flush draws

[gulebjorn]

Ok, i need to know the odds for being up against a flush draw on a 2-handed or 3-handed flop.
I cannot begin to count how many times i've been beaten by a flush today. It seemed like the boards on Pacific had a flush draw on it every single hand. I've seen more flushes than pairs today.

So what i am asking is, if the flop has two cards of the same suit, what are the odds that one of my opponents has a flush draw? And I'd like to know this for 1,2 and 3 opponents.

A steaming thank you

[gulebjorn]

bump

i'd still like to know

also, i wanna add this: if i am holding 32s and i make my flush, what are the odds that someone has a higher flush than me, and i'd also like to know this for 1,2,3 or more opponents

Thank You

[Cobra]

This type of question is difficult to answer. There is no way of measuring the probability that someone at your table being dealt two of a suit would stay to see the flop. What we can figure out is the probability that one of your opponents at the full table was dealt two of a suit after the flop. For the first answer I am assuming you have 0 of the suit in question. Using inclusion/exclusion the probability that one or more of your 9 opponents was dealt two of a suit given the flop is two of that same suit is as follows:

=9*(11c2)/(47c2)-(9c2)*(11c4)/(47c4)+(9c3)*(11c6)/(47c6)=

9 opponents = 39.49%
6 opponents = 27.84%
3 opponents = 9.77%

I can not answer whether they would have bet with those cards and stayed around to see the flop.

Your second question, you realize anyone with two of your suit would beat you if you have 23 suited. The probabiity that one or more of your opponents is dealt two of your suit given the flop has two is:

=9*(9c2)/(47c2)-(9c2)*(9c4)/(47c4)+(9c3)*(9c6)/(47c6)=

9 opponents = 27.49%
6 opponents = 18.94%
3 opponents = 9.77%

Cobra

[gulebjorn]

Something seems odd about those numbers:

[ QUOTE ]
=9*(11c2)/(47c2)-(9c2)*(11c4)/(47c4)+(9c3)*(11c6)/(47c6)=

9 opponents = 39.49%
6 opponents = 27.84%
3 opponents = 9.77%

I can not answer whether they would have bet with those cards and stayed around to see the flop.

Your second question, you realize anyone with two of your suit would beat you if you have 23 suited. The probabiity that one or more of your opponents is dealt two of your suit given the flop has two is:

=9*(9c2)/(47c2)-(9c2)*(9c4)/(47c4)+(9c3)*(9c6)/(47c6)=

9 opponents = 27.49%
6 opponents = 18.94%
3 opponents = 9.77%

[/ QUOTE ]

You get the same odds for both cases, when i'm up against 3 opponents. However, in the second case, there are 2 cards of that suit less in the deck. Shouldn't that make a difference?

Also, let's say for my limit, people play 30% of their starting hands. Let's say, for simplicity, they see 50% of all flops with their suited hands. Should i then divide all those percentages by two?

Thank you.

[Cobra]

Your right typo error. The first numbers should read.

9 opponents - 39.49%
6 opponents - 27.84%
3 opponents - 14.71%

You could still figure the above probabilities out if you put your opponents on a range of hand. I am not sure about just multiplying by 50%. To get a realistic number I think you would have to figure by hand all the different combo's and how each of them affects other hands. I will give this some thought, maybe someone smarter than me can answer this.

Cobra

[Cobra]

I have thought about this a little and came up with one way that I think might work. Let say you were to assume that your opponents would only call with the top seven clubs. That would be A-8 and all the combo's. Also assuming the flop came in with two clubs that were not A-8. The probability that one or more opponents out of nine were dealt A-8 combo is

= 16.78%.

Now lets assume you wanted to use the top six A-9 or that one of the flops was an A-8 card then the probability becomes

= 12.19%.

As you are aware if the ace comes out on the flop this eliminates alot of the hands your opponents might play where as if a 3 comes out in does not.

Cobra