dice roll of 20 what formula would be correct to find the odds of rolling 2 dice "X" number of times without a seven showing. There are 36 combinations when rolling 2 dice with six sides. Im wondering how often one could roll 20 times. One person suggested $\displaystyle 1 - \left(\frac{5}{6}\right)^{26}$. Another stated it would have to be $\displaystyle 1 - \frac{5}{6} \cdot \frac{1}{6}$. 
Also curious if this would be true... At times during the game the 'button' is 'off' and one could roll a seven without harm. If I wanted to count only the rolls when the button is 'on' wouldn't it be the same likelihood/formula? as in not counting the 'off' rolls. Trying to say a person rolled 6 times, then button was turned off for 3 rolls, then the person rolled another 14 times, wouldnt that be the same as odds of rolling 20 times at once?
 A: If you roll two dice the number of ways to get a total of seven is 6 as you could roll any of the following (1,6) (2,5) (3,4) (4,3) (5,2) (6,1).  Since there are 36 possible results then from one roll the probability of total not being 7 is
$$\frac{36 - 6}{36} = \frac{30}{36} = \frac{5}{6}$$
For no six from two rolls the probability of no 7 is
$$\left(\frac{5}{6}\right)^2 = \frac{25}{36} \approx 0.6944$$
And for X rolls it's simply 
$$\left(\frac{5}{6}\right)^X$$
Note: that because the outcome of a roll is random if you roll 6 times while counting the result then roll 3 times not counting the result (or even a million times) then roll another 14 times counting the result the probability is still
$$\left(\frac{5}{6}\right)^{20} \approx 0.026084$$
A: rolling 2 dice X times gives you $36^X$ unique combinations.  The chances of getting a 7 on a roll are $\frac{6}{36}=\frac{1}{6}$.  so the chances of not getting a 7 on a roll are $1-\frac{1}{6}=\frac{5}{6}$.  The chances of this happening X times is $(\frac{5}{6})^X$
