How to find distribution for roll of 2 dice 
You roll 2 ordinary dice. Let X denote the maximum of the two numbers you get. What is the distribution of X?

I did the problem as follows:
$$\begin{array}\\
X &= 1: (1, 1) \\
X &= 2: (1, 2), (2, 1), (2, 2) \\
X &= 3: (1, 3), (3, 1), (2, 3), (3, 2), (3, 3) \\
X &= 4: (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4) \\
X &= 5: [9 \; \text {possibilities}] \\
X &= 6: [11 \; \text {possibilities}]
\end{array}$$
and
$$\begin{array}\\
P(X = 1) &= 1/36\\
P(X = 2) &= 3/36\\
P(X = 3) &= 5/36\\
P(X = 4) &= 7/36\\
P(X = 5) &= 9/36\\
P(X = 6) &= 11/36 \end{array}$$
Is there a counting formula I can use to do this problem without enumerating all the possibilities? It was a bit tedious.
 A: Hint: Look for a pattern in the numerators of your probabilities!  See if you can derive the counting formula for yourself.
A: Break this down into cases:
If the dice both roll the same value that is 1 combination.
Otherwise, one of the die has to be lower than X and can have values from 1 to (X-1) and there are 2 dice which means there are $2*(X-1)=2X-2$ combinations in this case
Now, add these together and we get: $2X-1$ which is what you have there.
A: An approach that does not involve any counting follows from taking into account that the $z=\max(x_1,x_2)$ translates into the following cumulative distribution functions:
$$P(z\le k) = P(x_1\le k,x_2\le k) = P(x_1\le k)P(x_2\le k)$$
The last equality follows from the independence of the random variables $x_1$ and $x_2$. In order to obtain the distribution function of the random variable $z$, $P(z=k)$, and given that $z,x_1,$ and $x_2$ are discrete random variables we use:
$$P(z=k) = P(z\le k) - P(z\le k-1).$$
For the particular case of a two fair die with 6 sides:
$$P(z\le k) = \left( \frac{k}{6}\right)^2$$ 
and
$$P(z=k)= \left( \frac{k}{6}\right)^2-\left( \frac{k-1}{6}\right)^2.$$
Which is a different way of obtaining the above probabilities without counting.
I think this approach is better suited for generalisation to unfair die and die with different number of sides.
By the way, if there are $n$ die all with $p$ sides the solution is:
$$P(z=k)= \left( \frac{k}{p}\right)^n-\left( \frac{k-1}{p}\right)^n.$$
