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?

  • $\begingroup$ If Bob shoots in the air, then Carol will simply shoot him, and get a "win" probability of 2/3 (and Alice will have a win probability of 1/3, getting just one more chance to kill Carol). $\endgroup$
    – vadim123
    Feb 23 '14 at 16:03
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    $\begingroup$ It is not a given that Carol would shoot Bob then, depending on how much she values winning vs surviving. $\endgroup$
    – Arno
    Feb 23 '14 at 16:29
  • $\begingroup$ In the original problem, the goal is to win the duel, not to survive, which is why you get a different answer. $\endgroup$
    – DanielV
    Feb 23 '14 at 16:46

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.

  • $\begingroup$ Ok, but I think that her probability of surviving vs Bob with shot is 1/2 : if p is this probability and q the probability of surviving vs Bob without shot, p=1/3+2q/3, q=p/2, hence p=1/2 and q=1/4. $\endgroup$
    – Papagon
    Feb 25 '14 at 19:02
  • $\begingroup$ Why does q=p/2 though? I've added my calculations to the answer. $\endgroup$
    – Neil W
    Feb 25 '14 at 23:38
  • $\begingroup$ Ok, sorry I made my calculations with Bob success probability 1/2 and not 2/3. $\endgroup$
    – Papagon
    Feb 26 '14 at 10:51
  • $\begingroup$ @NeilW In your calculations of probability, why do you multiply by p on the right hand side of every equation? e.g. p = Pr(A hits B) + Pr(A misses B) * Pr (B misses A) * p. Why * p at the end? $\endgroup$
    – James
    Nov 20 '15 at 18:48
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    $\begingroup$ @James because after A misses B and B misses A we are back to the original position, so the probability of Alice surviing from that point is the same as was originally, p. $\endgroup$
    – Neil W
    Nov 21 '15 at 3:52

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.


The relevant question is whether Carol should shoot Alice, Bob or the air.

Its clear that shooting Alice while Bob is still alive is a bad idea. Shooting Bob means that Alice will attempt to shoot Carol next, and if Alice misses, Carol can kill her and win.

So Carol gets expected payoff $\frac{2}{3}W + \frac{1}{3}D$ in this course of action, where $W$ is her payoff for being the sole survivor, and $D$ her payoff for dying. If this number is greater than her payoff $S$ for a stalemate, she'd shoot Bob, which then means that Bob would attempt to kill her first.

So, let us assume that Carol is sufficently averse to dying compared to winning that Carol would prefer a stalemate to being shot at by Alice. Can we get a Nash equilibrium here?

Alice's strategy will be to fire in the air until someone else dies, and then shoot at the survivor. Carol's strategy will be to shoot in the air until either someone else dies, or someome shoots at her unsuccessfully, in which case she'll kill the perpetrator.

What does Bob do? Killing Alice is suicide, so either he shoots in the air, and we get the conjectured stalemate, or he tries to shoot Carol. If he does fail, he dies. If he succedes, Alice and he will keep taking turns until someone dies, giving him a risk of more than $\frac{1}{3} + \frac{2}{9}$ of dying. Depending on how Bob values winning vs surviving, there could be a stalemate-equilibrium or not.


Arno proved that if Alice, Bob and Carol want to survive, they will go to a slatemate, so as Daniel V points out, the goal should be to win the duel and not to survive. But assuming that the goal is to win the duel, if Alice and Bob shoot in the air, why Carol could not shoot in the air too? Because she will have to shoot Bob anyway. Indeed, if they keep shooting in the air, when Carol has two bullets left, she has to shoot someone because she prefers having a chance to win rather being sure not to win. As a matter of facts, if Bob does not shoot her dead in the begining, she will shoot him. For Alice, the probability to win a duel against Bob when it's her turn to shoot is 1/2, and the probability to win a duel against Carol is 1/3, hence her probability to win is 5/12.

  • $\begingroup$ Yes, their goal is to win, suppose they have deadly hates with each other and just want to make the others dead. $\endgroup$
    – fizis
    Oct 27 '16 at 2:55

Alice should miss because:

  1. Should Alice actually kill Carol OR Bob, then Alice is immediately the only remaining target.

  2. If she misses, Bob will surely target Carol. If Carol dies, it’s Alice’s turn again. If Bob misses, Carol will shoot at Bob, and it’s Alice’s turn again.

No math required. She should miss so the others can shoot at each other.


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.


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