Challenging a duel which favors neither shooter. 
A and B have challenged each other to a duel. They will take turns shooting all one another until one has been hit. A who can hit B only $q\,\%$ of time is the weaker shot so will be allowed to go first. They have determined that this duel favors neither shooter. What is the value of $q$?

As it says "this duel favors neither shooter" I thought that the probability that A can win in this duel is $\frac12$, and so is for B. So I wrote this equation that the probability that A will hit is q at first attempt, than plus $(1-q)q$ (here $1-q$ is the probability that I will not hit, and $q$ is the probability that B will not hit too).
$$\frac12 = q + (1-q)q$$
Solving this has got 
$$q_1 = \frac{2 + \sqrt 2}2 $$
$$q_2 = \frac{2 - \sqrt 2}2 $$
cause first one is greater than 1, I have picked as an answer the second one. 
Is this a correct solution or not? 
 A: Yet another approach, avoiding the geometric series:
Let $q$ be the probability that $A$ hits, and $p$ the probability that $B$ hits.
The following mutually exclusive events are possible:


*

*A hist on first shot

*A misses and then B hits

*A and B both miss their first shots


The first occurs with probability $q$ and leads to a win for $A$. The second occurs with probability $(1-q)p$ and leads to a win for $B$. The third occurs with probability $(1-q)(1-p)$ and is followed by a sequence that is equivalent to the original full duel. We know that in this case, $A$ wins with probability $\frac12$.
Hence $A$ wins the original duel with probability
$$ q\cdot 1+(1-q)p\cdot 0+(1-q)(1-p)\cdot \frac12$$
and again, we know that this is $\frac12$.
So we solve $$q+(1-q)(1-p)\cdot \frac12=\frac12$$
for $q$ and find
$$q=\frac p{1+p}.$$
A: As @HagenvonEitzen pointed out in the comment, your solution cannot be correct as if B is a perfect shooter then $q=1/2$. Actually you cannot have a unique numeric value which is independent of the hit rate of B.
So let $q$ be the hit rate of B and $p$ be the hit rate of A. You want to have 
$$ \frac{1}{2} = q + (1-q)(1-p)q + (1-q)^2(1-p)^2q + ... = q\sum_{i=0}^\infty (1-q)^i(1-p)^i$$
where $(1-q)$ is the probability that A misses and similarly $(1-p)$ is the probability that B misses. The sum on the right hand is a geometric sum so you have
$$ \frac{1}{2} = \frac{q}{1-(1-p)(1-q)}$$ 
hence solving for $q$
$$ q = \frac{p}{1+p}$$
Notice that if $p=1$ then you get $q=1/2$ as expected.
A: The probability that A hits B successfully is 
$$\begin{align}P(A) &= q + (1-q)(1-p)q+(1-q)^2(1-p)^2q+... \\
P(A) &= \frac{q}{1-(1-p)(1-q)}\end{align}$$
The probability that B hits A successfully is
$$\begin{align}P(B) &= (1-q)p + (1-q)^2(1-p)p+(1-q)^3(1-p)^2p+... \\
P(B) &= \frac{(1-q)p}{1-(1-p)(1-q)}\end{align}$$
Here I assumed the probability of B hitting A to be $p$. Equating both probabilities, we get
$$\begin{gather}P(A) = P(B) \\
q = (1-q)p\\
q = \frac{p}{1+p} \end{gather}$$
