
\problemdata     
{S01-PS8-3}           
{probability-intro}              
{} 		
{F98-PS9, S01-PS8}
{S01 Karger}


\begin{problem}


Three gunfighters, $A$, $B$, and~$C$, plan
to fight a three-cornered pistol duel. They all know that $A$'s chance
of hitting his target is $0.3$, $B$~never misses, and $C$'s chance of
hitting is~$0.5$. They will shoot one at a time in the sequence
$ABCABC\ldots$. On his turn, a gunfighter may target either of the
other two, or may fire harmlessly into the air. Assume that each
gunfighter uses the shooting strategy that maximizes his probability
of being the only one left standing. What should gunfighter $A$~do on
his first shot and what probability of survival does this give him?

\solution{
  
  First note that the optimal strategy for~$B$ is to shoot the most
  accurate gunfighter. So, if $B$~gets to shoot and both $A$ and~$C$
  are still alive, $B$~will shoot~$C$---this gives him a $1-0.3 = 0.7$
  chance of survival---and not~$A$---that would only give him a $1-0.5
  = 0.5$ chance of survival.
  
  Suppose that $A$ shoots into the air. Then $B$ shoots~$C$, and
  $A$~gets one shot to kill~$B$ before he himself is killed. Therefore
  $A$~has a $0.3$~chance of survival.
  
  Suppose that $A$ shoots at~$B$. If he hits, then he enters a duel
  with~$C$, where $C$ gets the first shot. In this case, let $p$~be
  the probability that $A$~wins the duel. For this to happen, either
  $C$~must miss and $A$~must hit (probability $0.5 \cdot 0.3$) or else
  $C$~must miss, $A$~must miss, and then $A$ must win a duel of the
  same type as before (probability $0.5 \cdot 0.7 \cdot p$).  As an
  equation:
  \begin{displaymath}
    p = \frac{1}{2} \cdot \frac{3}{10} + \frac{1}{2} \cdot \frac{7}{10} p,
  \end{displaymath}
  which has the solution $p = 3 /13$.
  
  Thus, If $A$ hits~$B$, his odds of survival are lower than if he
  fires into the air! If he does not hit, his probability of survival
  is the same as if he had started by firing in the air. Since
  $A$~hits with probability~$0.3$, the probability that $A$~survives
  when he begins with shooting at~$B$ is $0.3 \cdot 3 / 13 + 0.7 \cdot
  0.3 = 363 / 1300 < 0.28$.
  
  On the other hand, suppose that $A$ shoots at~$C$. If he hits, then
  $B$~immediately kills him. If he does not hit, his probability of
  survival is, again, the same as if he had started by firing in the
  air. Thus, his probability of survival in this case is $0.3 \cdot 0
  + 0.7 \cdot 0.3 = 0.21$.
  
  To conclude, $A$'s best strategy is to fire into the air; this gives
  him a $0.3$ chance of survival.}

\end{problem}
