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?


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


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

$$\left(\frac{5}{6}\right)^{20} \approx 0.026084$$


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 $(\frac{5}{6})^X$

  • $\begingroup$ to consider when the button is on, let X=number of rolls where the button is on. $\endgroup$ – f00d Jan 4 '14 at 20:06

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