Here I attempt to see the chances to see a double ended straight draw on the flop if holding small connectors

I will ignore made hands for the moment, And the double belly buster which I consider tackling a bit later

So, assuming one holds the 78. One gets a straight draw with the the 9,10, the 5,6, and the 6,9.
each with 4 suits so you have for every combination of thses two cards, a precentage of 4/50 *4/49. But with an extra possibility!

on the flop you get three cards, not just two

so you have to take into consideration these possibility above (if the flop goes 9,10,x 56x for example; in which case these above calulations apply) AND the situation of 9x10,; 5x6, which should be a different situation



then,

5,6,x = 4/50*4/49*48/48 =0.006531
5,x,6 = 4/50*49/49/4/48=0.006667
x,6,5 = 50/50*4/49*4*48=0.006803
6,5,x = 4/50*4/49*48/48=0.006531
6,x,5 = 4/50*49/49/4/48=0.006667
x,5,6= 50/50*4/49*4*48=0.006531

9,10,x = 4/50*4/49*48/48 =0.006531
9,x,10 = 4/50*49/49/4/48=0.006667
X,10,9 = 50/50*4/49*4*48=0.006803
10,9,x = 4/50*4/49*48/48=0.006531
10,x,9 = 4/50*49/49/4/48=0.006667
X,9,10= 50/50*4/49*4*48=0.006531

6,9,x = 4/50*4/49*48/48 =0.006531
6,x,9 = 4/50*49/49/4/48=0.006667
X,9,6 = 50/50*4/49*4*48=0.006803
9,6,x = 4/50*4/49*48/48=0.006531
9,x,6 = 4/50*49/49/4/48=0.006667
x,6,9= 50/50*4/49*4*48=0.006531


Odds: -sum- =0.11918 =11.918 %

For this I ignore the straight flush draw as being a different category, and included it in the straight draw

Are the thought process /math correct? What am I missing


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2 things:

#1
[9, 10, J] nor [6, 9, 10] are double ended straight draws, they are straights. So the last fraction shouldn't be 48 out of 48 but 40/48.
________________________________

#2. You missed [4,6,10]

That flop also gives you a double outer as well to make your straight. 5 or a 9 on the turn makes one.

Jay Shark gave this answer 8+ years ago and he’s pretty good.

"case 1: For 0-gap off-suit connectors, the probability that your best available draw is an open ended straight draw is:

3*(4*4*34 -2*8 + 2*4c2*4)/50c3 ~ 8.82%"
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ME: The first term inside the parenthesis is for all different ranks. The second term is subtracting out flush draws, The third term allows for dup ranks on the flop. The 3 multiplier is for how the draw is formed-

For 7- 8, need 5- 6, 6- 9 or 9- 10

You also missed [5,9, J]. That results in a double outer as well (4 or 10 on the turn for the straight).

"I will ignore made hands for the moment, And the double belly busters which I consider tackling a bit later"

Is the math correct otherwise?
Thanks for the 40/48..



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Sorry, I see that about made hands and double belly busters. Your math is still spotty, I'd have to see it all again with the correct numbers. The main problem is that [x] isn't the same value depending on the position it takes. For example:

5, 6, x -> x is 40/48 (4s & 9s)
5, x, 6 -> x is 37/49 (4s, 6s & 9s)
x, 5, 6 -> x is 34/50 (4s, 5s, 6s & 9s)

Mathematically speaking (not computationally), there has got to be a lot better way to calculate this than running through every single permutation of cards.

There is a better way explained in post 3.

Yes, learn how to use combinations. These are useful when the order of the cards (like on a flop) does not matter.

I suggest taking a look at the formula in post #3 and study it until you understand each term completely.


Edit: browni3141 said the exact same thing while I was typing.

.

Listen, I know combinations and yes one could apply them but it is really more intuitive to think about the way this works if you approach the matter like this.

You can get a feel about how the cards work

With combination and permutations at least for me it feels like you are using a tool to get to the result but you don't truly understand why the result is so

Which I guess is half the fun



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But yes, I see your point.I guess faster is most often better in any case

Can you share a link where this formula and the approach is discussed in depth?

Many thanks

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Suited Connectors, Implied Odds, and You (Theory/Math)