What defines a "description" of a probability distribution?

"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, $$\mathsf P(M=3) = \frac{13-2(3)}{36} = \frac{7}{36}.$$