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\begin{document}

\centerline{Notes for Feb 7 Tutorial}

{\Large The exercise involving Hershey's kisses is CANCELLED.
Remaining stock is on my desk in 203.}

There are four topics and an optional time sink:

\begin{itemize}
\item Team building.
\item Probability discussion.
\item Best-of-five.
\item GPA weirdness.
\item Strange coins.
\end{itemize}

\section{Team Building}

\begin{itemize}

\item Ask students for name, home town, living group, and some
nontechnical fact about them.

\item For students who have a hard time giving an additional fact,
prompt with: interesting place visited?  Hobby?  Sport played?
Musical instrument?  Most embarassing episode?

\item Ask a followup question about anything that intrigues you; this
shows that you're listening and establishes a conversational
atmosphere, which will be nice later!

\end{itemize}

Ask the students to suggest team names.  When every student but one
has suggested one, call on that student to pick among the suggestions.

You may want to point out that your job is to help your students get
the best grade they can.  Unless they're obvious slackers or
super-contributors, their grade will be determined by exam
performance, not you.  So you're on their side.

\section{Probability Discussion}

You may want to discuss probability in daily life.  When is it a
concern?  Why do we not worry much about auto accidents (40K die per
year), but focus our country to thwarting terrorism (3K died on 9-11)?
What risks are worth taking?  What \% chance of death would you take
for a million dollars?  (How does this compare with the risk of dying
in an auto accident this year, which is 1/6000.)  How about for the
experience of going into space?  (After all, 2/88 shuttle flights have
been fatal.)

\section{Best-of-Five}

We'll have some friendly inter-tutorial rivalry this term.  Suppose
that two tutorial teams, $A$ and $B$, are nearly equal in ability.
(Perhaps use your section's name and the course staff team, The King
Tuts).  We have:

\[
\Pr\{\text{$A$ beats $B$}\} = \frac{1}{2} + \epsilon
\]

If they play only one game, the weaker team might get lucky!  (You
might make reference to the intransitive dice demo in lecture, where
the professor lost despite having a $5/9$ chance of winning each
game.)  So suppose we have a best-of-five tournament to determine
which team is better.  What is the probability that the better team
wins a best-of-five tournament?

Guide the students through the four-step method.

\begin{itemize}

\item Find the sample space by drawing out the tree diagram.

\item Identify the event of intrest, the event that team A beats team B.

\item Find the outcome probabilities by labelling arc and multiplying
along paths.

\item Find the probability of the event of interest.

\end{itemize}

In step four, guide students to the solution of the form $\frac{1}{2}
+ \frac{15}{8} \epsilon$.  Explain why this is good enough when
$\epsilon$ is small.

Follow up question: what is the probability that the stronger team
wins if $\epsilon$ is small, say 1\%?

If your students are quick, you might ask the approximate probablity
that Team $A$ beats Team $B$ in a ``tournament of tournaments''.  This
is just two applications of the result just proved, giving
$\frac{1}{2} + \frac{15^2}{8^2} \epsilon$.

\section{GPAs}

There are two students, Alice and Bob.  In the fall semester, Alice's
grade point average is better than Bob's.  Inthe spring semester,
Alice's grade point average is again better than Bob's.  However,
Alice is horrified to discover that for the year as a whole, her
gradepoint average is {\em lower}.  How is this possible?

One solution is as follows:

\begin{tabular}{rcc}
       & Alice & Bob \\
Fall   & A     & AAAB  \\
Spring & BBBA  & B
\end{tabular}

\section{Time sink}

If you have extra time, this question should consume it!  This is a
polished version of the mangled question from the tutorial meeting.

There are three coins, $A$, $B$, and $C$.  Each coin has a number
written on either side.  They are NOT necessarily fair.  If you flip
all three, then:

\begin{itemize}

\item More than half the time, $A$ is greater than $C$.

\item More than half the time, $B$ is greater than $C$.

\item Among all three coins, $C$ is most likely to show the largest
number.

\end{itemize}

How is this possible?

One solution is as follows:

\begin{itemize}

\item Coin $A$ has 3 on both sides.

\item Coin $B$ has a 2 on one side and a 4 on the other.  It is fair.

\item Coin $C$ has a 1 on one side and a 5 on the other.  60\% of the
time it comes up 1, and 40\% of the time it comes up 5.

\end{itemize}

\end{document}
