Random Variable transformation problem I'm stuck with the following problem:
Problem:
If $X$ is a random variable defined as the sum of rolling a pair of dice, find the probability distribution of the remainder obtained when $X$ is divided by $3.$
Thoughts
Let say that $Y=$ the remainder obtained when $X$ is divided by 3
So $Y=X\bmod3$ which is defined as
$Y=X\bmod3=X-3\lfloor \frac{X}{3}\rfloor$ based on Wolfram MathWorld.
$Y=X-3\lfloor \frac{X}{3} \rfloor$
But honestly I'm pretty much stuck there, I'm not sure on how to represent the problem.
Any help would be highly appreciated.
 A: Considering a random variable:
\begin{eqnarray*}
    X:\Omega    & \to   &    \{2,3,\dots,12\}\\
      \omega    &\mapsto&    X(\omega) = n
\end{eqnarray*}
and
\begin{eqnarray*}
    Y:\Omega    & \to   &    \{0,1,2\}\\
      \omega    &\mapsto&    Y(\omega) = \frac{X(\omega)}{3}-\left[\frac{X(\omega)}{3}\right]
\end{eqnarray*}
where $[x]$ is the integer part of $x$. Then
\begin{eqnarray*}
    \mathbb{P}(Y=k)
    & = &    \mathbb{P}\left(\frac{X(\omega)}{3}-\left[\frac{X(\omega)}{3}\right]=k\right)
\end{eqnarray*}
But on the other hand, by total probability
\begin{eqnarray*}
    \mathbb{P}(Y=k)
    &=&    \sum_{n=2}^{12} \mathbb{P}(Y=k|X=n)\mathbb{P}(X=n)\\
\end{eqnarray*}
since you have a finite number of cases for $n$ and you know the distribution of $X$, then
\begin{eqnarray*}
    \mathbb{P}(Y=k)
    &=&    \sum_{n\in\{2,\dots,12\}\wedge k=0} \mathbb{P}(Y=k|X=n)\mathbb{P}(X=n)\\
    & &    +\sum_{n\in\{2,\dots,12\}\wedge k=1} \mathbb{P}(Y=k|X=n)\mathbb{P}(X=n)\\
    & &    +\sum_{n\in\{2,\dots,12\}\wedge k=2} \mathbb{P}(Y=k|X=n)\mathbb{P}(X=n)\\
\end{eqnarray*}
my notation is not very lucky, but the idea is to note that you can identify the elements $n$ that have remainder $0$,$1$ or $2$ as appropriate, since the probability over $Y$ will be equal taking advantage of the fact that $k$ is given from $n$.
