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Old 08-10-2005, 10:08 AM
vector vector is offline
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Join Date: Nov 2003
Location: UK
Posts: 97
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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