A three-way duel (probability puzzle) This puzzle is taken from Mathematical Puzzles: A Connoisseur's Collection [Peter Winkler]. I don't understand the solution.
Alice, Bob, and Carol arrange a three-way duel. Alice is a poor shot, hitting her target only 1/3 of the time on average. Bob is better, hitting his target 2/3 of the time. Carol is a sure shot. They take turns shooting, first Alice, then Bob, the Carol, then back to Alice, and so on until one is left. What is Alice's best course of action?

The solution is that Alice is better of missing than hitting Carol or Bob, so she should shoot into the air. Indeed, then Bob will shot Carol, and it can be shown that it gives the greatest probability of survival for Alice. But I wonder if Bob should not voluntary shoot into the air too, so that Carol will do the same, and no one be shot. If this is the case, Alice survival probability is 1. What do you think of it? What is Alice survival probability?
To maximise their chances the duellists prefer to be left with a weaker opponent. So Bob would not shoot at Alice in preference to Carol, and Carol will not shoot at Alice in preference to Bob. Therefore Alice will not be shot at until Bob or Carol is dead and she will either be left standing with Bob or Carol, with or without the shot.

Probability of Alice, with shot, surviving against Bob is given by:

p = Pr(A hits B) + Pr(A misses B) * Pr (B misses A) * p
p = 1/3 + 2/3 * 1/3 * p
p = 3/7

Probability of Alice, without shot, surviving against Bob is given by:

p = Pr(B misses A) * (Pr(A hits B) + Pr (A misses B) * p)
p = 1/3 * (1/3 + 2/3 * p)
p = 1/7

Probability of Alice, with shot, surviving against Carol is given by:

p = Pr(A hits C) + Pr(A misses C) * Pr (C misses A) * p
p = 1/3 + 2/3 * 0 * p
p = 1/3

Probability of Alice, without shot, surviving against Carol is given by:

p = Pr(C misses A) * (Pr(A hits C) + Pr (A misses C) * p)
p = 0 * (1/3 + 2/3 * p)
p = 0

So, her probability of surviving from each position is:

Bob, with shot: 3/7
Carol, with shot: 1/3
Bob, without shot: 1/7
Carol, without shot: 0

So Alice is best off not killing anyone since the advantage she gains by having the first shot exceeds any possible benefit of facing Bob rather than Carol. She should shoot into the air.

Given that Alice is neither going to shoot at them, or be shot at by them until one is dead, Bob and Carol are essentially in a two person duel, the winner to face Alice. They cannot improve their chances by forgoing a shot, so they shoot. Bob wins that 2/3 of the time, Carol 1/3.

Alice wins 2/3 * 3/7 + 1/3 * 1/3 = 25/63.
Bob wins 2/3 * 4/7 = 24/63.
Carol wins 1/3 * 2/3 = 14/63.
If Alice shoots in the air and under the assumption that Bob shoots Carol than Alice has a huge probability to survive the game.

BUT:
However if Bob shoots himself in the second round then Alice is basically death and Carol has a higher probability to win this game. So under the assumption that carol want to survive, she should shoot in the air. Under this thought I think your solution is much more justified.

If all the others shoot in the air and continue until infinity(I cannot see a restriction for bullets). Then thats her highest probability.