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Probability of Pairing

  
 
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G3O
Old 08-03-2005, 07:47 PM     Post subject: Probability of Pairing #1 (permalink)  

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My apologies if this has already been answered....

Question 1: How do I calculate the probability of pairing one of two random Hole Cards by the river. ie. If i have A7o, how do I calculate the probability % that I make a pair by the River?

Question 2: How do I calculate the probability that by the River, someone will make a pair higher than my pocket pair? ie. I hold 77, what are the chances that someone makes 88 or higher by the River?

Thanks in advance. I have been scratching my head over this for a while now.
"Due to circumstances beyond my control, I am the master of my fate, the captain of my soul." - Some dead guy with a great sense of irony.
 
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drmcboy
Old 08-03-2005, 07:51 PM #2 (permalink)  
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http://www.flopturnriver.com/essays_pairs.html

start here.
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G3O
Old 08-03-2005, 07:58 PM #3 (permalink)  

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G3O
Very Nice. Thanks.
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G3O
Old 08-04-2005, 02:14 PM #4 (permalink)  

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G3O
drmcboy (or anyone else who cares to enlighten me),

I have a couple more questions:

1) I assume that 'rnd' columns mean random? As in 'rnd o val' = percentage probability that a random card will be an overcard.

2) How can I use information on this chart to determine the % of at least 1 of 2 random cards pairing by the river?

Thanks for the help.
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biondino
Old 08-04-2005, 02:57 PM #5 (permalink)  
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It's 44%.
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G3O
Old 08-04-2005, 03:57 PM #6 (permalink)  

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G3O
Quote:
Originally Posted by biondino
It's 44%.
How did you calculate that? You didn't think I would let you off that easy did you?
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CrunchyNuts
Old 08-04-2005, 04:09 PM #7 (permalink)  
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Inversion...it's much easier to figure out the chance that neither will pair then taking the inverse of that.

There's 6 cards that pair any random 2 non-paired cards, and 50 unknown cards. So the chance of *not* pairing them is the odds of each card not being one of the 6 multiplied all together:
Code:
44/50 * 43/49 * 42/48 * 41/47 * 40/46
130320960/221088000
0.5894528875379939209726443768997
So the chance of pairing is 1-that, so roughly 1-.59 = 0.41 = 41%
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G3O
Old 08-04-2005, 06:39 PM #8 (permalink)  

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G3O
Thank you. You guys are great.


Crunchynuts,

I like your math. Now, if I am to express this in odds-

100 / 41 (chance from your Inversion example) = 2.439 or 1 : 1.439

I take this to mean that I will hit at least one pair once for every 2.439 times I try. Am I right or am I way off?
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EricE
Old 08-04-2005, 06:44 PM #9 (permalink)  
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Quote:
Originally Posted by G3O
Thank you. You guys are great.


Crunchynuts,

I like your math. Now, if I am to express this in odds-

100 / 41 (chance from your Inversion example) = 2.439 or 1 : 1.439

I take this to mean that I will hit at least one pair once for every 2.439 times I try. Am I right or am I way off?
...IF you pay all the way to the river. hehe, Thats not something I am going to do.
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CrunchyNuts
Old 08-04-2005, 06:48 PM #10 (permalink)  
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Yup.
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G3O
Old 08-04-2005, 06:54 PM #11 (permalink)  

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G3O
Thanks again CrunchyNuts.


Quote:
Originally Posted by EricE

...IF you pay all the way to the river. hehe, Thats not something I am going to do.
Come on EricE, don't you like to gamble? lol.
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EricE
Old 08-04-2005, 08:26 PM #12 (permalink)  
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hehe, playing when you have odds is not so much gambling as paying your dues... you know it will pay off in the long run. Im not much of a gambler.
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