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#1
12-16-2005, 12:49 AM
 Guest Posts: n/a
MATH!!!!AAARRRRGGGGGHHHHH!!!!!

Can someone please help me with these probability solutions. I do not want the answer. I want the solution to finding the answer. I want to find the differences in probability of flopping an open ended straight draw, flush draw, open ended straight flush draw, and flopping a flush draw OR an open ended straight draw with suited connectors, connectors, 1 gappers both suited and connected. Drawing to the nuts(staight), etc.....

I would love to chat with someone on how to go through these things. I can get started, but I hit bumps, that I can't figure out.
#2
12-16-2005, 04:34 AM
 Guest Posts: n/a
Re: MATH!!!!AAARRRRGGGGGHHHHH!!!!!

You might get more of a response if you post this in the Probability Forum under the General Gambling heading.
#3
12-16-2005, 07:40 AM
 AaronBrown Senior Member Join Date: May 2005 Location: New York Posts: 505
Re: MATH!!!!AAARRRRGGGGGHHHHH!!!!!

The general idea is simple. You hold two cards in your hand, so there are 50 unseen cards. These can flop in:

50*49*48/(3*2*1) = 19,600

ways. To solve the kinds of problems you mention, you have to figure out how many flops will qualify.

For example, if you hold two suited cards, there are 11 matching cards unseen. There are:

11*10*9/(3*2*1) = 165

ways to flop a flush. So the probability is 165/19,600 = 0.84%.

There are:

(11 * 10)/2 * 39 = 2,145

ways to flop a flush draw. The 11*10/2 is the number of ways to get two flush cards, then you could pair them with any of the 39 cards of other suits.

Counting for the straights is a bit tougher because it depends on the gap between your cards as well as the level of the cards. It's not hard to figure how many ways to flop a straight, you just have to count the number of possible straights (if you hold AK or J7, for example, there is only one straight to flop, while if you hold JT there are four). Each straight can come in 4*4*4 = 64 different ways. Open-ended and gut-shot straight draws get a bit more complicated, but it just takes careful accounting.
#4
12-16-2005, 07:54 AM
 Guest Posts: n/a
Re: MATH!!!!AAARRRRGGGGGHHHHH!!!!!

Hold'em odds book by Mike Petriv shows you how to calculate the "probability of flopping an open ended straight draw, flush draw, open ended straight flush draw, and flopping a flush draw OR an open ended straight draw with suited connectors, connectors, 1 gappers both suited and connected"...

It is difficult too find and relatively expensive (39.99 USD), considering the quality of the binding and the printing.

I searched far and low for a book that would answer all my questions on probabilities. It is the only one I found.

There is another book, Texas Hold'em odds by Catalin Barboianu. The reviews have been lukewarm, so I haven't purchased it yet. The author is a romanian mathematician, who is semi-literate in english.

Considering the avalanche of books on poker on the market, it is rather surprising to only be able to recommend a 1996 book, which is now out of print. People must simply be math-phobic.

The maths in Hold'em odds is basic and very simply explained, though not always rigorously enough to my taste. It is written by a poker player so all the subjects in your post are covered.

All basic poker probabilities boil down to simple combinatorics and basic logic.

Basically, you just need to know the combinatorics formula: C(n,k) = n!/(n-k!)k!. If you have a calculator, you don't even need to know that. If you don't have one yet, get a calculator where the combinatorics formula is available withou having to sift through the menu. I bought the Casio fx-991MS (20 CAD) just for that purpose.

Here is a simple way of expressing the flush draw probabilities which Aaron Brown has already done. I suspect he read Petriv's book, since he uses his method.

<u>Probability of having 2 suited cards</u>

C(13,2)/C(52,2) * 4 = 23.53%

<u>Probability of flopping a flush when holding 2 suited cards</u>

C(11,3)/C(50,3) = 0.84%

<u>Probability of flopping a flush draw when holding 2 suited cards (not a flush)</u>

C(11,2)*C((50-11),1)/C(50,3) = 10.9%

You can have a blind faith in BruceZ's posts in the Probabilities and Statistics forum. Brian Alspach is a mathematics professor who has published extensively on the subject. You can find his articles on the web. I wouldn't really recommend it, but Daniel Kinberg's has published a few articles on CardPlayer, which are geared towards the beginner but still enjoyable.

I'm still slogging through Petriv's book. Give me a month or so and I'll be able to answer your questions as competently as BruceZ.

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