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.

I will take a less formal approach, in order to illustrate my thinking.
My first instinct was to visualize the usual $6 \times 6$ array of outcomes $(X,Y)$ of $A$'s dice rolls, and looking at when the larger of the two values is less than or equal to some value:
$$\begin{array}{cccccc}
(1,1) & (2,1) & (3,1) & (4,1) & (5,1) & (6,1) \\
(1,2) & (2,2) & (3,2) & (4,2) & (5,2) & (6,2) \\
(1,3) & (2,3) & (3,3) & (4,3) & (5,3) & (6,3) \\
(1,4) & (2,4) & (3,4) & (4,4) & (5,4) & (6,4) \\
(1,5) & (2,5) & (3,5) & (4,5) & (5,5) & (6,5) \\
(1,6) & (2,6) & (3,6) & (4,6) & (5,6) & (6,6) \\
\end{array}$$
It's intuitively clear from this diagram that the number of ordered pairs whose maximum is at most $k$ is $k^2$, for $k \in \{1, 2, 3, 4, 5, 6\}$.  This is because geometrically, the set of such outcomes are arranged in a series of nested squares in the array.  So it follows that for each of the six equiprobable outcomes for $B$'s die roll $Z \in \{1, 2, 3, 4, 5, 6\}$, $A$ will lose with probability $z^2/6^2$, hence the total probability of $A$ losing to $B$ is simply $$\frac{1^2 + \cdots + 6^2}{6^2(6)} = \frac{6(7)(13)}{6} \cdot \frac{1}{6^3} = \frac{7(13)}{6^3}.$$  Hence $A$ wins with probability $1 - \frac{7(13)}{6^3} = \frac{125}{216}$.
This line of reasoning is what I would use if I had no access to pencil or paper and had to answer the question mentally, reserving the computational part to the very end.
