
\problemdata     
{S01-PS9-2}           
{probability-intro}              
{} 		
{S01-PS9}
{S01 Karger}


\begin{problem}



After their arrest and sentencing for
unlawful use of firearms, and short hospital stays, the three
gunfighters, $A$, $B$, and $C$ applied for parole.  The parole board
decided to release two of the three, chosen uniformly at random, but
has not released their names.

Gunfighter $A$ figures that he will be released with probability
$\frac{2}{3}$.  A guard offers to tell gunfighter $A$ the name of one
of the other gunfighters who will be released (either $B$ or $C$).
However, gunfighter $A$ declines the offer.  He reasons that if the
guard says, for example, ``$B$ will be released'', then his own
probability of release will drop to $\frac{1}{2}$.  This is because he
will then know that either he or $C$ will also be released, but these
events are equally likely.

Either prove that gunfighter $A$ has reasoned correctly or prove that
he is wrong.  Assume that if the guard has a choice of saying $B$ or
$C$ because both are to be released, then he picks one uniformly at
random.

\solution{Gunfighter $A$ has reasoned incorrectly.  We must compute

\begin{eqnarray*}
\Pr(\text{$A$ released} \mid \text{guard says ``$B$''})
        & = &   \frac{\Pr(\text{$A$ released and guard says ``$B$''})}
                        {\Pr(\text{guard says ``$B$''})}.
\end{eqnarray*}

These probabilities are worked out in the tree diagram below.

\graphicspath{{/toc/a/class/6.042/spring02/freshRepository/figures/}}

\bigskip
\centerline{\resizebox{!}{3.0in}{\includegraphics{sol9-prisoner}}}
\bigskip

Plugging these probabilities into the preceding equation gives
\begin{displaymath}
  \Pr(\text{$A$ released} \mid \text{guard says $B$ released})
  = \frac{1}{3} \biggm/ \frac{1}{2} = \frac{2}{3}\,.
\end{displaymath}
If the guard says that $B$ will be released, then $A$ still has a
$\frac{2}{3}$ chance of release.
}

\end{problem}
