Shooting with Probability, Game Theory Two people are standing in front of each other in a rail with distance of $2$ meters. Player $A$ stands on point $-1$ and Player $B$ stands on point $1$.
They each have only one gun with one bullet.
Player A can fire from points -1 or -0.5 or 0 (only if alive by then). Also, Player B can fire on points 1 or 0.5 or 0 (only if alive by then).
Definition: $<a, b>$, $a$ is the place of Player $A$ and $b$ is the place of Player $B$.
First of all they are at $<-1, 1>$, everybody can have shot. The shot will hit the other player with some probability, for instance $1/8$.
Second, if both of them are still alive, they are at the place of $<-0.5, 0.5>$ and they can take another shot. It means that the rail moves them to the place of $<-0.5, 0.5>$, here also we have a probability of hitting, but not the same as in $<-1, 1>$, for example $1/4$.
Finally at the same time they reach to the point $<0, 0>$; if they did not die in previous steps here they are dead.
How model it to find Nash equilibria?
How many mixed strategy do we have?
 A: Suppose $p$ is the probability of hitting from far ($F$, $-1$ or $1$ in your notation) and $P$ is the probability of hitting from close ($C$, $-0.5$ or $0.5$ in your notation). In your example $p=1/8$ and $P=1/4$.
The only possible actions are to shoot from far or to shoot from close. Let's assume that the payoff of the agents is their respective probability of survival. 
Then we can write their payoff matrix as follows:
$$ \begin{matrix}  & {\bf F} & {\bf C}\\ {\bf F}& p,p & p,p P\\ {\bf C}& p P,p & P,P \end{matrix} $$
If both shoot from far or from close, the only chance they have of surviving is that of hitting the other ($p$ and $P$ respectively). 
Now, say $A$ shoots from far while $B$ shoots from close. The only chance of $A$ surviving is that of hitting the other from far ($p$). For $B$ to survive, $A$ must miss and $B$ must hit (which happens with probability $pP$).
Now it is easy to solve for the Nash equilibria. There are three: $(-1,1)$, $(-0.5,0.5)$, and a mixed equilibrium.
The mixed equilibrium must be such that, given that the other player is playing it, the other must be indifferent between playing $F$ or $C$. Suppose $\pi$ is the probability they will play $F$, then
$$ p = \pi pP + (1-\pi)P \Rightarrow \pi = \frac{P-p}{P(1-p)} .$$
