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.

Because the point to an interview question is to demonstrate your thinking, I want to emphasize two things:

*

*Finding a simple, clear, analysis using minimal calculation and straightforward notation.


*Explicitly identifying the rules of probability that justify the solution.
Thus, in the following, you will see repeated references to simplicity and to key concepts of probability: independence, additivity, and complementation.  The hope is that following such a principled, efficient analysis and explaining it well will help get you the job in addition to merely finding the answer.

Focus first on the simplest part of the situation: B's throw.  To win, A has to beat B's value and A has two independent tries to do so.
This framing of the question suggests we break down the analysis according to B's possibilities.  Again, in pursuit of the simplest approach, consider the chance that $A$ loses after B throws some value $b.$  To lose, A must throw two values between $1$ and $b$ inclusive.  Since each of those events has a chance of $b/6$ and because the events are independent, A loses with probability $(b/6)^2$ in such a case.
Because the chance of throwing $b$ was $1/6$ and A's throws are independent of B's throw,

the event "B throws $b$ and A loses" has a probability of $1/6\times (b/6)^2 = b^2/6^3.$

This is why computing the chance of a loss is simpler than the chance of a win: all we need do is multiply.
All that remains to do is sum these probabilities, because these events exhaust all the possible outcomes and are non-overlapping.  The answer therefore can be expressed as
$$\Pr(\text{A loses}) = \frac{1}{6^3}\left(1^2+2^2+3^2+4^2+5^2+6^2\right) = \frac{91}{216}.$$
Do not lose sight of the question: it asks for the chance that A wins.  That, of course, is the complement of the chance of losing,
$$\Pr(\text{A wins}) = 1 - \frac{91}{216} \approx 57.87\%.$$
