Three Cornered Duel -- Different Solution?

Question

So there is the famous Three-Cornered Duel problem from the book "Fifty Challenging Problems in Probability with Solution" by Frederick Mosteller. It was answered in the thread provided but I approached it in a different way and reached the same answer. I was wondering if this is a correct approach.

First of all, the question and answer (quoted from the book):

The problem

A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be?

Mosteller's solution:

A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly shoots the more dangerous C first, and A gets one shot at B with probability 0.3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits B. Then C and A shoot alternately until one hits. A's chance of winning is $$(.5)(.3)+(.5)^2(.7)(.3)+(.5)^3(.7)^2(.3)+…$$ Each term corresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get $$(.5)(.3)+\{1+(.5)(.7)+[(.5)(.7)]^2+… \}= \frac{(.5)(.3)}{1-(.5)(.7)}= \frac{0.15}{0.65}= \frac{3}{13} < \frac{3}{10}$$ Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck.

My Approach

All these answer think of the probability of winning given if A succeeds or fails at who they target. The stack exchange thread linked solves for $P(A_{win} | A \ hits\ B)$ and $P(A_{win} | A \ misses\ B)$.

Shouldn't we be thinking in terms of $P(A_{win} | A \ targets\ B)$? It's the same logic more or less, but in this case we take into account the probability that A can both hit and miss B vs. the probability that he just waits until B kills C and then tries to kill him directly.

Some probabilities I will use for the below proof (from Mosteller's book)

$P(AvsC) = P(A_{win} | \text{infinite shootout with C}) $

$P(AvsC) = P(C_{miss}) * (P(A_{hit}) + P(A_{miss})P(AvsC))$

solving for $P(AvsC)$, we have $P(AvsC) = \frac{3}{13}$

So that would be:

                            -- P(hits B AND AvsC) = P(hits B) * P(AvsC) = 0.3 * 3/13 = ~0.069
                           /
                          /
P (A_win | A targets B) =
                          \
                           \
                            -- P(miss B AND hits B the 2nd try) = P(A_miss) * P(A_hit) =0.7 * 0.3 = 0.21

So if A targets B on his first shot, he will win with probability $\approx 0.279$

But if he just fires at the fround, he will hit him the second time around with probability = 0.3

So he should fire at the ground.

Is that a valid approach? Isn't that a better approach? I get that Mosteller's Solution basically normalizes for A firing at B, but isn't that a wrong approach? I dunno... my logic just makes more sense to me. I'd never be able to think like the book solution if I were given this problem with no answer.... trying to learn new ways to think!