Probability - Tales Game A dice game played by two players is like this: each player throws two dice and sum their results; that is the number of points the player scored. Whoever scores more, wins.
One additional detail is that if the numbers of both dice of a player are equal, the player can roll the two dice again and the sum of these points will be added to the previous sum - and so on, indefinitely.
  A) A player has k points. Calculate his probability of victory.
  B) A group of friends decided to play the same game with n players.
       Find the winning probability for a player who scored k points.

I've tried for some time to do this, but it seems impossible to me. I don't know much of this kind of probability. Does anyone know a way to solve this problem?
 A: When you throw 2 dice, the sum is one of the numbers $2,3,\dots,12$. Do you know the probability of each of these sums? If you do, then you can figure out the probability of winning if you get $k$; it's the probability that your opponent gets a number less than $k$, plus half the probability that your opponent gets exactly $k$. 
EDIT: For example, suppose you roll a 4. You win if your opponent rolls a 2 or a 3; the total probability of that is 1/12 (do you know why?). With probability 1/12 your opponent rolls a 4, and in that case, your probability of winning is 1/2, since you are essentially back to square one, and it is a symmetric (thus, fair) game. So your total probability of winning if you throw a 4 is $$(1/12)+(1/2)(1/12)=1/8$$  
MORE EDIT: Note that the question has been edited several times since I posted the above, so the above may not speak to the current version of the question. 
A: The probability of winning is $1-p(k)$, where $p(k)$ is the probability of losing, given that the other player rolled $k$. If a player rolled k, it is enough to compute all possible ways one can roll a total that is less than $k$, and there are finitely many possibilities, which can be easily computed.
