Probability of a dice game with an emphasis on the number $4$ If I were to roll a set of $4\,$ four-sided dice $4\,$ times, what is the probability that on each of the $4\,$ dice rolls, the sum of the dice will be divisible by $4\,$?
 A: It will be the $4$-th power of the probability that it happens when we toss the $4$ dice once.
Presumably we are to assume that the $4$ sides are numbered $1$ to $4$. Since we are working modulo $4$, we might as well assume it is $0$ to $3$. 
We want to find the sum of the coefficients of terms in the expansion of $(1+x+x^2+x^3)^4$ that are of the shape $x^{4t}$. Let the expansion be 
$$f(x)=a_0+a_1 x+a_2x^2+\cdots +a_{12}x^{12}.$$
Substitute in turn $x=1,-i,i,-i$. We get
$$f(1)=a_0+a_1+a_2+a_3+\cdots+a_{12}=4^4.$$
$$f(-1)=a_0-a_1+a_2-a_3+\cdots +a_{12}=0.$$
$$f(i)=a_0+ia_1-a_2-ia_3+\cdots +a_{12}=0.$$
$$f(-i)=a_0-ia_1-a_2+ia_3+\cdots+a_{12}=0.$$
Add up. The only thing that survives is coefficients divisible by $4$, each multiplied by $4$. The sum of these coefficients is therefore $\frac{4^4}{4}$.  
It follows that the probability that when we toss once the sum is divisible by $4$ is $\frac{\frac{4^4}{4}}{4^4}$. Thus the required probability is
$$\left(\frac{\frac{4^4}{4}}{4^4}\right)^4.$$
We could simplify, but then we would lose a lot of $4$'s. 
