Two six-sided dice are rolled. Find the probability distribution of $\max[X_1,X_2]$ I am not sure if I am understanding this problem correctly. It says;

Two six-sided dice are rolled. Let X1 and X2
  denote the outcomes of the first and second die roll respectively.
  Define a new random variable Y such that: 
Y = max{X1 , X2}

Does this mean I am supposed to roll the two dice and find the largest roll? 
From my own understanding, if I am supposed to find the largest roll from the two dice roll, is this example of mine correct?
ex: the outcomes are {2,3} then Y=3
Moving on to what is asked of me:

Find the probability distribution of Y

For this part, do I need to find all the possible outcomes of the two dice? For example; 
{1,1} {1,2} {1,3} {1,4} {1,5} {1,6}
{2,1} {2,2} {2,3} {2,4} {2,5} {2,6}
{3,1} {3,2} {3,3} {3,4} {3,5} {3,6}
etc....
And do I need to find the probability of the sum of two dice? For example;
Sum = 2 , Prob = 1/36
Sum = 3 , Prob = 2/36
Sum = 4 , Prob = 3/36
etc....
After this, what do I do from here to find the probability distribution of Y? 
If I am totally off track, explanation from you guys will be great. Sorry if it looks like I'm asking too much. Please feel free to correct me on anything!
 A: Notice that the mirror image of every L-shaped event corresponds to the outcome $Y$
As an example, consider the L-shaped event corresponding to the outcome $Y=3$
$$\begin{align}
                  \{1,3\} \\
                  \{2,3\} \\
\{3,1\}, \{3,2\}, \{3,3\} 
\end{align}$$
We can easily see that $$\mathbb{P}(Y=3)=\frac{5}{36}$$
In general, you can generalize that $$\mathbb{P}(Y=y)=\frac{2y-1}{36}$$ where $y=1,2,3,4,5,6$
A: It may be useful to construct a table, with $X_1$ down the left, and $X_2$ along the top, and each cell gets the maximum of the two values.  The table below has been started for you; fill in the rest of the values.
$$
\begin{array}{|c|c|c|c|c|c|c|} \hline
  & \textbf{1} & \textbf{2} & \textbf{3}
  & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline
\textbf{1} & 1 & 2 & 3 & & & \\ \hline
\textbf{2} & 2 & 2 & 3 & & & \\ \hline
\textbf{3} & 3 & 3 & 3 & & & \\ \hline
\textbf{4} & & & & & & \\ \hline
\textbf{5} & & & & & &  \\ \hline
\textbf{6} & & & & & & \\ \hline
\end{array}
$$
The probability distribution for $Y = \max(X_1, X_2)$ can then be obtained by counting up the proportion of entries that are $1, 2, 3, \ldots$.
