Runner runner straight maths

Work one case, say KQ. By symmetry, the probability of the other case (98) will be the same. For KQ, you have 8 possible turn cards that keep you alive, so 8/47 is the probability that you still could hit the river. On the river, assuming you hit a turn out, you now only have 4 outs, so 4/46 probability of completing the straight assuming you hit a turn out. Runner/runner KQ probability then is 8/47 x 4/46 = 32/2162 = 16/1081 = ~1.48%. The probability of the other case is the same, and the two cases are exclusive - both can never occur on the same runout, so we can add the probabilities, or equivalently multiply by 2 to get 32/1081 = ~2.96%

Your attempted strategy works better for “OR” events, like “I have 8 outs; what is the probability that I will hit one on the turn OR the river?” This case is an “AND” event — “What is the probability that I will hit one of eight outs on the turn AND one of four on the river?” No need to do the probability of the event not occurring; we can get the probability directly by multiplying the individual event probabilities.

Thank you so much for this reply! You really helped!

If there was only one straight draw available you wouldn’t x 2 at the end right?
That’s just because of the two separate combos that could come (Kq and 89)

Thank you again

Correct. A runner runner with only one possible combo to hit (like AT2 on the board when you have a Q and no K or J - you can only hit the straight with KJ) would be half as likely. The x2 is due to the two possible runner runner combos