[BruceZ]

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so would it still be more likely to hit a set if:
1.) instead of specifically just hitting a set, you include any time the rank of ur pair hits the flop icluding quads and full houses.

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1 - C(48,3) / C(50,3) = 1 in 8.5

That is 1 minus the probability that your pair does not hit.


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2.)you include all gutshot straight draws and flush draws flopped with the suited connector,not just open end straight draw and flush draw.

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6*(39-3-6) + 6*6*(10-1) - 18 - 36 = 450 flops with gutshot + flush draw

This is similar to the open end + flush draw case except that there are now 6 pairs of ranks that give a gutshot draw (97, 87, Q8, K9, AK, AQ). There are 6 of these that are suited, and these combine with the remaining 39 off suit cards minus 3 which complete the straight and minus 6 that make pair which were already counted. Then there are 6*6 of these with only one suited card, and these combine with 10-1 suited cards which don't complete the straight. Now there are 18 possible double gutshot flops with 2 suited cards (9 each of AQ8 and K97). The first term above counts 12 of these, and the second term counts 24, so we must subtract off 18 which are counted twice. Then there are 36 flops which have both an open ended straight draw AND a gutshot (9 each of K98, Q97, KQ8, and AQ9, each with two flush cards), and since these have already been counted with the open ended straights, they must be subtracted off. With these additional 450 flops, the probability comes to

(330 + 225 + 450) / C(50,3) = 1 in 19.5.

So the set is still more than twice as likely.