What are the chances of rolling 2 dice 43 times without rolling a 7?

Question

Forgive my very long explanation. I play craps and I attempt to make my throws less random.
I track my throws and collect data from sets of 900 throws. Obviously, a random thrower is likely to throw around 150 sevens every 900 rolls. My last set of 900 rolls I rolled 113 sevens or a seven every 7.96 rolls. The 900 rolls prior to that I rolled 123 sevens or a seven every 7.32 rolls.
Obviously. the math will not work out to be the odds of rolling 43 times in a single turn as a player may throw a seven on a come out roll after making the point. I am asking specifically about 43 rolls with no sevens as I recently went to the casino for the first time since the COVID restrictions began. My first turn I threw 43 times with my first seven on my 44th throw. I hope you made it through my long explanation. Thank you.

Answer 1

The chance to roll 7 is $1/6$ which I assume you already know from your description. So the chance to not roll a 7 is $5/6$. Assuming independent rolls this means not getting any 7 in 43 rolls has probability $(5/6)^{43} \approx 0.0004$, so less than 0.04%