Why is my intuition wrong that one of two archers win in a tournament? The probability of Robin and Tuck hitting a target respectively are $0.45$ and $0.38$.  For each round of the tournament, each archer must shoot simultaneously at their targets. A player wins the tournament if in a round, one archer hits the target while the other does not. If both or neither archers hit the target, then the tournament is continued with another round and this process is repeated. Intuitively, I feel that the probability that Robin wins the tournament is $p=\frac{0.45}{0.38+0.45}\approx0.5421686%$, but to be on the safe side, I did the below:

Consider the two cases in which Robin wins the tournament: 
Either a) Robin wins on his first shot, or b) Robin ties on his first shot but wins on the $n^{\text{th}}$ round, $n > 1$. 
Let $p$ denote the probability that robin wins.
Case a: If Robin wins on his first shot, that means he hit the target and Tuck missed the target. The probability of this is $(0.45)(1-0.38)=0.279$
Case b: The probability that no one wins on the first round is the probability that no one hits the target plus the probability that both hit the target, which is $(1-0.45)(1-0.38)+(0.45)(0.38)=0.512$. After the first round, when the second round starts, it is as if the entire tournament has just started. Therefore, after a tie in the $1^{\text{st}}$ round, Robin still has the same probability $p$ of winning on the $n^{\text{th}}$ round where $n>1$. Thus the probability that Robin ties the 1st round and wins some round after that is $0.512p$.
Adding cases a and b gets that $p = 0.279 + 0.512p$. Solving for $p$ yields around  $0.5717213115$.

What accounts for the discrepancy between my intuition and my reasoning above? Is my reasoning flawed?
 A: Take an example where the probability of Robin and Tuck hitting a target respectively are $0.9$ and $1.0$. Your method would give the probability of Robin winning being $\dfrac{0.9}{0.9+1.0}$ when the true probability is $0$.
Going back to your original probabilities of hitting of $0.45$ and $0.38$, you would do better saying the probability of Robin winning overall might be $\dfrac{0.45\times(1-0.38)}{ 0.45\times(1-0.38) + 0.38\times(1-0.45)}$ by looking at the decisive and mutually exclusive events of one hitting and the other not. This gives your $0.5717\ldots$
A: Whatever the probability distribution of the number of rounds, on any round Robin wins with probability $\propto r\bar t$ and loses with $\propto \bar rt$, hence the ratio
$$\frac{r\bar t}{r\bar t+\bar rt}=\frac{0.45\cdot0.62}{0.45\cdot0.62+0.55\cdot0.38}\approx0.572$$ as the game ends with probability $1$.
A: The easiest way to get there is to say the chance that exactly one of them hits the target is $0.45(1-0.38)+(1-0.45)0.38=.488$  The chance that Robin wins given that somebody did is $\frac {0.45(1-0.38)}{0.488}\approx 0.57172$ which matches your second calculation.  
The error in your intuitive calculation is that Tuck loses more of his hits to ties because Robin is more accurate, so his chance of winning should be less than $1-0.542$
A: I thought visual illustration might be helpful. 

