\documentclass[11pt,twoside]{article}

\usepackage{latex-macros/course}

\begin{document} 
\inclassproblems{13, Fri.}

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\begin{problem}
There are three fair, but strangely-numbered dice:

\bigskip
\centerline{\resizebox{!}{0.75in}{\includegraphics{intransitive-dice}}}
\bigskip

Each hidden face is numbered the same way as the opposite, visible
face.  Answer the three questions below.  Work out each part with a
tree diagram and the four-step method.  (We are trying to transform
you into mindless, drooling, tree-diagramming automatons.)

\bparts

\ppart What is the probability that die $A$ is greater than die $B$?

\solution[\newpage]{
The probability that die $A$ is greater than die $B$ is $5 / 9$:

\bigskip
\centerline{\resizebox{!}{3in}{\includegraphics{intransitive-AB}}}
\bigskip
}

\ppart What is the probability that die $B$ is greater than die $C$?

\solution[\vspace{4in}]{
The probability taht die $B$ is greater than die $C$ is $5 / 9$:

\bigskip
\centerline{\resizebox{!}{3in}{\includegraphics{intransitive-BC}}}
\bigskip
}

\ppart What is the probability that die $C$ is greater than die $A$?

\solution{
The probability that die $C$ is greater than die $A$ is $5 / 9$:

\bigskip
\centerline{\resizebox{!}{3in}{\includegraphics{intransitive-CA}}}
\bigskip
}

\eparts

\end{problem}

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\instatements{\newpage}

\begin{problem}
Suppose that the rules of Ultimate Frisbee were revised as follows:

\begin{quotation}
\noindent \textit{Representatives of the two teams each flip a disc.
The representative of one team predicts the orientation of his or her
own disc by calling ``up'' or ``down'' while the discs are in the air.
However, if both discs land the same way, then the call does not count
and the process starts over.}
\end{quotation}

Assume that a disc lands face-up with probability $p$.  What is the
probability that the caller wins by always saying ``up''?

Suggestions: The tree diagram is infinite, so draw only enough to see
a pattern.  Summing all the winning outcome probabilities directly is
difficult.  However, a neat trick solves this problem and many others.
Let $s$ be the sum of all winning outcome probabilities in the whole
tree.  Notice that {\em you can write the sum of all the winning
probabilities in certain subtrees as a function of $s$}.  Use this
observation to write an equation in $s$ and then solve.

\solution{In the tree diagram below, the small triangles represent
subtrees that are themselves complete copies of the whole tree.

\bigskip
\centerline{\resizebox{!}{3in}{\includegraphics{fair-frisbee}}}
\bigskip

Let $s$ equal the sum of all winning probabilities in the whole tree.
There are two extra edges with probability $p$ on the path to each
outcome in the top subtree.  Therefore, the sum of winning
probabilities in the upper tree is $p^2 s$.  Similarly, the sum of
winning probabilities in the lower subtree is $(1-p)^2 s$.  This gives
the equation:

\begin{equation*}
s = p^2 s + (1-p)^2 s + p (1-p)
\end{equation*}

The solution to this equation is $s = 1/2$, for all $p$ between 0 and
1.}

\end{problem}

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\instatements{\newpage}

\begin{problem}
Suppose we flip a coin repeatedly, generating a sequence of heads and
tails:

\[
TTHTT\underline{HTH}
\]

If the pattern $HTH$ appears (as in the example above), then we stop
and you're declared the winner.  If the pattern $HHT$ appears, then we
stop and I'm declared the winner.  What is the probability that you
win?  Use a tree diagram and the four-step method.  You'll need the
trick described in the preceding problem again.

\solution[\vspace{4.5in}]{A tree diagram is shown below.  The
triangles denote subtrees that are complete copies of the whole tree.

\bigskip
\centerline{\resizebox{!}{3in}{\includegraphics{hth}}}
\bigskip

Let $s$ be the sum of all your winning outcome probabilities.
Reasoning as in the preceding problem, we get the equation:

\begin{align*}
s & = \frac{1}{2} s + \frac{1}{8} s +
      \frac{1}{8} \\
  & = \frac{5}{8} s + \frac{1}{8} \\
\end{align*}

The solution is $s = 1 / 3$.
}
\end{problem}

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\begin{problem}
\textbf{(challenge) } Find a way to number the faces of dice $A$, $B$,
and $C$ so that:

\begin{enumerate}

\item The probability that $A$ is greater than $C$ is more than $\frac{1}{2}$.

\item The probability that $B$ is greater than $C$ is more than $\frac{1}{2}$.

\item If all three dice are rolled, then $C$ is most likely to show
the greatest number.

\end{enumerate}

\solution{One solution is to number opposite faces the same way as
follows:

\begin{align*}
A & = 4, 4, 2 \\
B & = 4, 4, 2 \\
C & = 5, 3, 1
\end{align*}

Now $\pr{A > C} = \frac{5}{9}$, $\pr{B > C} = \frac{5}{9}$, and $\pr{C
> A \text{ and } C > B} = \frac{10}{27}$.
}
\end{problem}

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\end{document}
