$Pr(A+B+C=X+Y+Z)$ 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 $G_{A+B+C}(t)$?
 A: Since all six variables are independent. $$\mathsf G_{A+B+C-X-Y-Z}(t)=\mathsf G_A(t)\,\mathsf G_B(t)\,\mathsf G_C(t)\,\mathsf G_X(t^{-1})\,\mathsf G_Y(t^{-1})\,\mathsf G_Z(t^{-1})$$
And since they are identically distributed therefore
$$\mathsf G_{A+B+C-X-Y-Z}(t)=\mathsf G_A^3(t)\,\mathsf G_A^3(t^{-1})$$
Now... what is the PGF of a uniform discrete random variable over $[0;9]\cap\Bbb N$ and how do you plan to deal with the complication at $t=0$?
$$\mathsf G_A(t)=\tfrac 1{10}\sum_{x=0}^9 t^x$$
So... ?

via:
$\begin{split}\mathsf G_{\sum_k a_kX_k}(t) &=\mathsf E(t^{\sum_k a_kX_k})\\[1ex]&=\mathsf E(\prod_k t^{a_kX_k}) \\[1ex]&\overset{ind.}= \prod_k\mathsf E((t^{a_k})^{X_k}) \\[1ex]&= \prod_k\mathsf G_{X_k}(t^{a_k})\end{split}$

PS:  You are not looking for the $t^0$ term. The probability mass function is recovered from the derivatives of $\mathsf G$.$$\mathsf P(S=k)= \mathsf G_S^{(k)}(0)/k!$$
A: 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 
$\mathbb{P}(A+B+C=X+Y+Z)=\displaystyle 2\cdot \sum_{k=0}^{13}\mathbb{P}(A+B+C=k)^2=\\ \displaystyle = 2 \cdot \frac{1^2+3^2+6^2+10^2+15^2+21^2+28^2+36^2+45^2+55^2+63^2+69^2+73^2+75^2}{1000^2}=  \\ =5.5252\%.$
A: Hint: If $X_1$ and $X_2$ are iid, then $P(X_1=X_2)=\sum_{n}P(X_i=n)^2$. If $X_1,X_2,X_3$ are iid uniformly on $\{1,\ldots, n\}$, then $$P(X_1+X_2+X_3=k)=\sum_{i,j} P(X_1=i)P(X_2=j)P(X_3=k-i-j)=n^{-3}\#\{(i,j)\mid k-n\leq i+j<k\}.$$
The number of pairs $(i,j)$ such that $i+j<k$ is...?
