\problemdata                                                             
        {probability-intro-1} % latex-friendly label for the prob.
        {Basic probability, Monty Hall, tree diagrams} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 CP9M; F01; S99 TUT10} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
Suppose the game \emph{Let's Make A Deal} is changed slightly.  Instead of
having 3 doors with 1 grand prize, in the new game there are 4 doors with
2 grand prizes.  Otherwise, the rules of the game remain the same:
\begin{enumerate}
\item The contestant picks one of the doors.

\item Assistant Carol opens a different door that has a goat behind it.

\item The contestant can then \textbf{stick} with his original pick or
\textbf{switch} to an unopened door.  He wins a prize only if his final
pick is the door with the prize.

\end{enumerate}

Assume that the prizes are equally likely to be placed behind each of the
doors, and that Carol and the contestant are also equally likely to pick
each door among their possible choices.

\begin{problemparts}

\problempart In the new game, what is the probability of winning with the ``stick''
strategy?  How about the ``switch'' strategy?


\solution{The player wins with a ``stick'' strategy iff he chooses the
``correct door'' in the first place.  Therefore the probability of winning
with the ``stick'' strategy is $1/2$.

To analyze the switch strategy we form the usual tree, as shown in
Figure~(\ref{mont-mod}).  
%CAK/ASHM INSERT ONE HERE.

\begin{figure}[htb]
\begin{center}
%\includegraphics[width=4in]{figures/mont-mod}
\input{figures/mont-mod.latex}
\end{center}
\caption{Coding Numbered Trees}
\label{mont-mod}
\end{figure}

By symmetry, we may assume that the prizes are behind doors 1 \& 2.  The
possible outcomes are then of the form:
\begin{center}
(contestant-initial-choice, Carol-choice, contestant-switch-choice)
\end{center}
with probabilities:
\begin{center}
\begin{tabular}{ccc}
doors & probability & win-with-switch\\
\hline\\
(1,3,2) & 1/16 & win\\
(1,3,4) & 1/16 & lose\\
(1,4,2) & 1/16 & win\\
(1,4,3) & 1/16 & lose\\
(2,3,1) & 1/16 & win\\
(2,3,4) & 1/16 & lose\\
(2,4,1) & 1/16 & win\\
(2,4,3) & 1/16 & lose\\
(3,4,1) & 1/8 & win\\
(3,4,2) & 1/8 & win\\
(4,3,1) & 1/8 & win\\
(4,3,2) & 1/8 & win
\end{tabular}
\end{center}
So the switch strategy wins 3/4 of the time.}

\problempart
In the original Monty Hall game, the probabilities of the
winning-with-stick and the winning-with-switch strategies summed to one,
but here they don't.  How come?

\solution{ In the original Monty Hall game with one prize, the switch
strategy would win iff the stick strategy would not win, so
winning-with-stick and winning-with-switch were complementary events whose
probabilities had to sum to one.

This time the two probabilities do not need to sum up to 1 because they
are no longer disjoint events.  It is possible for both the ``switch'' and
the ``stick'' strategy to pick doors that both have a prize behind them,
because there are two prizes.  So the same outcome can be a win for both
strategies.}

\end{problemparts}

\end{problem}
                                 