How to achieve the best outcome by a single statement in this game? This game is taken from Schelling's Game Theory: How to Make Decisions by R.V. Dodge, in which contenders practice brinksmanship to their own advantages. It goes as follows:

Anderson, Barnes, and Cooper are to fight a gun duel. They will stand close to one another, so that each can kill one of the others in one shot or deliberately miss. The first to fire will be chosen at random, and they will rotate in the order Anderson, Barnes, and Cooper, each firing one shot at a time.

If there is more than one survivor after a number of rounds, one of the contenders will be chosen at random and forced to kill one of the others, and this will be repeated if there is still more than one alive after a few more rounds.

Before the duel starts, Anderson may make any statement, followed by a statement from Barnes, and finally one from Cooper. They will adhere to the following rules:


Statements are irrevocable. A contender may not act to contradict his statement.
He will act in his own best interest when it does not conflict with Rule One.
He will act randomly when it does not conflict with Rules One and Two.


There are referees to ensure that the rules are followed. If a contender commits himself to a mixed strategy (for example, to miss with a probability of 1/3), the probability will be determined objectively (by tossing dice, etc.).

Q1: What statement will Anderson make? What's his best strategy and his probability of surviving?

Q2: If three contenders are to make their statements in the order of ACB, what would be the best statement for Anderson?

Q3: If there're more than three contenders, does this game become simpler or more difficult? Can we say anything about the $N$ contenders case for $N\gt 3$?



Notice that if no one makes any statement, no one will shoot, and everyone has a surviving chance of 1/3. If only Anderson is allowed to make a statement, he can guarantee near certain survival by making this statement to B and C: "If you don't kill each other at your first opportunities, I will kill the first of you who fail to do so at my first opportunity; otherwise, I'll shoot at the survivor of you with 1% chance of missing."

In the book, a suggested best solution for Q1 is for A to say:"If B does not commit to unconditionally shooting C, I will shoot him." The argument is that B has no choice but to accept, because refusing A's proposal would result in certain death for B. But that is clearly erroneous! Because B can say: "If C doesn't promise to kill A at his first opportunity, I will kill C at mine; if C does, I will kill A at my first opportunity and shoot at C with 1% chance of missing if he kills A first." By refusing B's proposal C has 2/3 chance to survive (1/3 for A shooting first and 1/3 for C shooting first); by accepting he has 2/3+0.33% chance (1/3 for A shooting first, 1/3 for B shooting first and 0.33% for C shooting first). So C will accept. Then A is doomed, bound by his own inconvenient statement.
A caveat, this is not really an answer, more of an extended comment--that's why I've put it as community wiki. This is a cool game, one of commitment.

I'll try to formalize the problem. This is a first pass, so people should edit as they see fit.

It's probably easiest to solve the case where, $T$, the number of rounds before a forced execution is $1$. Let each player get $1$ if they shoot someone else and $-1$ if they are themselves shot.

There are two stages of the game, the commitment stage and the shooting stage. The sequential stage is (fittingly) sequential move and the commitment stage is not really a game persay, since the players are automata (their strategies are predetermined). I'll return to this shortly.

An easy result: there is an equilibrium in which rules 2 and 3 are rendered superfluouous since each player can commit to every possible contingency. 

There is a source of exogenous randomness, a random state of the world $\Theta = \Theta_{1} \times \Theta_{2}$, $i = 1,2$, where $\Theta_{i}$ is the set of permutations of $\left\{A,B,C\right\}$. These are the possible orders in the shooting portion.

The set of actions for player $A$ is $\mathcal{A} = \mathcal{A}_{1} \times \mathcal{A}_{2}$ where $\mathcal{A}_{i} = \left\{B,C,\emptyset\right\}$ for $i = 1,2$ (corresponding to "shoot $B$," "shoot $C$," and "shoot nobody," respectively). The sets of actions for the other two players, $\mathcal{B}$ and $\mathcal{C}$, are defined analogously.

A (mixed) commitment strategy of $A$ is a mapping $$\sigma_{A}: \Theta \times \Sigma_{B} \times \Sigma_{C} \to \Delta\left(\mathcal{A}\right)$$
where $\Sigma_{i}$, $i = A, B, C$ is the set of all (feasible) mixed strategies. An example of an infeasible strategy for player $A$, say, is one that does not have $\emptyset$ played with probability $1$ when the state and mixed strategies of the other players are such that $A$ has been shot for certain.

Note that I am using the fact that the shooting portion of the game can be viewed as a instantaneous (random) outcome of the commitment portion, which is the real game.

I'm not quite sure about the commitment strategies for $B$ and $C$, due to the sequential move nature of the commitment stage. I suspect that that does not matter, and that they are defined analogously. I'm not completely sold on this and open to being convinced otherwise. 

Each player's payoff is defined in the obvious way.