Probability of a player winning in a multi round shooting game. I've been trying to compute the answer to this problem analytically, but I can't quite get to the answer. (The problem below is from Characteristics of Games Exercise 2.9)
Three players A, B, and C, each have a balloon. They each simultaneously choose an opposing balloon and throw a dart at it. Anyone whose balloon is popped is out (if all three hit or all three miss, everyone goes again). If only one person is knocked out the remaining two play for the win, repeating if necessary until one of the two wins.
Suppose A has a 60 percent chance to hit his target, B a 50 percent chance, and C a 40 percent chance. The "obvious" strategy is for each person to go for the biggest target (A aims at B, B and C both aim at A). This gives B a 19 percent chance to win. If B switches to C, though, then B has a 23 percent chance to win.
I've created a javascript program and can get the win rate for B statistically, but I'm at a loss analytically.
 A: Straightforward, but lots of steps.
intuitively:  $B$ should prefer the second strategy because the "obvious" one leaves no quick path to victory.  Of course there are tradeoffs ($A$ has a better survival rate) and it isn't obvious how it sorts out.
Well, let's start with the "strategy free cases", i.e. those states in which we only have two players.  (in that case, of course, they just shoot at each other so there is no need for strategy). 
(A,C):  A survives with probability $.6$, C survives with probability $.4$ hence with  probability $.24+.24=.48$ the game restarts; A wins with probability $.36$, $C$ wins with probability $.16$.  Let $P[A,C]_A$ be the probability that $A$ wins this game (eventually).  Then we have: 
$$P[A,C]_A=.48^*P[A,C]_A+ .36^*1+.16^*0\;\Rightarrow\,P[A,C]_A=.6923$$
(B,C):  B survives with probability $.6$, C survives with probability $.5$ hence with  probability $.3+.2=.5$ the game restarts; B wins with probability $.3$, $C$ wins with probability $.2$.  Let $P[B,C]_B$ be the probability that $B$ wins this game (eventually).  Then we have: 
$$P[B,C]_B=.5^*P[B,C]_B+ .3^*1+.2^*0\;\Rightarrow\,P[B,C]_B=.6$$
(A,B):  A survives with probability $.5$, B survives with probability $.4$ hence with  probability $.2+.3=.5$ the game restarts; A wins with probability $.3$, $B$ wins with probability $.2$.  Let $P[A,B]_A$ be the probability that $A$ wins this game (eventually).  Then we have: 
$$P[A,B]_A=.5^*P[A,B]_A+ .3^*1+.2^*0\;\Rightarrow\,P[A,B]_A=.6$$
Ok.  Now let's consider the "obvious strategy".  Let $\phi$ be the probability that $B$ eventually wins via this strategy.  on the first round, $A$ survives with probability $.3$, $B$ survives with probability $.4$ and $C$ definitely survives. Thus the game restarts with probability $.12$, we move to $(B,C)$ with probability $.28$ and otherwise $B$ is gone.  Thus $$\phi=.12\phi+.28^*.6\;\Rightarrow\;\phi=.1909$$
Alternatively, let's consider the strategy wherein A fires on B, but B fires on C (and C still fires on A) Let $\psi$ be the probability $B$ eventually wins this way.  In this version, $A$ survives the first round with probability $.6$, $B$ survives with probability $.4$, $C$ survives with probability $.5$.  thus the game restarts with probability $.12+.12=.24$, $B$ wins immediately with probability $.08$, we move to $(A,B)$ with probability $.12$ and we move to $(B,C)$ with probability $.08$.  Otherwise the game goes on without poor $B$.  Thus 
$$\psi=.24\psi+.08^*1+.12^*.4+.08^*.6\;\Rightarrow\;\psi=.2316$$
