2023-10-06 09:07:27 -07:00
|
|
|
\documentclass[
|
2024-05-25 10:11:24 -07:00
|
|
|
nosolutions,
|
2023-10-06 09:07:27 -07:00
|
|
|
hidewarning,
|
|
|
|
singlenumbering,
|
|
|
|
nopagenumber
|
|
|
|
]{../../resources/ormc_handout}
|
2023-10-17 18:32:10 -07:00
|
|
|
\usepackage{../../resources/macros}
|
|
|
|
|
|
|
|
|
2023-10-06 09:07:27 -07:00
|
|
|
|
|
|
|
\usepackage{tikz}
|
|
|
|
\usetikzlibrary{arrows.meta}
|
|
|
|
\usetikzlibrary{shapes.geometric}
|
|
|
|
|
|
|
|
% We put nodes in a separate layer, so we can
|
|
|
|
% slightly overlap with paths for a perfect fit
|
|
|
|
\pgfdeclarelayer{nodes}
|
|
|
|
\pgfdeclarelayer{path}
|
|
|
|
\pgfsetlayers{main,nodes}
|
|
|
|
|
|
|
|
% Layer settings
|
|
|
|
\tikzset{
|
|
|
|
% Layer hack, lets us write
|
|
|
|
% later = * in scopes.
|
|
|
|
layer/.style = {
|
|
|
|
execute at begin scope={\pgfonlayer{#1}},
|
|
|
|
execute at end scope={\endpgfonlayer}
|
|
|
|
},
|
|
|
|
%
|
|
|
|
% Arrowhead tweak
|
|
|
|
>={Latex[ width=2mm, length=2mm ]},
|
|
|
|
%
|
|
|
|
% Nodes
|
|
|
|
main/.style = {
|
|
|
|
draw,
|
|
|
|
circle,
|
|
|
|
fill = white,
|
|
|
|
line width = 0.35mm
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
\title{Warm Up: Odd dice}
|
2024-04-01 21:50:50 -07:00
|
|
|
\uptitler{\smallurl{}}
|
|
|
|
\subtitle{Prepared by Mark on \today}
|
2023-10-06 09:07:27 -07:00
|
|
|
|
|
|
|
|
|
|
|
\begin{document}
|
|
|
|
|
|
|
|
\maketitle
|
|
|
|
|
|
|
|
\problem{}
|
|
|
|
|
|
|
|
We say a set of dice $\{A, B, C\}$ is \textit{nontransitive}
|
|
|
|
if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
|
|
|
|
In other words, we get a counterintuitive \say{rock - paper - scissors} effect.
|
|
|
|
|
|
|
|
\vspace{2mm}
|
|
|
|
|
|
|
|
Create a set of nontransitive six-sided dice. \par
|
|
|
|
\hint{All sides should be numbered with positive integers less than 10.}
|
|
|
|
|
|
|
|
\begin{solution}
|
|
|
|
One possible set can be numbered as follows:
|
|
|
|
\begin{itemize}
|
|
|
|
\item Die $A$: $2, 2, 4, 4, 9, 9$
|
|
|
|
\item Die $B$: $1, 1, 6, 6, 8, 8$
|
|
|
|
\item Die $C$: $3, 3, 5, 5, 7, 7$
|
|
|
|
\end{itemize}
|
|
|
|
|
|
|
|
\vspace{4mm}
|
|
|
|
|
|
|
|
Another solution is below:
|
|
|
|
\begin{itemize}
|
|
|
|
\item Die $A$: $3, 3, 3, 3, 3, 6$
|
|
|
|
\item Die $B$: $2, 2, 2, 5, 5, 5$
|
|
|
|
\item Die $C$: $1, 4, 4, 4, 4, 4$
|
|
|
|
\end{itemize}
|
|
|
|
\end{solution}
|
|
|
|
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
\problem{}
|
|
|
|
Now, consider the set of six-sided dice below:
|
|
|
|
\begin{itemize}
|
|
|
|
\item Die $A$: $4, 4, 4, 4, 4, 9$
|
|
|
|
\item Die $B$: $3, 3, 3, 3, 8, 8$
|
|
|
|
\item Die $C$: $2, 2, 2, 7, 7, 7$
|
|
|
|
\item Die $D$: $1, 1, 6, 6, 6, 6$
|
|
|
|
\item Die $E$: $0, 5, 5, 5, 5, 5$
|
|
|
|
\end{itemize}
|
|
|
|
On average, which die beats each of the others? Draw a graph. \par
|
|
|
|
|
|
|
|
\begin{solution}
|
|
|
|
\begin{center}
|
|
|
|
\begin{tikzpicture}[scale = 0.5]
|
|
|
|
\begin{scope}[layer = nodes]
|
|
|
|
\node[main] (a) at (-2, 0.2) {$a$};
|
|
|
|
\node[main] (b) at (0, 2) {$b$};
|
|
|
|
\node[main] (c) at (2, 0.2) {$c$};
|
|
|
|
\node[main] (d) at (1, -2) {$d$};
|
|
|
|
\node[main] (e) at (-1, -2) {$e$};
|
|
|
|
\end{scope}
|
2024-04-01 21:50:50 -07:00
|
|
|
|
2023-10-06 09:07:27 -07:00
|
|
|
\draw[->]
|
|
|
|
(a) edge (b)
|
|
|
|
(b) edge (c)
|
|
|
|
(c) edge (d)
|
|
|
|
(d) edge (e)
|
|
|
|
(e) edge (a)
|
|
|
|
|
|
|
|
(a) edge (c)
|
|
|
|
(b) edge (d)
|
|
|
|
(c) edge (e)
|
|
|
|
(d) edge (a)
|
|
|
|
(e) edge (b)
|
|
|
|
;
|
|
|
|
\end{tikzpicture}
|
|
|
|
\end{center}
|
|
|
|
\end{solution}
|
|
|
|
|
|
|
|
\vfill
|
|
|
|
|
|
|
|
Now, say we roll each die twice. What happens to the graph above?
|
|
|
|
|
|
|
|
\begin{solution}
|
|
|
|
The direction of each edge is reversed!
|
|
|
|
\end{solution}
|
|
|
|
|
|
|
|
\vfill
|
|
|
|
\pagebreak
|
|
|
|
|
|
|
|
\end{document}
|