
\begin{problem}

\section*{Is Professor Leighton Right?}

In the Monty Hall problem, there are three doors.  There is a car
behind one, and there are goats behind the other two.  A contestant
picks a door.  Then the hostess, Carol, is required to open a
different door, revealing a goat.  The contestant can either stay with
the door she chose originally or switch to the remaining, unopened
door.  We saw that the contestant can wins the prize with probability
$2/3$ by always switching.

Our analysis assumed that when Carol had a choice about which door to
open, she selected one uniformly at random.  Professor Leighton
asserted that if Carol's choice were not random, then the contestant
could obtain extra information about the location of the car and with
greater probability.

Repeat the analysis done in class assuming that Carol always opens the
\textit{lowest numbered door possible} with the constraint that she
can not open the contestant's door or the door concealing the car.  

\end{problem}



\newpage

\section*{The Four-Door Deal}

\begin{problem}
Suppose that {\em Let's Make a Deal} is played according to different
rules.  Now there are \underline{four} doors, with a prize hidden
behind one of them.  The contestant is allowed to pick a door.  The
host must then reveal a different door that has no prize behind it.
The contestant is allowed to stay with his or her original door or to
pick one of the other two that are still closed.  If the contestant
chooses the door concealing the prize in this second stage, then he or
she wins.

\bparts

\ppart Contestant Stu, a sanitation engineer from Trenton, New Jersey,
stays with his original door.  What is the probability that he wins
the prize?

The tree diagram is awkwardly large.  This often happens; in fact,
sometimes you'll encounter \textit{infinite} tree diagrams!  Try to
draw enough of the diagram so that you understand the structure of the
remainder.

\solution[\newpage]{Let's make the same assumptions as in the original
problem:
%
\begin{enumerate}
\item The prize is equally likely to be behind each door.
\item The contestant is equally likely to pick each door initially,
regardless of the prize's location.
\item The host is equally likely to reveal each door that does not
conceal the prize and was not selected by the player.
\end{enumerate}
%
A partial tree diagram is shown below.  The remaining subtrees are
symmetric to the full-expanded subtree.
%
\begin{center}
\includegraphics[width=6in]{four-door}
\end{center}
%
The probability that Stu wins the prize is:
%
\[
\pr{\text{Stu wins}} = 4 \cdot \left(\frac{1}{48} + \frac{1}{48} + \frac{1}{48}\right) = \frac{1}{4}
\]
%
We multiply by 4 to account for the four subtrees of which we've only
drawn one.}

\ppart Contestant Zelda, an alien abduction researcher from Helena,
Montana, switches to one of the remaining two doors with equal
probability.  What is the probability that she wins the prize?

\solution{A partial tree diagram is worked out below.
%
\begin{center}
\includegraphics[width=6in]{four-doorb}
\end{center}
%
The probability that Zelda wins the prize is:
%
\[
\pr{\text{Zelda wins}} = 4 \cdot \left( \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} \right) = \frac{3}{8}
\]
}
\eparts

\end{problem}
