Duel question, why shoot when $P_1 (s)+P_2 (s)=1$ There are two duelists with one shot each, their probability to hit from a distance $x$ is given by two functions $P_1 (x)$, $P_2 (x)$ that are both continuously and strictly decreasing from $P(0)=1, P(L)=0$. They walk towards each other from $L$ to $0$, and are each free to shoot whenever. 
Our payouts: hit is 1, both miss or both die is 0 and die is -1.
We can logically come to the conclusion that they will shoot simultaneously, but we're having trouble showing that they will shoot from a distance $s$ when $P_1 (s) + P_2 (s) = 1$. Any help here would be greatly appreciated.
Assuming P1 is the better player we believe that P1 will shoot when when his payout is maximized (forcing P2 to shoot), this occurs at $max(P_1 (s) - P_2 (s))$. 
 A: Note what happens if shooter $1$ decides to fire.  This gives shooter $2$ a choice.  Either he can return fire or he can wait. Of course, if he waits (and survives) he is guaranteed a payout of $1$ at distance $0$. If he returns fire then shooter $1's$ expected payout is $$(P_1(x))(1-P_2(x))-(1-P_1(x))(P_2(x))=P_1(x)-P_2(x)$$
If shooter $1$ holds off then shooter $1's$ expected payout is $$P_1(x)-(1-P_1(x))=2P_1(x)-1$$
If equilibrium occurs when these expectations coincide then we have $$P_1(x)-P_2(x)=2P_1(x)-1$$ which is equivalent to $$P_1(x)+P_2(x)=1$$
All that remains is to show that when the two expectations do not coincide, both players should wait.
Toward that end:   Consider the difference of the two expectations $$F(x)=P_1(x)-P_2(x)-(2P_1(x)-1)=1-P_1(x)-P_2(x)$$
Then $F(0)=1$ and $F(x)$ is strictly decreasing.  
Case I:   $F(x)>0$.  Then if shooter $1$ fires, shooter $2$ should hold off so shooter $1's$ expectation is $2P_1(x)-1$. That function is strictly increasing so shooter $1$ should wait.  
Case II: $F(x)<0$ Then, if shooter $1$ is holding off, shooter $2's$ expectation is $2P_2(x)-1$ which is increasing.  It follows that shooter $1's$ expectation must be decreasing in that region, so shooter $1$ should shoot when $F(x)=0$.
