A three dice roll question I got this question from an interview.
A and B are playing a game of dice as follows. A throws two dice and B throws a single die. A wins if the maximum of the two numbers is greater than the throw of B. What is the probability for A to win the game?
My solution.
If $(X,Y)$ are the two random throws of A and $Z$ is the random throw of B, then the problem asks (if I guessed correctly) to compute $P(\max(X,Y)>Z)$. But how to do it? I highly appreciate your help.

Consider the probability of its complementary event $\{\max(X, Y) \leq Z\}$:
\begin{align}
P(\max(X, Y) > Z) = 1 - P(\max(X, Y) \leq Z) = 1 - P(X \leq Z, Y \leq Z).   \tag{$*$}
\end{align}
The term $P(X \leq Z, Y \leq Z)$ can be evaluated by the law of total probability and the independence assumption:
\begin{align}
 & P(X \leq Z, Y \leq Z) = \sum_{i = 1}^6P(X \leq Z, Y \leq Z, Z = i) \\
=& \sum_{i = 1}^6 P(X \leq i, Y \leq i, Z = i) \\
=& \sum_{i = 1}^6 P(X \leq i)P(Y \leq i)P(Z = i) \\
=& \sum_{i = 1}^6 \frac{i}{6} \times \frac{i}{6} \times \frac{1}{6} = \frac{1}{216} \times \frac{1}{6} \times 6 \times 7 \times 13 
= \frac{91}{216}. \\
\end{align}
Plugging this back to $(*)$ to get
\begin{align}
P(\max(X, Y) > Z) = 1 - \frac{91}{216} = \frac{125}{216}.   
\end{align}
