Probability of eight dice showing sum of 9, 10 or 11 Suppose we roll eight fair dice. What is the probability that:
1) The sum of the faces is $9$
2) The sum of the faces is $10$
3) The sum of the faces is $11$
I'm thinking that we start with $8$ dice, each showing $1$. Then we think of the problem as assigning one $1$, two $1$'s or three $1$'s to the $8$ dice.
So that would give us:   
1) $P(\sum = 9)$ = $8(\frac{1}{6})^8$
2) $P(\sum = 10)$ = $8^2(\frac{1}{6})^8$
3) $P(\sum = 11)$ = $8^3(\frac{1}{6})^8$
Is this right?

In all possible orders:
1) $21111111$. Probability: $\frac{8!}{1!7!}\times6^{-8}$
2) $31111111$ or $22111111$. Probability: $\left(\frac{8!}{1!7!}+\frac{8!}{2!6!}\right)\times6^{-8}$ 
3) $41111111$or $32111111$ or $22211111$. Probability: $\left(\frac{8!}{1!7!}+\frac{8!}{1!1!6!}+\frac{8!}{3!5!}\right)\times6^{-8}$
