Brinksmanship: how to achieve the best outcome by a single statement 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 randomly 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."
 A: If only one contender is allowed to make a statement, we have the following Theorems:
Theorem 1: If $N\gt 3$, then survival probability for that contender must be strictly less than $\frac{N-1}N$.
Proof: 
WLOG, suppose only player $1$ can make a statement and players $2$~$N$ must keep silent. Notice that by killing player $1$ in their turns, players $2$~$N$ will bring the game to a subgame where $N-1$ players are alive and no one has made any statement. In this subgame each player has $\frac1{N-1}$ survival probability. So in the original game players $2$~$N$ each have a survival probability at least $\frac1{N(N-1)}$. This means player $1$ has survival probability strictly less than $1-\frac{N-1}{N(N-1)}=\frac{N-1}N$.  QED 
Theorem 2: For $N=4$, Theorem 1 gives the least upper bound, i.e. $\forall \varepsilon\gt 0$, there's a statement for player $1$ such that his surviving probability $P_{1}^{4}=3/4-\varepsilon$.
Proof: Similar as the proof for Theorem 3. 
Theorem 3: For $N=5$ we can prove there's a statement for player $1$ such that $P_{1}^{5}=3/5-\varepsilon$, $\forall \varepsilon\gt 0$. The statement for player $1$ will be too long to put into actual words, but it works as below.
Proof: 
Name the players $A_i$ for $i=1$ to $5$. $A_1$'s strategy is to persuade whoever is currently shooting to shoot a player after him, by promising him a surviving probability of $\frac1{M-1}+\varepsilon$, until only $3$ players remain, where $M$ is the number of players alive. When players are reduced to three, $A_1$ can then allocate the probabilities he promised to the other two players. There're 5 possibilities corresponding to who's chosen to shoot first by the referees. We'll illustrate with the case where $A_2$ is first to shoot. 
$A_1$ will have stated that in this case, $A_2$ must kill $A_3$, after that, $A_4$ must kill $A_5$. For this to happen, he must promise them surviving probabilities $\frac1{4}+\varepsilon$ and $\frac1{3}+\varepsilon$ respectively, otherwise they will be better off killing $A_1$ (if promised probabilities are less than $\frac1{4}$ and $\frac1{3}$), or by Rule 3 randomize (if promised probabilities are equal to $\frac1{4}$ and $\frac1{3}$). The promised probabilities can be realized in the three way duels between $A_1$, $A_2$, $A_4$ by some appropriate commitment such as this: "$A_2$ shall shoot at $A_3$ with $\frac1{4}+\varepsilon$ probability of killing. If he kills, I'll miss and let him kill me too; if he misses, $A_3$ must kill him, then I'll shoot at $A_3$ with $\frac{\frac1{3}+\varepsilon}{\frac3{4}-\varepsilon}$ probability of missing. I shall kill anyone who first violates this proposal at my first opportunity." This leaves $A_1$ with surviving probability $\frac5{12}-2\varepsilon$.
Other 4 possibilities are calculated similarly, yielding surviving probabilities for $A_1$ as $\frac5{12}-2\varepsilon$, $\frac3{4}-2\varepsilon$, $\frac3{4}-2\varepsilon$, $\frac2{3}-2\varepsilon$ corresponding to $A_3$, $A_4$, $A_5$, $A_1$ shooting first respectively. Summing them up and dividing the result by 5 yields $P_{1}^{5}=3/5-2\varepsilon$. QED 
