$Pr(A+B+C=X+Y+Z)$ with $A, B, C, X, Y, Z$ being iid random variables $A, B, C, X, Y, Z$ are iid random variables and they are uniformly distributed on the set of integers $0$ to $9$ (inclusive). What is $\Pr(A+B+C=X+Y+Z)$?
I am kind of lost and my attempt is to use generating functions such that $$G_{A+B+C}(t)\cdot G_{X+Y+Z}(1/t)$$ because i can transform $\Pr(A+B+C=X+Y+Z)$ into $\Pr((A+B+C)-(X+Y+Z)=k)$ and $k=0.$
So I need to find the coefficient of $t^0$ in the function $$G_{A+B+C}(t) \cdot G_{X+Y+Z}(1/t).$$ But how do i find $G_{A+B+C}(t)$?

Let $A,B,C$ and $X,Y,Z$ are iid uniformly distributed on $\{0,1,\dots,9\}$, split the below probability by the value of the sums, 
$\mathbb{P}(A+B+C=X+Y+Z)=\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k,\ X+Y+Z=k).$
Since $A+B+C$ and $X+Y+Z$ are independents, for all $k\in \mathbb{Z}$: 
$\mathbb{P}(A+B+C=k,X+Y+Z=k)=\mathbb{P}(A+B+C=k)\cdot \mathbb{P}(X+Y+Z=k)$ 
and $\mathbb{P}(A+B+C=k)\cdot \mathbb{P}(X+Y+Z=k)=\mathbb{P}(A+B+C=k)^2$, because $A+B+C$ and $X+Y+Z$ has the same distribution. It means
$\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k,\ X+Y+Z=k)=\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k)^2$.
There are 28 possible value for $A+B+C$ from $0$ to $27$, please note that distribution of the sum is symmetric, so $\mathbb{P}(A+B+C=k)=\mathbb{P}(A+B+C=27-k)$, which means if one calculates their probabilities, the sum of squares will be 
$\mathbb{P}(A+B+C=X+Y+Z)=\displaystyle 2\cdot \sum_{k=0}^{13}\mathbb{P}(A+B+C=k)^2=\\ \displaystyle = 2 \cdot \frac{1^2+3^2+6^2+10^2+15^2+21^2+28^2+36^2+45^2+55^2+63^2+69^2+73^2+75^2}{1000^2}=  \\ =5.5252\%.$
