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.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ You see in the below answer that you hit a very rare run of consecutive non-sevens. $\endgroup$– PeterCommented Mar 15, 2022 at 8:28
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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%
