A three-way duel (probability puzzle)

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 they all agree to shoot int the air, their survival probability is $1$. They will all run out of bullets. Implicit in the problem is that only one can survive, but there is no mechanism to make sure the duel terminates.