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!
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 $y=1,2,3,4,5,6$
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.

$$
\begin{array}{|c|c|c|c|c|c|c|} \hline
  &amp; \textbf{1} &amp; \textbf{2} &amp; \textbf{3}
  &amp; \textbf{4} &amp; \textbf{5} &amp; \textbf{6} \\ \hline
\textbf{1} &amp; 1 &amp; 2 &amp; 3 &amp; &amp; &amp; \\ \hline
\textbf{2} &amp; 2 &amp; 2 &amp; 3 &amp; &amp; &amp; \\ \hline
\textbf{3} &amp; 3 &amp; 3 &amp; 3 &amp; &amp; &amp; \\ \hline
\textbf{4} &amp; &amp; &amp; &amp; &amp; &amp; \\ \hline
\textbf{5} &amp; &amp; &amp; &amp; &amp; &amp;  \\ \hline
\textbf{6} &amp; &amp; &amp; &amp; &amp; &amp; \\ \hline
\end{array}
$$

The probability distribution for $Y = \max(X_1, X_2)$ can then be obtained by counting up the proportion of entries that are $1, 2, 3, \ldots$.