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stokken
04-22-2005, 08:01 PM
Holding ak suited. What are the odds of flopping a pair, two pairs, or a flush draw? A pair with a flush draw, and a pair with a backdoor flushdraw? And what are the odds of completing a pair or better by the river? Much appreciated if any math wizz could help out. And explain simply if possible

Stokken

closer2313
04-22-2005, 09:49 PM
Pair
1 - ((44 choose 3) / (50 choose 3)) * 100 = 32.4285714%

Two Pair
(((3 choose 1) * (3 choose 1) * (44 choose 1)) / (50 choose 3)) * 100 = 2.02040816%

Flush Draw
(((11 choose 2) * (39 choose 1)) / (50 choose 3)) * 100 = 10.9438776%

A Pair with a Flush Draw
(((11 choose 2) * (6 choose 1)) / (50 choose 3)) * 100 = 1.68367347%

A Pair and a Backdoor Flush Draw
(((11 choose 1) * (6 choose 1) * (33 choose 1)) / (50 choose 3)) * 100 = 11.1122449%

Completing a pair or better by river
(1 - ((44 choose 5) / (50 choose 5))) * 100 = 48.7432272%

I believe those are correct, but if someone would double check my work, it would be great!

Cobra
04-22-2005, 10:00 PM
You have 50 unseen cards 6 are AK and 11 are of your suit. We will figure out some probabilities and then answer your questions.

Pair of AK( some of these are two pair with a pair on the board. = 6*(44c2) = 5676 flops

Two pair or three of a kind = (6c2)*44 = 660 flops

Flush draw = (11c2)*39 = 2145 flops

Pair and flush draw = (11c2)*6 = 330 flops

Gutshot flush and a pair = 6*11*33 = 2178 flops

Now we can answer some questions:

Two pair or three of a kind = 660/19600 = 3.37%

Pair and flush draw = 330/19600 = 1.68%

Pair and gutshot flush = 2178/19600 = 11.11%

Pair only = 5676-330-2178 = 3168/19600 = 16.16%

Flush draw only = 2145-330 = 1815/19600 = 9.26%

The probability of getting to the river with a pair or better is.

=1-(44c5)/(50c5)= 48.74%

Cobra

stokken
04-23-2005, 02:30 AM
Great! And nicely explained
Thank you /images/graemlins/smile.gif

Stokken

Megenoita
04-23-2005, 04:52 PM
What is all this "44 choose 5" stuff? What does that mean on my calculator? Do I need a special calculator to perform this? Am I the most undereducated person ever?

M

carsten
04-24-2005, 01:17 AM
See http://mathworld.wolfram.com/BinomialCoefficient.html, but don't try to read past equation number (1) if you have a weak stomach /images/graemlins/wink.gif

What it means on your calculator depends on what kind of calculator you have. Some calculators have an nCr function, but if yours doesn't, you can calculate binomial coefficients by performing the necessary multiplications and division. For example, 44 choose 5 is (44*43*42*41*40)/(5*4*3*2*1).

Hope this helps,

Carsten.

Student
04-24-2005, 12:28 PM
Appreciate these formulas! As a beginner in poker, and with an awakened interest in the probabilities of the game, I'm but vaguely aware of the kinds of formulas that work with these sorts of questions.

Do you mean by "44 choose 3" 44 things taken 3 at a time? And if so, is this 44 factorial divided by (41 factorial times 3 factorial)? It's been more 40 years since I worked with this sort of things, but I have kind of a probabilistic mind, and I'd like to restore my earlier capability as quickly as possible. I have a feeling you are one who understands these things rather thoroughly!

Thanks, very much!

Dave

Cobra
04-24-2005, 01:52 PM
That is correct. You are choosing three cards out of 44 when order does not matter. Your factorial equation is correct this would simplify to 44*43*42/1*2*3. Most stastistical calculators have an nCr function that works. You can also use the following equation in excel.

Combin(44,3)

Cobra

binions
04-24-2005, 02:49 PM
Buy Holdem's Odd Book by Mike Petriv. Has all the math you need to know, and explains the formulas.

[ QUOTE ]
Holding ak suited. What are the odds of flopping a pair, two pairs, or a flush draw? A pair with a flush draw, and a pair with a backdoor flushdraw? And what are the odds of completing a pair or better by the river? Much appreciated if any math wizz could help out. And explain simply if possible

Stokken

[/ QUOTE ]

Student
04-24-2005, 02:50 PM
I think I get it, but perhaps you could clarify. 44 choose 5 is 44 factorial divided by ((44 - 5) factorial times 5 factorial), hence 44!/(39! x 5!). So 119 choose 72 would be 119 factorial divided by ((119 - 72) times 72 factorial), which is 119!/(47! x 72!).

I intend to do actual Monte Carlo calculations, which I'd done 25 years ago concerning blackjack, but I'd like to know how to do the theoretic probability computations first, as a check of my results.

Thanks much!

Dave

carsten
04-24-2005, 08:43 PM
I'll assume you want me to clarify why (44 choose 5) = (44*43*42*41*40)/(5*4*3*2*1).

You are correct in stating that (44 choose 5) = 44!/(39!*5!). Now observe that you can write 44! as 44*43*42*41*40*(39!), since 39! is the product of all integers from 1 to 39, and the term 44*43*42*41*40 completes the expression to the product of all integers from 1 to 44, which is 44!.

With this in mind, you get (44 choose 5) = 44!/(39!*5!) = (44*43*42*41*40*39!)/(39!*5!), which yields my original expression by writing 5! as 5*4*3*2*1 and cancelling out the 39!. The end result is an expression that's much more practical to compute.

-Carsten

TonyBlair
04-25-2005, 05:28 AM
[ QUOTE ]
The probability of getting to the river with a pair or better is.

=1-(44c5)/(50c5)= 48.74%

[/ QUOTE ]

I'm pretty sure you'll end up with a pair or better more times than this when all-in with AKs.