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  #1  
Old 08-10-2005, 10:08 AM
vector vector is offline
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Default method for determining possible flop outcomes - reasonable?

Hi.

I'm trying to calculate the odds of flopping a range of hands, using a decision tree where a node is a flop card and the verticies are the chances of a given card(s) falling.

For example to work out the chances of flopping 2 pair or better with any 2 cards, I make node 1 the first flop card, and construct 2 vertices from here labelled with the chances of hitting one of my six outs on this card.

(1) --> hit (6/50; 12%)
[
[ --> miss (44/50; 88%)

From each of these outcomes construct another node for flop card 2 and so on, adjusting the outs based on remaining cards and whether I have already hit one of my cards or not.

Tracing the three possible "good" outcomes (hit, hit; hit, miss, hit; miss, hit, hit) the chance of each is determined by calculating the outcomes on each arm, and the overall odds of flopping 2 pair (or better) is obtained from adding the three percentage outcomes.

This gave me the result that there is a 3.46% chance of flopping 2 pair or better with any two unpaired cards.

My first question is, is this method correct?

I then did a similar thing for determining the odds of flopping at least a flush draw with any two suited, to come up with the result of 11.79%.

My next question is: is it correct to say that the chance of flopping at least two pair OR at least a flush draw with any two suited = 3.46% + 11.79% = 15.25%? Is it as simple as adding them like this, or is this missing something?

Finally I would like to do the same thing to determine the chances of flopping (at least) an OESD with two connected cards, with the ultimate goal being to determine the chance of flopping at least two pair OR a flush draw OR an OESD with suited connectors.

But the straight draw is more complicated to do this way due to the number of straights initially possible, and the constraints as each of the cards fall. It seems definitely possible to do, just more complicated. I was wondering if someone could suggest a simpler method?

Thanks a lot.
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  #2  
Old 08-10-2005, 10:55 AM
BruceZ BruceZ is offline
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Default Re: method for determining possible flop outcomes - reasonable?

[ QUOTE ]
Hi.

I'm trying to calculate the odds of flopping a range of hands, using a decision tree where a node is a flop card and the verticies are the chances of a given card(s) falling.

For example to work out the chances of flopping 2 pair or better with any 2 cards, I make node 1 the first flop card, and construct 2 vertices from here labelled with the chances of hitting one of my six outs on this card.

(1) --> hit (6/50; 12%)
[
[ --> miss (44/50; 88%)

From each of these outcomes construct another node for flop card 2 and so on, adjusting the outs based on remaining cards and whether I have already hit one of my cards or not.

Tracing the three possible "good" outcomes (hit, hit; hit, miss, hit; miss, hit, hit) the chance of each is determined by calculating the outcomes on each arm, and the overall odds of flopping 2 pair (or better) is obtained from adding the three percentage outcomes.

This gave me the result that there is a 3.46% chance of flopping 2 pair or better with any two unpaired cards.

My first question is, is this method correct?

[/ QUOTE ]

Yes, but there are easier ways, such as

[C(6,2)*44 + C(6,3)] / C(50,3) =~ 3.47%.


[ QUOTE ]
I then did a similar thing for determining the odds of flopping at least a flush draw with any two suited, to come up with the result of 11.79%.

[/ QUOTE ]

11.76% if you exclude the straight flush.


[ QUOTE ]
My next question is: is it correct to say that the chance of flopping at least two pair OR at least a flush draw with any two suited = 3.46% + 11.79% = 15.25%? Is it as simple as adding them like this, or is this missing something?

[/ QUOTE ]

You can add these probabilities because the flush draw and the 2-pair/full house/quads are mutually exclusive (can't have them both).



[ QUOTE ]
Finally I would like to do the same thing to determine the chances of flopping (at least) an OESD with two connected cards, with the ultimate goal being to determine the chance of flopping at least two pair OR a flush draw OR an OESD with suited connectors.

But the straight draw is more complicated to do this way due to the number of straights initially possible, and the constraints as each of the cards fall. It seems definitely possible to do, just more complicated. I was wondering if someone could suggest a simpler method?

[/ QUOTE ]

Start with this post and follow the links. They should have everything you need.
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  #3  
Old 08-10-2005, 11:16 AM
vector vector is offline
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Default Re: method for determining possible flop outcomes - reasonable?

Thanks a lot!
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  #4  
Old 08-10-2005, 01:26 PM
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Default Re: method for determining possible flop outcomes - reasonable?

Just use hypergeometric formula.
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  #5  
Old 08-10-2005, 02:41 PM
VivaLaViking VivaLaViking is offline
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Default Re: method for determining possible flop outcomes - reasonable?

If you are asking if you could access probabilites in a tree like manner, I would assume that you could but the number of nodes may be surprisingly cumbersome.

