A and B are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability $p_{A}$, and each shot of B will hit A with probability $p_{B}$.
(Book: Sheldon, Ross M. A First Course in Probability, Problem 3.63, p.107.)
Now I was wondering what the probability that the final result is that A hits B at all. Thus, letting $A$ be the event that A hits B at all, it is clear that we can condition on the number of rounds that occur after someone is shot. Letting $N_{n}$ denote that neither is hit on the nth round,
$$ P(A) = \sum_{n=1}^{\infty}P(N_{1}...N_{n-1})p_{A}=p_{A}\sum_{n=1}^{\infty}(1-(1-p_{A})(1-p_{B}))^{n-1}$$ $$=\frac{p_{A}}{1-(1-p_{A})(1-p_{B})}$$
Similarly, if $B$ is the event that B hits A at all,
$$P(B)=\frac{p_{B}}{1-(1-p_{A})(1-p_{B})}$$
And the probability that both shoot each other is
$$P(AB)=\frac{p_{A}p_{B}}{1-(1-p_{A})(1-p_{B})}$$
Now, it is surprising to me that the events $A$ and $B$ are not independent. Namely,
$$P(AB)=\frac{p_{A}p_{B}}{1-(1-p_{A})(1-p_{B})}\neq\frac{p_{A}p_{B}}{(1-(1-p_{A})(1-p_{B}))^2}=P(A)P(B)$$
The fact that B does or doesn't shoot A should not influence the probability that A shoots B, right? What is the intuitive explanation that the events $A$ and $B$ aren't necessarily independent?
P.S: A question about the same problem in Ross's book, but doesn't answer mine: Dual probability question
Wrong, unless either shooter's shot is sure to hit (or sure to miss).
If the duel ends without B having shot A, then A will certainly have shot B to end the duel.
$\qquad\mathsf P(A\mid B^\complement)=1$
If the duel ends with B having shot A, then A may only have shot B on the final turn.
$\qquad\mathsf P(A\mid B)=p_A$
Thus, without invoking infinite series:
$\qquad\begin{align}\mathsf P(A)&=\mathsf P(A\mid B^\complement)\,\mathsf P(B^\complement)+\mathsf P(A\mid B)\,\mathsf P(B)\\[1ex]&= 1\cdot(1- \mathsf P(B))+p_A\,\mathsf P(B)\\[1ex]&= 1-(1-p_A)\mathsf P(B)\\[1ex]&=1-q_A\,\mathsf P(B)&&\text{letting }q_A:=1-p_A, q_B:=1-p_B\\[2ex]\mathsf P(B)&=1-q_B\,\mathsf P(A)&&\text{symmetrically}\\[1ex]&=1-q_B-q_Aq_B\mathsf P(B)\\[1ex]&=\dfrac{1-q_B}{1-q_Aq_B}&&\text{if the denominator is non-zero}\\[1ex]&=\dfrac{p_B}{1-(1-p_A)(1-p_B)}\\[2ex]\mathsf P(A)&=\dfrac{p_A}{1-(1-p_A)(1-p_B)}&&\text{symmetrically}\end{align}$