<s>[INST] 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?
 [/INST] 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
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