Probability that $n$ fair six-sided dice has a sum of $9$ The probability that eight fair six-sided dice has a sum of $9$ is $8\left(\frac16\right)^8.$
But if we use this formula with two dice, the answer $2\left(\frac16\right)^2$ is incorrect.
Why is there is a coefficient $n$ in the first answer but not with the other cases? Why does the formula fail for $n\ne8\,?$
 A: This is because if we want $k$ dice to be $k+1$ sum, There must be exactly one of them be two and the others are one. We can choose one from $k$ and there are exactly $k$ ways. So the answer is $k/6^k$, and put $k=8$ gives the result.
Of course you can not do it with two dice with sum $9$. But if you want to have the same kind of answer, you can do it with sum $3$.
A: Obtaining a sum of $9$ from eight dice requires that there is at most one number greater than $1$ (otherwise the sum will be at least $10$). It also requires that there is at least one number greater than $1$ (otherwise the sum will be $8$). Thus, we require exactly seven $1$s, and consequently exactly one $2.$ If we wish the first die to be the $2$, then the required probability is $\left(\frac16\right)^8.$ Since we don't care which die of the eight lands $2$, the required probability is $$8\left(\frac16\right)^8.$$
On the other hand, obtaining a sum of $9$ from just two dice requires obtaining one of the following cases from thirty-six: $(3,6),(4,5),(5,4),(6,3);$ so, the required probability is $$\frac4{36}.$$
