# The sum of the points obtained by rolling a perfect die

A perfect die is rolled n times. The sum of all points should follow a multinomial distribution, and the pmf is

$$P(S_n=k)=\sum_{a_1+...+a_n=k}{n\choose{a_1,a_2,...,a_n}}(\frac{1}{6})^n$$

So when n=100, how can I find $P(330\leq S_n \leq 380)$? It seems that I should apply the normal approximation of multinomial distribution, but I don't know why the approximation is vaild, neither do I know how to apply it.

I know why, how and when I can approximate the binomial distribution with Poisson or Normal distribution.

The distribution is not multinomial. The detailed probability distribution would be quite painful to calculate.

Let $X_i$ be the number obtained on the $i$-th roll. Then the total number of points $X$ obtained is given by $$X=X_1+X_2+\cdots +X_{100}.$$ So $X$ is a sum of $100$ independent identically distributed "nice" random variables. The distribution of this sum should be well approximated by a suitable normal.

Which normal? Calculate the mean $\mu$ and variance $\sigma^2$ of $X_i$. Then $X$ has mean $100\mu$ and variance $100\sigma^2$. So these are the mean and variance of the appropriate normal $W$.

It might improve accuracy to use a continuity correction, and calculate $\Pr(W\le 380.5)-\Pr(W\le 329.5)$.

• Thanks! Could you fill me in with some background information about why the sum of a lot of uniformly distributed random variables follows the normal distribution? – arax Nov 30 '13 at 7:32
• This is a very important theorem called the Central Limit Theorem. All that is needed is for the mean and variance to exist. However, knowing whether $100$ is "large enough" and the $X_i$ are "nice enough" for the approximation to be good enough is a combination of guesswork and experience. – André Nicolas Nov 30 '13 at 7:32
• Just saw it in the textbook. It's in the next chapter and I was only rummaging in this chapter, which is mainly about the less-generalized De Moivre-Laplace Theorem. – arax Nov 30 '13 at 7:38
• For a discussion of the CLT, please see this. Note that the CLT tells you, roughly, that the probability can be computed with increasingly good accuracy by the procedure I described as $n\to\infty$ ($n$ the number of rolls). The number $100$ is well short of $\infty$, hence the vagueness about "good enough." – André Nicolas Nov 30 '13 at 7:40

Let $p=\frac{1}{6}(x+x^2+x^3+x^4+x^5+x^6)$.

Then the coefficient of $x^n$ in $p^{100}$ yields the probability of a sum of $n$ after 100 rolls of a fair, six-sided die.

Summing these yields the probability you seek, exactly. No approximation needed. (Roll the die a million times? Well ...)

For instance, the probability that the sum is between 340 and 360, inclusive, is exactly $$\frac{8362812942456116380825281817303173384065140369245183388396883580028163482771}{18147739541668636280463618532168272792698436402026524209529776843597142818816}$$

This is all easily done with any computer algebra system.