Hi, I'm examining some probabilities using the deck missing 2s, 3s, 4s, and 5s so a 36 card deck

JT and T9 have same 18.9% probability of making an open ended straight draw. However JT is better since all of its straights are the nuts.

I'm trying to calculate a "Nut OESD %" for each of the hands to clarify this. To do this I think I simply need a "weighting" of how many combos three specific cards on flop (for a Double Belly Buster" vs a more obvious open ender e.g. for JT a 98X flop where X is not a Q or 7, X = not straight making card.

For example for JT, it's % is 100% when it hits a straight making card

JT OESD

DBB K97 nuts
DBB Aq8 nuts
KQx nuts
98X nuts
Q9X nuts

Now for T9

T9 OESD

DBB q86 nuts
DBB kj7 Q nut nuts, 8 is nuts
QJX K not nuts, 8 is nuts
87X nuts
J8X nuts

Some of the draws will not hit the nuts, thus T9 is not as strong. But I would like to quantify this specifically with a "Nut %" for comparison of hands. As I mentioned if I could get some type of weighting for DBB which is likely vs something like QJX I think it will be relatively easy....thanks

No need, just figured it out on my own using algebra..