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.
Best Answer
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