Suppose that 8 dice are rolled. What is the probability that the sum of the eight dice is 9?

I would interpret this question as: What is the probability that we get exactly 7 ones and one 2. We have 8 possible indices and out of them we choose 7 for the ones. The remain index will go for the remaining two.


$P(sum = 9) = \binom{8}{7} (\frac{1}{6})^8 (\frac{5}{6})^0$

Is this correct? Thanks in advance!

  • $\begingroup$ What is that $(\frac{5}{6})^0$ for? $\endgroup$ – Jorge Fernández Hidalgo Feb 26 '15 at 14:24
  • $\begingroup$ I was using the general formula of $\binom{n}{k} \theta^k (1-\theta)^{n-k}$ $\endgroup$ – geomquestion Feb 26 '15 at 14:25

There are eight possible outcomes (result sequences) that get you $7$ ones and one $2$. Each outcome has probability $\frac{1}{6^8}$ .Hence the probability is $\frac{8}{6^8}$

  • $\begingroup$ Thanks! That's another way of looking at it, but much more straightforward. $\endgroup$ – geomquestion Feb 26 '15 at 14:25

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.