\documentclass[
	nosolutions,
	hidewarning,
	singlenumbering,
	nopagenumber
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}



\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}
\uptitler{\smallurl{}}
\subtitle{Prepared by Mark on \today}


\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}

			\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}