Verify my straight draw calc?

[gergery]

If I’m dealt 89, what are the chances I will specifically get an open ended straight draw on the flop?

I calculate it as follows: I can get an OESD 3 ways, with 67, 7T, or TJ and all are equally likely. So the chances of 67 is

C(4,1) * C(4,1) * C(40,1) (the 40 is 52 minus 4 5’s, 4 T’s, my 89 and first two 6,7)
---------------------------
C(50,3)

= 3.26%
So 3* 3.26% = 9.79% in total
Did I do this right?

[Indiana]

close enough but not exact. the problem is in saying that there are 40 left to choose from after the 6,7 are removed. Cannot simply throw them back into the remaining cards like that....I remember this to be around 9% though...

[Indiana]

actually, why not just say an OESD without the 6 or 7 being paired? then it is easy, 4*4*C(34,1)/C(50,3) and then u mult by 3 to get 8.3%. This way you don't allow the 6,7 to be chosen and then try to lump them back into the remaining cards. Doesn't it make sense anyway that the board should not be paired for an OESD to be really good to draw on?

Indy

[Lost Wages]

4*4*40/C(50,3) isn't quite right because it double counts flops where you have two 6's or 7's. I.e. it counts 6[img]/images/graemlins/diamond.gif[/img]7[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/spade.gif[/img] as different from 6[img]/images/graemlins/spade.gif[/img]7[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/diamond.gif[/img].

Here is the correct way. First, take the 67X combinations where X isn't a 5,6,7,T or one of the two cards in your hand. That is simply 4*4*34 = 544 combinations. Then add the 667 combinations, C(4,2)*4 = 24 and 677 combos, also 24. Total combinations = 592. Probability = 3*592/19600 = 9.06%.

Don't forget about the 8 out double gutshots as well. For 89 you can flop 6TQ or 57J. That adds another 2*4*4*4 = 128 combos. Total probability of flopping an 8 out straight draw = (592+128)/19600 = 9.71%.

If your cards were suited you would have to subtract those times that you flopped a flush.

Lost Wages

[Indiana]

I know you are right about the double counting. But does it really treat the to 676 combos differently as you said in your first sentence? How so?

Indiana

[Lost Wages]

4*4*40 cycles through the 67X combinations where X is not a 5,T, the first 2 cards on the flop or the 2 cards in your hand. So going through the combinations we would get a pattern like:

6 [img]/images/graemlins/spade.gif[/img]7 [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/spade.gif[/img]
6 [img]/images/graemlins/spade.gif[/img]7 [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/heart.gif[/img]
6 [img]/images/graemlins/spade.gif[/img]7 [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/club.gif[/img]
6 [img]/images/graemlins/spade.gif[/img]6 [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/diamond.gif[/img]
6 [img]/images/graemlins/spade.gif[/img]7 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/spade.gif[/img]
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6 [img]/images/graemlins/spade.gif[/img]7 [img]/images/graemlins/spade.gif[/img]6 [img]/images/graemlins/heart.gif[/img]
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6 [img]/images/graemlins/heart.gif[/img]7 [img]/images/graemlins/heart.gif[/img]2 [img]/images/graemlins/spade.gif[/img]
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6 [img]/images/graemlins/heart.gif[/img]7 [img]/images/graemlins/heart.gif[/img]6 [img]/images/graemlins/spade.gif[/img]
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The 676 and 677 combinations will get counted twice. Hope that helps cuz I'm not quite sure how else to explain it.

Lost Wages