What defines a "description" of a probability distribution? Say you take a dice and you roll it twice so that you have a pair (X,Y) where X represents the first roll, Y represents the second. 
When you have a distribution like max(X,Y), what type of distribution is that? It's not uniform, obviously, because the outcomes of max(X,Y) are clearly more favorable towards the higher numbers. I'm being asked to "describe" the distribution. I know that we can look at each distinct outcome, like:
max(1,1) = 1
max(1,2) = 2
max(1,3) = 3
max(1,4) = 4
max(1,5) = 5
max(1,6) = 6
max(2,1) = 2
max(2,2) = 2
max(2,3) = 3
max(2,4) = 4
max(2,5) = 5
max(2,6) = 6
max(3,1) = 3
max(3,2) = 3
max(3,3) = 3
max(3,4) = 4
max(3,5) = 5
max(3,6) = 6
max(4,1) = 4
max(4,2) = 4
max(4,3) = 4
max(4,4) = 4
max(4,5) = 5
max(4,6) = 6
max(5,1) = 5
max(5,2) = 5
max(5,3) = 5
max(5,4) = 5
max(5,5) = 5
max(5,6) = 6
max(6,1) = 6
max(6,2) = 6
max(6,3) = 6
max(6,4) = 6
max(6,5) = 6
max(6,6) = 6
But is that the most effective way to look at a distribution like this? Is there a faster and more concise way to describe this distribution other than saying the probability of each outcome?
i.e.,
P(max = 1) = 1/36
P(max = 2) = 3/36 = 1/12
P(max = 3) = 5/36
P(max = 4) = 7/36
P(max = 5) = 9/36 = 1/4
P(max = 6) = 11/36
So, obviously, I understand how to LOOK at a discrete distribution and see what the probabilities are, but not until I look at each possible outcome. But what does it mean to "describe" a distribution? This isn't uniform or any sort of discrete distribution we're taught. Can someone provide clarity?
 A: "Describe" is a vague term in this case. There are two possibilities that I assume they mean. They either will accept a distribution table, which would amount to writing down all possibilities and probabilities like you say. The other way is to give the pmf. To get the pmf of $Z = \max\{X,Y\}$, you want to do the following:
$$\begin{align*}
\mathsf P(Z=z) &= \mathsf P(Z< z+1)-\mathsf P(Z<z)
\\[1ex]
&=\mathsf P(X<z+1,Y<z+1)-\mathsf P(X<z,Y<z)
\\[1ex]
&=\mathsf P(X<z+1)\mathsf P(Y<z+1)-\mathsf P(X<z)\mathsf P(Y<z)\tag 1
\\[1ex]
&=\left(\frac{z}{6}\right)\left(\frac{z}{6}\right)-\left(\frac{z-1}{6}\right)\left(\frac{z-1}{6}\right)
\\[1ex]
&=\left(\frac{z}{6}\right)^2-\left(\frac{z-1}{6}\right)^2\\
&= \frac{2z-1}{36}
\end{align*}$$
where $(1)$ is true by independence.

Let's suppose now that we want the minimum $M$ of two dice rolls. What is the probability of $M=3$. Well, if you draw out a $X\times Y$ table, then the minimum is $3$ where the cells are dotted, and
it appears to be $\frac{7}{36}$. There are a number of ways to count this. 

One way is to take the number of cells in the larger blue box and subtract the number of cells in the smaller red box. 
This is
$$\frac{4^2}{36}-\frac{3^2}{36} = \frac{16-9}{36} = \frac{7}{36}.$$
Notice that is precisely
\begin{align*}
\mathsf P(M = m) &= \mathsf P(M>m-1)-\mathsf P(M>m)\\
&=\mathsf P(X>m-1)\mathsf P(Y>m-1)-\mathsf P(X>m)P(Y>m)\\
&=\left(\frac{6-(m-1)}{6}\right)\left(\frac{6-(m-1)}{6}\right)-\left(\frac{6-(m)}{6}\right)\left(\frac{6-(m)}{6}\right)\\
&=\left(\frac{7-m}{6}\right)^2-\left(\frac{6-m}{6}\right)^2\\
&= \frac{13-2m}{36}
\end{align*}
So,
$$\mathsf P(M=3) = \frac{13-2(3)}{36} = \frac{7}{36}.$$
A: Most likely, they want you to give the probability mass function (PMF) of $Z=\max \{ X,Y \}$. You can get that by writing down the values for each outcome $(x,y)$, then count up the number of $1$s, $2$s, etc. and divide by the number of outcomes. So $p(1)=1/36,p(2)=2/36$, and so forth.
