Circle of prisoners I do not know if the following game is well-defined:
Suppose that there are $n$ prisoners standing in a circle one after other $n\geq 2$ everyone holding a silent gun.
When a bulb lights all either kill the one standing before him or not in some time interval of a few seconds. You are the $1st$, w.l.o.g.W hat are all Nash Equlibria of this game on $n$ players the strategy for each $p\in[0,1]$ killing the one before you with probability $p$. If $n=2$ the one NE is $(1,1)$ i.e. hoping that you will kill your enemy first. If $n=3$ it is good not to kill the one in front of you hoping that she will kill the one behind you. So I think one NE is $(0,0,0).$ What are all NE in general for $n$ arbitrary (in mixed strategies) ?
If the definition of the game leaks somewhere someone may fix it.
EDIT
A less macabre way:
All $n$ free people have hats on their heads. Everything goes like above but with taking off their heads instead of shooting. If your hat remains you win 1$. If someone has n hat she is not allowed to take off the hat in front of him.
 A: I’m not sure the discussions in the comments have really clarified the time element. I’ll interpret the question to mean that the players shoot at different times and the order in which they shoot is random. The exact distribution of the possible shooting orders is irrelevant as long as there’s a non-zero probability for each of the $n$ orders where the shooting goes around the table in the “forward” direction.
Everyone not shooting is always a Nash equilibrium, as nothing is to be gained if no one else shoots. In fact, any strategy profile in which two consecutive players don’t shoot is a Nash equilibrium, as neither of them can gain from shooting instead (one because she won’t get shot anyway, and the other because his target won’t shoot anyway) and no one can influence their own chance of being shot if these two break the chain.
For odd $n$, the strategy profiles in which a single player shoots with probability $0$ are also Nash equilibria, since that player would only be making it worse for themselves (making it worse? how could it be worse?!) if they increase that probability, and no one else can influence their own chance of being shot because of the break in the chain. If all players shoot with non-zero probabilities, they can all gain from lowering their probability, so such a strategy profile isn’t a Nash equilibrium.
However, for even $n$, a single player with probability $0$ gains by increasing that probability, and once they close the chain, everyone gains from increasing their probability, so in this case the only Nash equilibrium in addition to the ones with two consecutive non-shooters is the one where everyone shoots with probability $1$.
