Probability that 100 dice rolls sum to 400? I would like to find the probability that the results of 100 dice rolls sum to 400. I think to do this exactly would take a lot of work unless there's some trick I am not privy to?
I believe the sum is distributed with a multinomial distribution with 100 trials. I need to figure out the number of ways 100 dice rolls can sum to 400, which seems to be a tall task. Any hints?
The other thought I have is typically for these kind of problems, I can use CLT and approximate the sum using a normal distribution. But typically in those kind of problems, I am trying to find the probability that a variable falls within a range of values. Here I am trying to find that the sum is exactly 400, and it's not clear to me how I can apply CLT and the normal distribution. One thought I did have is I could use a normal distribution with mean 350 and variance 291, and then integrate the area around the curve right around 400, e.g., the bounds / limits of integration would be $400 \pm \epsilon$, but it's not clear to me what $\epsilon$ should be.

I would definitely go here for generating function and then for complex derivative. Now the number of possible outcomes is listed in the coefficients of
$$ G(x)=(x+x^2+x^3+x^4+x^5+x^6)^ {100} $$
You need the coefficient at $x^{400}$. If you find $400$-th derivative you just need to multiply it by $400!$.
But this is all what Cauchy integral formula is about
$$G^{(n)}(0) = \frac{n!}{2\pi i} \oint_\gamma \frac{G(z)}{z^{n+1}}\, dz$$
which comes to calculating
$$N(100,400)=\frac{1}{2\pi} \int_0^{2\pi} \frac{(e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}+e^{5ix}+e^{6ix})^{100}}{e^{400ix}}\, dx$$
which can be shortened to
$$N(100,400)=\frac{1}{2\pi} \int_0^{2\pi} \frac{(e^{6ix}-1)^{100}}{(e^{ix}-1)^{100} e^{400ix}}\, dx$$
Now this integral can be solved in many ways, starting from using different approximations to going for solving this integral directly.
Dividing the result by $6^{100}$ you have got close to $0.000317214$
