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vector
08-10-2005, 10:08 AM
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) --&gt; hit (6/50; 12%)
[
[ --&gt; 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.

BruceZ
08-10-2005, 10:55 AM
[ 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) --&gt; hit (6/50; 12%)
[
[ --&gt; 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 (http://forumserver.twoplustwo.com/favlinker.php?Cat=&amp;Entry=120568&amp;F_Board=genpok&amp;Thr ead=2948276&amp;partnumber=&amp;postmarker=) post and follow the links. They should have everything you need.

vector
08-10-2005, 11:16 AM
Thanks a lot!

08-10-2005, 01:26 PM
Just use hypergeometric formula.

VivaLaViking
08-10-2005, 02:41 PM
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).

BruceZ
08-10-2005, 02:54 PM
[ 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

VivaLaViking
08-10-2005, 06:01 PM
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.

vector
08-10-2005, 09:16 PM
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.

VivaLaViking
08-11-2005, 10:34 AM
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.

BruceZ
08-12-2005, 05:55 AM
[ 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.

VivaLaViking
08-12-2005, 08:47 AM
Bruce, Good Morning and Congrats.

[ QUOTE ]

, or the curl (vector) when we take the cross product of del with a vector.

[/ QUOTE ]

If you don't call the cross product a mathematical operation, all I can say is some people say tomato. And yes, the dot product always results in a scalar but the cross product always results in a vector. That post was limited to the topic of vectors.

[ QUOTE ]

which gives a sum of second partial derivatives in Cartesian coordinates.

[/ QUOTE ]

I take issue with this in that Vector math is not restricted to any specific coordinate system; Cartesian (x, y,z), Cylindrical (r, theta, z) or Spherical (R, theta, phi). I wish that it were (it would have saved months in my life) lol.

[ QUOTE ]

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

[/ QUOTE ]

These are all divergences in String Theory which shows that the subject matter is a "work in progress". I was just trying to point out as simply as possible how reasonable it is to think in multiple and different dimensions because Combinatorial math may not answer all our questions and we are subject to these limitations..

BruceZ
08-12-2005, 09:58 AM
[ QUOTE ]
Congrats

[/ QUOTE ]

For?

[ QUOTE ]
If you don't call the cross product a mathematical operation, all I can say is some people say tomato.

[/ QUOTE ]

Huh? Where did I say anything about it not being a mathematical operation? Of course it is.

[ QUOTE ]
And yes, the dot product always results in a scalar but the cross product always results in a vector.

[/ QUOTE ]

The Laplacian is not a cross product. The Laplacian is the dot product of del with the gradient. The Laplacian operates on functions not vectors, and it gives a scalar, not a vector.

[ QUOTE ]
That post was limited to the topic of vectors.

[/ QUOTE ]

Then why bring up the Laplacian?

[ QUOTE ]
[ QUOTE ]

which gives a sum of second partial derivatives in Cartesian coordinates.

[/ QUOTE ]

I take issue with this in that Vector math is not restricted to any specific coordinate system; Cartesian (x, y,z), Cylindrical (r, theta, z) or Spherical (R, theta, phi). I wish that it were (it would have saved months in my life) lol.

[/ QUOTE ]

I said that it only gives a sum of second partial derivatives in Cartesian coordinates. In cylindrical coordinates, there is a first partial with respect to r multiplied by 1/r, and the second derivative with respect to theta is multiplied by 1/r^2. In spherical coordinates, there are factors of R^2 and sin(theta) which are multiplied by the first derivative with respect to R and theta respectively, before taking another first derivative. Also, the second derivative with respect to phi^2 is multiplied by 1/sin^2(theta), the theta term is multiplied by 1/sin(theta), and all terms are multiplied by 1/R^2.

[ QUOTE ]
I was just trying to point out as simply as possible how reasonable it is to think in multiple and different dimensions because Combinatorial math may not answer all our questions and we are subject to these limitations..

[/ QUOTE ]

What kind of poker problem do you think can't be solved with combinatatorial math that can be solved with vector math?

wdbaker
08-13-2005, 07:06 PM
BruceZ,
You crack me up, I love reading your stuff even though I don't have a clue what your talking about most of the time, I keep hoping something will click someday... /images/graemlins/grin.gif

You really like mixing it up with these guys and I think that's what makes it fun to read

Now as I have not contributed anything of value to this discussion I shall dismiss my self /images/graemlins/smile.gif

One Street at a Time
wdbaker Denver, Co