How do I calculate odds of straight draw with connectors on the flop?

[Stefan Prodan]

Title pretty much says it all. I haven't taken any class in combinatorics in awhile, and I'm having a hard time thinking about how to go about this.

So, of course, the total possibilities on the flop are (50c3), I realize that. The denominator is the easy part [img]/images/graemlins/smile.gif[/img].

I'm having a hard time figuring out the numerator here.

Take, for example, 89. If I have 89, I will need a flop containing either J7, 76, or JT.

Now, I can't decide if I should do this by subtracting the odds of NOT flopping a combination of two of the above cards from 1 or if I should do it by just adding up the possibitilies of flopping each one.

If possible, could you guys just give me some sort of "hints" instead of just explaining it to me outright at first?

I really wanna see if I can get this on my own, but it doesn't seem like I really can, and I do need a little help, but I'd like to see how much I can do.

If I get this, I'm trying next to calculate the odds of flopping at least a gutshot (including a flush draw).

[aloiz]

First I assume you just want odds for flopping an open-ended straight draw, and do not want to include the times you flop a double belly buster. Also I assume you're only dealing with connectors 45 - JT.
My route would be to figure out how many possible ways to flop the draw and divide that by total number of flops. You are on the right track in that you defined all the different ways of flopping draw. First hint I would give is that there the three ways you mentioned each occurs an equal number of times, so if calculate the combinations with just one of the possiblities then multiply by three.

aloiz

[Stefan Prodan]

Yeah, I already figured that out. I'm just having trouble finding the exact numbers I need to put together here.

[aloiz]

Oh, well how many ways can you flop say 76x. You choose one seven, one six, and one card that doesn't complete the straight. Don't forget you then need to adjust for the times you double count a board of say 7h 6s 7d. I'm not sure what else to say without actually telling you the answer.

aloiz

[Stefan Prodan]

I think I got it at:

1 - ((42c3)+2*(4*(39c2)+6*(38c1)+4c3)/19600) = about 8.81%

I actually had some help from a math major friend of mine, I don't think I would have thought about doing it this way if it weren't for him. Oh well.

Is that right?

How would you do it the way you're describing? I can't figure that out for the life of me.

I was thinking:

4 (sevens) * 4 (sixes) * 38 (cards which are not sevens, sixes, or tens)

That gives me a value which is different from the other way, and is about 9.3%. Is that correct, or is the other way correct?