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What is the probability that it takes more than seven rolls of a fair $6$-sided die to roll a six?

The probability of rolling a six per roll is $1/6$. Therefore, the probability of rolling something other than a 6 is $5/6$.

So wouldn't the probability that you don't roll a six within the first $7$ rolls just be $(5/6)^7$?

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    $\begingroup$ Yes, that is correct. $\endgroup$ – N. F. Taussig Apr 30 '18 at 21:28
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    $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 30 '18 at 21:29
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Your answer is correct.

Let $X$ be number of rolls to get the first $6$, then $X$ follows geometric distribution.

$$P(X > 7)=1-P(X \le 7)=(1-p)^7$$

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