Let's suppose 8 players are playing a deadly game. Each of them has a revolver with 2 bullets and you know for a fact that organizer made all of them spin the bullet chamber for at least 30 seconds, there is NO way for a player to cheat. Also, no teaming is possible because players don't know each other and even can't speak to each other, not even a single word or a hand signal.
Each player can aim at anyone. This is where it gets tricky. Let's say we are number 8. and this is what we see:
Who should we aim for? Obviously, our goal should be to manipulate and aim our gun in a way so that it makes it eliminate as much as players in a single round, right? Lets look at number 2. Odds of him surviving is 2/3 * 2/3 * 2/3 = %29,629... That guy is probably dead, but if we aim at him too, we can be almost sure that he is a goner. *2/3 * 2/3 * 2/3 * 2/3 = %19,758...*Even a worse odd. A clean kill. One guy out of equation
But! If we aim at player 5, game has chance of eliminating 2 players in a single round, thus doing an ultimately good thing for us in the long term. Right now, number 5 has 2 players on his neck, *2/3 * 2/3 = %44,444... * If we aim at him too, his odds shrink to 2/3 * 2/3 * 2/3 = %29,629
Maybe? We can aim at number 7? Or number 3?
You get the idea. My question is, in this scenario, which strategy is the optimal one for us, number 8?
Note: Players pull the trigger at the exact millisecond, think of it like robots.
If the objective is to maximize the expected number of eliminations in a single round, the shot should be directed towards the person with the "maximum marginal probability of being shot". With that, I mean: the person who adding one more target increases his total probability of being shot the most.
Let $I$ denote the set of player indexes. The probability $P(n_i)$ of any player $i \in I$ being shot, as a function of the number of targets $n_i$ directed at him is $$P(n_i) = 1 - p^{n_i}$$ where $p = 5/6$ in our example (the probability of not being shot after a single target). By concavity of this expression, which is bounded above by one, the marginal probability of being shot is decreasing in the number of targets. Therefore, the best target would be the individual with the least amount of shots directed at him, $\arg\min_{j \in I} n_j$.
If the types of all players are equal, then the unique pure strategy Nash Equilibrium is one where every player is only targeted by one player. However, there are other mixed strategy Nash Equilibria. Let $\mu_i(j)$ be the probability that player $i$ will target player $j$. Then a family of mixed strategy Nash Equlibria will be given by: $$\forall i \in I, \mu_i(j): \sum_{i\in I} \mu_i(j) = 1$$ $$\sum_j \mu_i(j) = 1$$
Notice that the pure strategy Nash Equilibrium is a particular case of this more general family of mixed strategies.