Dice rolling game Grallace and Womit are debating who should get to eat the last piece of Wensleydale in the fridge. To decide the issue, they agree to have a "roll-off", using the dice from their old MouseTrap set. 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.
a. What is the probability that Womit wins the game on the first roll? (Give your answer as a fraction in its lowest terms.)
b. What is the probability that Grallace wins the game, but not on the first roll? (Give your answer as a fraction in its lowest terms.)
c. Suppose Grallace rolls first, and gets a '$2$'. At that moment, what is the probability that Womit will win the game? (Give your answer as a fraction in its lowest terms.)
Here is my attempt, have no idea if they are correct so please help.
a. The only way W can win is if he doesn't get a $1$ and there are no draws. So if he gets a 2, then G has to get a 1. The chances of these are 1/6*1/6=1/36
If W gets a 3 then G has to get a 1 or 2 so 1/6*2/6=2/36
W gets 4, then G gets 1 or 2 or 3 so 3/36
W gets 5, then 4/36 and if W gets 6, then 5/36. Add all these up gives 15/36. 
b. So the first game has to be a tie which is 1/6 chance. Then since W winning is 15/36, then G winning is 11/36. So 1/6*11/36=11/216.
c. If G gets 2 (which is 1/6 chance), then for G to lose, W has to get 3 or 4 or 5 or 6 (which is 4/6 chance) so 1/6*4/6=2/3
Feel free to chance the tags because I wasn't sure which one to put.
 A: $a)$ Your answer is correct but you can get it easier. Out of the 36 cases, 6 cases are a tie, 15 G wins, 15 W wins (by symmetry). So the answer is $$\frac{15}{36}=\frac{5}{12}.$$
$b)$ The first roll is a tie with probability $\frac{1}{6}$. After that both have the same chance so the answer is $$\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.$$
Another way of seeing this is to compute the probability that G wins in round $i$. We must have a tie in rounds $1,2,\dotsc,i-1$ and G winning in round $i$:
$$\left(\frac16\right)^{i-1}\frac5{12}.$$
Now we sum this up for $2\le i\le \infty$ to get the probability that G wins but not in round 1:
$$\sum_{i=2}^\infty \left(\frac16\right)^{i-1}\frac5{12}=\frac1{12}$$
$c)$ Suppose G gets a $2$. The probability of a tie is then $\frac{1}{6}$ and after that W wins with probability $\frac{1}{2}$. So the probability that W wins and the first round is a tie is $\frac{1}{12}$. The probability that W wins without a tie is $\frac{4}{6}=\frac{2}{3}$. So in total the probability of W winning is $$\frac{1}{12}+\frac{8}{12}=\frac{9}{12}=\frac{3}{4}.$$
Note that here we are looking for a conditional probability given that G gets a 2. This is already given as something that has happened. So we do not need to consider its probability in our calculations.
A: The answer to a is correct, though the calculation can be simplified: after taking away the 1/6 probability of a draw on the first round you can divide the remaining probability equally between the W and G. And 15/36=5/12.
The answer for b can be deduced from similar considerations. The probability of having more than one round is 1/6, of all games longer than 1 G wins 1/2, which is 1/12. What you solved is for W not to win on the second roll, but that does not mean G wins.
The probability you gave in c is for W to win on the first round, but you neglected the possibility of a tie and W winning later.
