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\usepackage{light}

\hidesolutions
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\begin{document}

\recitation{17}{November 10, 2004}

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\section*{The Four-Step Method}

This is a good approach to questions of the form, ``What is the
probability that -------?''  Intuition \textit{will} mislead you, but
this formal approach gives the right answer every time.

\begin{enumerate}
\item Find the sample space.  (Use a tree diagram.)

\item Define events of interest.  (Mark leaves corresponding to these
events.)

\item Determine outcome probabilities:

\begin{enumerate}

\item Assign edge probabilities.

\item Compute outcome probabilities.  (Multiply along root-to-leaf
paths.)

\end{enumerate}

\item Compute event probabilities.  (Sum the probabilities of all
outcomes in the event.)

\end{enumerate}

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\newpage

\section*{A Baseball Series}

\begin{problem}
The New York Yankees and the Boston Red Sox are playing a
two-out-of-three series.  (In other words, they play until one team
has won two games.  Then that team is declared the overall winner and
the series ends.)  Assume that the Red Sox win each game with
probability $3/5$, regardless of the outcomes of previous games.

Answer the questions below using the four-step method.  You can use
the same tree diagram for all three problems.

\bparts

\ppart What is the probability that a total of 3 games are played?

\ppart What is the probability that the winner of the series loses the
first game?

\ppart What is the probability that the \textit{correct} team wins the
series?

\eparts

\end{problem}

\solution{
A tree diagram is worked out below.
%
\begin{center}
\includegraphics[width=6.25in]{series}
\end{center}
%
From the tree diagram, we get:
%
\begin{align*}
\pr{\text{3 games played}}
    & = \frac{12}{125} + \frac{18}{125} + \frac{12}{125} + \frac{18}{125}
      = \frac{12}{25} \\
\pr{\text{winner lost first game}}
    & = \frac{18}{125} + \frac{12}{125}
      = \frac{6}{25} \\
\pr{\text{correct team wins}}
    & = \frac{18}{125} + \frac{18}{125} + \frac{9}{25} = \frac{81}{125}
\end{align*}
}

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\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}

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\newpage

\section*{Professor Leighton Lies!  LIES!}

\begin{problem}
Let's consider another variation of the four-doors problem.  Suppose
that Carol always opens the \textit{lowest-numbered} door possible
with the restriction that she can neither reveal the prize nor open
the door that the player picked.

This gives contestant Mergatroid--- an engineering student from
Cambridge, MA--- just a little more information about the location of
the prize.  Suppose that Mergatroid always switches to the
lowest-numbered door, excluding his initial pick and the one Carol
opened.  What is the probability that he wins the prize?

(Interestingly, in the three-door problem, the contestant gains
\textit{no} advantage if Carol always opens the lowest-numbered door
available.  Professor Leighton misspoke on this point in lecture.)
\end{problem}

\solution{
A tree diagram is worked out below.
%
\begin{center}
\includegraphics[height=3.5in]{carol}
\end{center}
%
The probability that Mergatroid wins is:
%
\[
\pr{\text{win}} = 8 \cdot \frac{1}{16} = \frac{1}{2}
\]
}


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\end{document}

