can rescue me (another K) so that is 2:24 or I am a 21-1 underdog?

Part 3... When I am ahead, I will win, 78.5% of the time. When I am behind, I will win 5% of the time... so I have a 83.5% chance of winning?

Is that correct? Am I doing this right? If so, is there a way I can get there quicker? Some system that I can program into my brain? Or am I only going to get better at it by sitting down with a pencil and paper and running through scenarios?

This is a simple scenario, I can't imagine calculating flushes, straights, or other scenarios beyond pairs when I am sitting at the table...

I've searched for other posts, but apparently my search terms aren't hitting the mark.
Any help is appreciated!

Thanks!

[mosdef]

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and there are 6 ways of being dealt KK (of which my hand prevents 3 of them)

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your KK prevents 5 of them. there is only one KK hand your opponent can have if you have KK

/images/graemlins/blush.gif

Yeah... very true.

See? I need some skillz.

[BruceZ]

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For example I hold KK and I put the opponent on another High pocket pair... well there are 6 ways of being dealt AA and there are 6 ways of being dealt KK (of which my hand prevents 3 of them)

[/ QUOTE ]

Your hand prevents 5 of them, there is only 1 left.


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and 6 ways of being dealt QQ, JJ, 10-10

So...

Part 1) I am ahead of 18 possibilities, tied with 3 possibilities and behind 3 possibilities.

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You are tied with 1, and behind 6 (AA).


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Part 2 (Where I fall down) Of the hands that I am ahead, there are 6 possible cards that can come to beat me (a card to make trips) ... so that is 6:48 or I'm an 8-1 favorite?

[/ QUOTE ]

There are only 2 cards that can beat you if you only have 1 opponent. You are against one of JJ, QQ, or TT, not all 3. There are 48 cards remaining, so you are a 46:2 = 23:1.


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Of the hands I am behind, there are two cards that can rescue me (another K) so that is 2:24 or I am a 21-1 underdog?

[/ QUOTE ]

46:2 = 23:1 underdog.


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Part 3... When I am ahead, I will win, 78.5% of the time. When I am behind, I will win 5% of the time... so I have a 83.5% chance of winning?

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You have to multiply the 78.5% by the chance that you are ahead, and the 5% by the chance you are behind. Taking the numbers from above, where you are ahead 18 out of 22, behind 3 out of 22, and tied 1/22:

18/22 * 78.5% + 3/22 * 5% = 64.9%.

Oh man... I need a book.

Thanks though. This experiment was a rousing success. I know the probability of me getting my next calucation correct.

oh yeah... and I should proof the OP a little better next time. There are some lousy typoes in there.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Part 2 (Where I fall down) Of the hands that I am ahead, there are 6 possible cards that can come to beat me (a card to make trips) ... so that is 6:48 or I'm an 8-1 favorite?

[/ QUOTE ]

There are only 2 cards that can beat you if you only have 1 opponent. You are against one of JJ, QQ, or TT, not all 3. There are 48 cards remaining, so you are a 46:2 = 23:1.


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Of the hands I am behind, there are two cards that can rescue me (another K) so that is 2:24 or I am a 21-1 underdog?

[/ QUOTE ]

46:2 = 23:1 underdog.

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BTW, this is for just a single card to come, which doesn't make any sense since if you are preflop, you have to figure the chance of hitting on one of 3 cards, not just 1.

Thanks Bruce... You're really making me feel better about my journey. :-P

Seriously though, Thanks. Right now, the only info. I have on this topic is my old prob&stats class (circa 97) and the few pages in Harrington. I try to read Matt Mattros' stuff, but I get lost REAL easily there.

Any good source you can recommend? Think I need another prob&stats course?

[LetYouDown]

Intro to Probability (http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/amsbook.mac.pdf)

This will probably help, although it does assume a certain level of math competence and some prior knowledge.