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?
Best Answer
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 $7\choose1$.