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.

Let $V =\max(X,Y)$ and note that $V$ takes values in $\{1,2,\ldots,6\}$. We see that $V=1$ if A realises $(1,1)$, $V=2$ if it realises $(1,2), (2,2), (1,2)$ and so on. Because the dice are uniformly distributed, you can check that $P(V=1) = \frac{1}{36}$, $P(V=2) = \frac{3}{36}$, and so on, $P(V=6) = \frac{11}{36}$. To compute $P(V>Z)$ use conditional probability
$$
P(V>Z) = \sum_{z=1}^6 P(V>z, Z=x) = \sum_{z=1}^6 P(V>z| Z=z)P(Z=z) = \cdots = \frac{125}{216}.
$$
Note: $P(V>z | Z=z) = P(V>z)$ for all $z$, since $V$ is independent of $Z$.
