Basketball - two players with no turns 
Two players are shooting free throws. The game ends when one of them
  scores and the other one misses, and the winner is the one who scores.
  Suppose player A has a probability of $p$ to score and player B has a
  probability of $q$. All throws are independent.
  
  
*
  
*What is the probability that player A wins?
  
*What is the probability that player A wins, given he scored in the
      first round?
  

As for part 1, I denoted $X,Y$ as the random variables for the final score of  players A,B, respectively. So according to the description of the game, A wins iff $X-Y=1$, thus:
$\Pr (X-Y=1)=\sum_{k=0}^{\infty}\Pr (X=k+1,Y=k)\\=\sum_{k=0}^{\infty}\Pr (X=k+1)\Pr(Y=k)=p\cdot \sum_{k=0}^\infty(pq)^k = {p \over 1-pq} $
As for part 2, it doesn't seem that conditioning $\Pr (X-Y=1|X\geq1)$ would lead to a different result.
So I guess my approach is problematic - I can't help to think that I'm not considering the rounds where they both missed. 
On the other hand, any method in which I tried to incorporate the total number of games as a geometric variable with prob. $p(1-q)+q(1-p)$ evantually turned out to be very messy.
Would appreciate some guidance.
 A: 1)
Think of it as if they keep on going also if someone has won already.
Let $D_n$ denote the event that at the $n$-th round both players score or both players do not score.
Let $A_n$ denote the event that at the $n$-th round $\mathbf A$ scores and $\mathbf B$ does not.
Let $A$ denote the event that $\mathbf A$ wins the game.
Then: $$A=A_1\cup(D_1\cap A_2)\cup(D_1\cap D_2\cap A_3)\cup\cdots$$
So that: $$P(A)=P(A_1)+P(D_1)P(A_2)+P(D_1)P(D_2)P(A_3)+\cdots=$$$$P(A_1)[1+P(D_1)+P(D_1)^2+\cdots]=\frac{P(A_1)}{1-P(D_1)}$$
Note that this equals $P(A_1\mid D^{\complement}_1)$ so matches with the hint of lulu.
More directly you can argue that: $$P(A)=P(A_1)+P(D_1)P(A)$$leading to the same result.
(If $\mathbf A$ does not win in the first round then with probability $P(D_1)$ he gets a second (and equal) chance on winning.)
2)
Here we find likewise:
$$P(A\mid \mathbf A\text{ scores at first round})=$$$$P(\mathbf B\text{ does not score at first round})+P(\mathbf B\text{ scores at first round})P(A)$$
Where $P(A)$ has been calculated at 1).
A: Here is a related scenario that will help you think about your problem. A is the probability that player 1 will win a particular round of a game, B is the probability that player 2 will win a round, and C is the probability that the two players tie and have to repeat the round.  Then the probability that player 1 wins the round is the infinite series: $(A + AC + AC^2 + AC^3 + ...)$.  So factor A out and get $A(1 + C + C^2 + ...)$.  For 0 < C < 1, the infinite series $(1 + C + C^2 + ...)$ = $\frac{1}{1 - C}$.  
The first question of your problem is worded slightly differently, but it is still the same.  You just have to compute A based on P and Q.  You have to compute C based on P and Q.
For the second question, let [Ans] be the answer to the first question.  Then the answer to the second question is (1 - Q) + Q*[Ans].  I.e. (1 - Q) being the probability that player 2 misses and the game is decided.  Q*[Ans] being the probability that player 2 makes his shot, and from there, the probability of victory from this point forward is the same as before.
