A and B are involved in a duel. The rules of the duel are that they are allowed to pick up their guns simultaneously. If one or both are hit, the duel ends. 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 probabilty $p_A$ and each shot B will hit A with probability $p_B$.
d) The conditional probability that the duel ends after the nth round of shots given that A is not hit.
e) The conditional probability that the duel ends after the nth round of shots given that both duelists are hit.
I know the question has already been asked here Duel probabilty question but I'm still confused with the answers:
Firstly, it's easy to compute that $P$(A is not hit)$=\frac{p_A\,(1-p_B)}{1-(1-p_A)(1-p_B)}$
So that $P$(d)=$P$(the duel ends after the nth round|A is not hit)*$P$(A is not hit)$=(1-p_A)^{n-1}(1-p_B)^{n-1}(1-p_B)p_A/\frac{p_A\,(1-p_B)}{1-(1-p_A)(1-p_B)}=(1-p_A)^{n-1}(1-p_B)^{n-1}(1-(1-p_A)(1-p_B))$
Similarly, we have $P$(e)=$P$(the duel ends after the nth round|both are hit)*$P$(both are hit)$=(1-p_A)^{n-1}(1-p_B)^{n-1}p_Bp_A/\frac{p_A\,p_B}{1-(1-p_A)(1-p_B)}=(1-p_A)^{n-1}(1-p_B)^{n-1}(1-(1-p_A)(1-p_B))$
My confusion is: how's $P$(d)=$P$(e), what's more, they both equal to $P$(the duel ends after the nth round).