Consider some of the possibilities:
<font class="small">Code:</font><hr /><pre>
Non-pair pocket. Flopping only one pocket card without pairing the board.
C(3,1) * C(44,1) * C(40,1) 5280
------------------------- = ------ x 100 ~ 26.94%
C(50,3) 19600

Non-pair pocket. Flopping both pocket cards for 2 pair without trips
C(3,1) * C(3,1) * C(44,1) 396
------------------------- = ----- x 100 ~ 2.02%
C(50,3) 19600


Non-pair pocket. Flopping trips without a full house.
C(3,2) * C(40,1) 120
---------------- = ----- x 100 ~.61%
C(50,3) 19600

Non-pair pocket. Flopping exclusive full house (You hold both the values flopped).
C(3,2) * C(3,1) 9
---------------- = ----- x 100 ~.06%
C(50,3) 19600

Suited Pocket Flopping a flush
(11 C 3) 165
------- = ----- x 100 ~ .842%
C(50,3) 19600

</pre><hr />

The math seams reasonable to me but I am just learning combinatorial math and would ask the pros approval and/or for them to point out the discrepancy(ies).
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  #6  
Old 08-10-2005, 02:54 PM
BruceZ BruceZ is offline
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Default Re: method for determining possible flop outcomes - reasonable?

[ QUOTE ]
Non-pair pocket. Flopping only one pocket card without pairing the board.
C(3,1) * C(44,1) * C(40,1) 5280

[/ QUOTE ]

Correct, but actually C(6,1)*44*40/2 to pair either card.


[ QUOTE ]
Non-pair pocket. Flopping both pocket cards for 2 pair without trips
C(3,1) * C(3,1) * C(44,1) 396

[/ QUOTE ]

OK


[ QUOTE ]
Non-pair pocket. Flopping trips without a full house.
C(3,2) * C(40,1) 120

[/ QUOTE ]

2*C(3,2)*44


[ QUOTE ]
Non-pair pocket. Flopping exclusive full house (You hold both the values flopped).

C(3,2) * C(3,1) 9

[/ QUOTE ]

2*C(3,2)*C(3,1) for either full house


[ QUOTE ]
Suited Pocket Flopping a flush
(11 C 3) 165

[/ QUOTE ]

OK
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  #7  
Old 08-10-2005, 06:01 PM
VivaLaViking VivaLaViking is offline
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Default Re: method for determining possible flop outcomes - reasonable?

Thank you for the correction. If it provides any satisfaction I agree with you and see my mistakes. It seems Vector is doing work similar to myself in that we are trying to reduce the complexity of a large data set. I am interested if Vector is considering the applicability of Vector math eg the LaPlacian. Thank you again.
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  #8  
Old 08-10-2005, 09:16 PM
vector vector is offline
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Default Re: method for determining possible flop outcomes - reasonable?

No, I don't even know what the LaPlacian is. The name is just a coincidence.

I've started using probability trees lately to calculate the expectation of various all-in moves. This is new to me, but it occured to me I could use something similar for determining the chances of flopping various hands. But thanks to BruceZ I see now there are better ways to do it.

I'm not writing any code or anything, this is all just scribbles on paper.

Thanks for your response.
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  #9  
Old 08-11-2005, 10:34 AM
VivaLaViking VivaLaViking is offline
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Default Re: method for determining possible flop outcomes - reasonable?

Vectors, Dimension and Poker

A vector only lives in a world of dimensions within a sperical, cylindrical or rectangular coordinate system.
The LaPlacian is a mathematical operation performed on a vector. It's symbol is an upside down triangle and pronounced "del".

21st Century math, called String Theory, is said to have upward of twelve dimensions.

At first appearance it seems like "Space Shot" material and that is because most people through familiarity try to map these multiple dimension to length, width and height. That would be "Space Shot" stuff. Over-simplifying, if these added dimensions were temperature, mass, velocity . . . it conceivably would start to make sense.

Now, back to vectors and it's dimensions. I could conceive of a vector having dimensions HoleCardOne, HoleCardTwo, Flop . . . . Where this math could lead I don't know but I would never dismiss it. If everyone is using the same math we will all undoubtably end reaching the same conclusion. Is it the best way? I don't know but I would remain open minded to different thought processes.
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  #10  
Old 08-12-2005, 05:55 AM
BruceZ BruceZ is offline
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Default Del and the Laplacian

[ QUOTE ]

The LaPlacian is a mathematical operation performed on a vector. It's symbol is an upside down triangle and pronounced "del".

[/ QUOTE ]

That's incorrect. The Laplacian (scalar) is written as del^2, and it is not an operation performed on a vector, but on a function. It is the dot product of del and the gradient, which gives a sum of second partial derivatives in Cartesian coordinates. Del is the gradient (vector) when it acts on a function, or the divergence (scalar) when we take the dot product of del with a vector, or the curl (vector) when we take the cross product of del with a vector.


[ QUOTE ]
21st Century math, called String Theory, is said to have upward of twelve dimensions.

[/ QUOTE ]

String theory had 10 (9 spatial and 1 time), membrane theory has 10 or 11. Bosonic string theory has 26.
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