probability - A basic game of "roll-off", finding probabilities of events

A and B play a "roll-off". The rules of the game are simple: they'll each roll the dice (which is a standard, fair, six-sided dice with faces numbered 1 to 6), and whoever obtains the higher number wins. In the event of a tie, they'll both roll again, until there's a winner.

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a. What is the probability that B wins the game on the first roll? (Give your answer as a fraction in its lowest terms.)

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Here we want whatever player B gets to be greater than whatever player A gets. So we want $\Pr(A>B)$ how do I find this though?

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b. What is the probability that A wins the game, but not on the first roll? (Give your answer as a fraction in its lowest terms.)

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Here I want $\Pr(A=B)*\Pr(A>B)$ but again I'm not really sure how to find it.

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c. Suppose A rolls first, and gets a '2'. At that moment, what is the probability that B will win the game? (Give your answer as a fraction in its lowest terms.)

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Here I want to find \Pr((A>B)|(B=2)) and you guessed it, I don't really know how to.

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Answer

(a) In this case you can just enumerate the outcomes:

$$(2,1),(3,1),(4,1), (5,1), (6,1), \ldots$$
There are $15$ outcomes with $B>A$ and $36$ possible outcomes from rolling two dice, so the proability is $$\frac{15}{36} = \frac 5{12}.$$

(b) It's clear by symmetry that both players have an equal chance to win. So this probability is $1/2$ times the probability that the first roll is a tie (since the first roll must be a tie in order for $A$ to win but not on the first roll). So the probability is

$$\frac12\cdot\frac16=\frac1{12}.$$

(c) The probability that $B$ wins on the first roll, given that $A$ rolled a 2, would be $4/6=2/3$ (since $B$ could roll $3$, $4$, $5$, or $6$). We also need to add to this $1/2$ times the probability that the first roll is a tie (which is $1/6$, i.e. when $B$ rolls a $2$ on the first roll). So the probability is

$$\frac23 + \frac12\cdot\frac16=\frac34.$$