PMF for the maximum of two four-sided dices 
Suppose we roll two four-sided dice, that is, each die has four sides, numbered $1, 2, 3, 4$. Let $X_1$ and $X_2$ be the numbers that appear on the first and second die respectively, and let $Z = \max\{X_1, X_2\}$, that is $Z$ is the larger of the two numbers rolled. Find the probability mass function of $Z$.

The tricky part for me is that it takes the $\max$ of two numbers. Therefore, my initial approach was listing all the possible combinations where there could be a $\max$ number, e.g.: $(1,2, 2,1, 2,2)$ for side $2$. But I assumed that, because there is no $0$ side, $1$ can't be max. I'm struggling to get the probabilities $f_z(z) = 12/16$, $f_z(z) = 4/16$, $f_z(z) = 3/16$. $f_z(z) = 5/16$. What $z$ values would they have to be to get to these probabilities? Note that I used $f_z(z)$, for $Z$. I initially thought the $z$ would be an interval of values but I was completely wrong.
 A: Assuming the die is fair, there are $4^2 = 16$ equally likely outcomes.
The maximum of a set, if it exists, is the largest element of the set.  
\begin{array}{c c}
\text{outcome} & \text{maximum}\\ \hline 
(1, 1) & 1\\
(1, 2) & 2\\
(1, 3) & 3\\
(1, 4) & 4\\
(2, 1) & 2\\
(2, 2) & 2\\
(2, 3) & 3\\
(2, 4) & 4\\
(3, 1) & 3\\
(3, 2) & 3\\
(3, 3) & 3\\
(3, 4) & 4\\
(4, 1) & 4\\
(4, 2) & 4\\
(4, 3) & 4\\
(4, 4) & 4
\end{array}
By inspection, a maximum of $1$ occurs once, a maximum of $2$ occurs three times, a maximum of $3$ occurs five times, and a maximum of $4$ occurs $7$ times.  Hence, we obtain the probability mass function
\begin{align*}
\Pr(Z = 1) & = \frac{1}{16}\\
\Pr(Z = 2) & = \frac{3}{16}\\
\Pr(Z = 3) & = \frac{5}{16}\\
\Pr(Z = 4) & = \frac{7}{16}
\end{align*}
As a sanity check, notice that we have accounted for all possible outcomes and that $$\Pr(Z = 1) + \Pr(Z = 2) + \Pr(Z = 3) + \Pr(Z = 4) = \frac{1}{16} + \frac{3}{16} + \frac{5}{16} + \frac{7}{16} = 1$$
as required
A: $$\mathbb P(Z\leqslant t)=\mathbb P(\max \{X_1,X_2\}\leqslant t)=\mathbb P(X_1\leqslant t , X_2\leqslant t)=\mathbb P(X_1\leqslant t)\mathbb P(X_2\leqslant t)=\frac t4\times\frac t4$$
So,
$$\mathbb P(Z= t)=\frac{t^2}{16}-\frac{(t-1)^2}{16}=\frac{2t-1}{16}.$$
