craps "pass/dont pass" probability

hey guys have a math question for craps in regards to the probability of rolling 9 "pass" bets in a row or on the flip side 9 "dont pass" rolls in a row. I dont know if there is a formula to figure this out or not due to all the different variables but, I figured if there was, this would be the place to find it. I have read, to hit 5 rolls in a row is around 33-1 and 6 rolls in a row is 70-1. but nothing for 7-9 rolls in a row. any truth to this? thanks in advance for any help that you may have.

The probability of winning a passline is 244/495. Winning atleast 3 in a row would be (244/495)^3 . Just type (244/495)^9 into google and it will give you the answere. If you want to know 1 in how many just type 1/(244/495)^9 . It's about 1 in 582.
For winning don't pass use 2847/5940 if you count 12 as not winning or 2847/5775 if you ignore 12.

you might explain whether you are trying to compute odds of winning or losing the come-out roll, or if for some reason you want to know the odds of it going so many times in a row resolved immediately with a 7, 11, or craps roll.

Quote: miplet

The probability of winning a passline is 244/495. Winning atleast 3 in a row would be (244/495)^3 . Just type (244/495)^9 into google and it will give you the answere. If you want to know 1 in how many just type 1/(244/495)^9 . It's about 1 in 582.
For winning don't pass use 2847/5940 if you count 12 as not winning or 2847/5775 if you ignore 12.

thank you man, i figured this was the place for the answer