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+\documentclass[
+	solutions,
+	hidewarning,
+	singlenumbering,
+	nopagenumber
+]{../../resources/ormc_handout}
+
+\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}
+\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}
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