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.

We need to know two things to calculate Player A's overall chances of winning: 1) his odds of rolling each score and 2) his odds of winning with that score.
If you visualize player A's possible outcomes on a 6x6 grid, it is easy to see that they form a series of bands (for lack of a better word), and we can count them and divide by the total number of outcomes to get the probabilities for each.

It's also fairly easy to see that the number of occurrences of each outcome can be represented as 2n-1 (because it's two sides of a square of size n with the overlapping corner being -1), and the total number of outcomes is 6 squared, so the probability of each of A's possible outcomes is

Player B's outcomes are evenly distributed: 1/6 (or 16.7%) for each number 1 through 6.  Since Player A has to BEAT Player B's roll, we can find his chance of winning by summing the probabilities of each of B's possible outcomes that are less than his outcome.  That is, if he scores 1, none of B's outcomes are less, meaning that he can beat 0% of B's possible scores.  A score of 2 can beat just one of B's possible scores, or 16.7%, and a score of 3 wins in two out of six scenarios (33.3%).  This can be represented as


We can multiply these probabilities together (A's chance of rolling a specific number vs his chance of winning with that number) to obtain his overall chances of winning:

57.9%, or 125/216.
The formula for the overall chance of winning can be derived by multiplying the two previously given formulae:
 or 
giving the same table of values:

We can represent the summation like this:

and this does yield the expected 125/216.  Just a slightly different way to think about it, and the demonstration of it can quickly be setup in a spreadsheet in just a minute or two, if they were letting you use a computer, or sketched out just as quickly.
