Probability of Pairing

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.

http://www.flopturnriver.com/essays_pairs.html

start here.

Very Nice. Thanks.

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.

It's 44%.

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?

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%

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?

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.

Yup.

Thanks again CrunchyNuts.

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.

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.