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 $(5/6)^7$?

  • 2
    $\begingroup$ Yes, that is correct. $\endgroup$ – N. F. Taussig Apr 30 '18 at 21:28
  • 1
    $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 30 '18 at 21:29

Your answer is correct.

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.