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?
 A: Your solution for $\Sigma=9$ is correct, but the other two are not.  In place of the $8^2$ and $8^3$, you should have ${8\choose2}+{8\choose1}$ and ${8\choose3}+{8\choose1}{7\choose1}+{8\choose1}$, respectively.
The idea is this:  For $\Sigma=10$, you need to either turn two of your initial eight $1$'s into $2$'s, or one of them into a $3$.  For $\Sigma=11$, you need to either turn three $1$'s into $2$'s, or one to a $2$ and one to a $3$, or one to a $4$.  It's sometimes helpful to think systematically with binomial coefficients, even though it might be quicker to write $8$ and $7$ instead of $8\choose1$ and $7\choose1$.
A: I'm assuming that you're talking about 6-sided dice since you used $\frac16$ in your equations.  
The generating function of a six-sided die is $g_6(x) = \frac16(x + x^2 + x^3 + x^4 + x^5 + x^6)$, so the generating function of rolling 8 of them is $(g_6(x))^8$.  The probability of rolling $n$ can be determined by looking at the coefficient of $x^n$ in the generating function.  
I get $P(\Sigma = 9) = \dfrac1{209952}$,  $P(\Sigma = 10) = \dfrac1{46656}$, and $P(\Sigma = 11) = \dfrac5{69984}$, which don't match your numbers.
A: 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: $\left(\frac{8!}{1!7!}+\frac{8!}{1!1!6!}+\frac{8!}{3!5!}\right)\times6^{-8}$
A: Sum of 9 :  1, 1, 1, 1, 1, 1, 1, 2 = $$\frac{8!}{7!} = 8$$
Sum of 10:  1, 1, 1, 1, 1, 1, 2, 2 =$$\frac{8!}{6!2!}  = 28$$
1, 1, 1, 1, 1, 1, 1, 3 =$$\frac{8!}{7!} = 8$$
$$ 8+28  = 36$$
Sum of 11: 1, 1, 1, 1, 1, 1, 1, 4 :$$\frac{8!}{7!} = 8$$
1, 1, 1, 1, 1, 1, 2, 3 : $$\frac{8!}{6!} = 56$$
1, 1, 1, 1, 1, 2, 2, 2 : $$\frac{8!}{5!3!} = 56$$
$$ 8+56*2  = 120$$
Thus the probabilities for sums of 9, 10, 11 are $$8.\frac{1}{6^8}, 36.\frac{1}{6^8}, 120.\frac{1}{6^8}$$ respectively
