Modular Dice

Consider $P(x)=(x+x^2+x^3+x^4+x^5+x^6)^{1000}$

In the expansion, for every roll combination with a sum of $k$ there is a single $x^k$ .

Let $s_k$ be the number of roll combinations with a sum congruent to $k$ modulo $4$ .

Then $P(i)=s_0-s_2+(s_1-s_3)i$ since for each sum congruent to $k$ modulo $4$ there is an extra $1,i,-1, -i$ respectively in the expansion.

$P(i)=(i-1)^{1000}=(\sqrt{2} \mathrm{cis}( 3\pi/4))^{1000}=2^{500} \mathrm{cis}( 750 \pi) = 2^{500}$

Thus $s_1=s_3$ and $s_0-s_2=2^{500}$

Now let's relate the number of odd-sum permutations to the number of even-sum permutations.

There are a variety of ways to see why they are equal, but keeping with the theme, we see that $P(-1)=0=(s_0+s_2)-(s_1+s_3)$ .

Since $s_0+s_1+s_2+s_3=6^{1000}$ , we have $s_0+s_2=6^{1000}/2$ .

Therefore $s_0=\dfrac{6^{1000}/2+2^{500}}{2}$

$=\dfrac{6^{1000}}{4}+2^{499}$

$\dfrac{1}{n}=\dfrac{2^{499}}{6^{1000}}$

$\ln{n}=1000\ln{6}-499\ln{2}=1445.88$