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+\chapter{Classes}
+
+\section*{Basic Class Structure}
+
+\subsection*{Combat}
+
+\paragraph{Stats}
+Each class grants +1 in three separate stats; no two classes grant the exact same stats.
+
+Most combat classes grant bonuses to physical stats, aiming for the class to be self-sufficient, such as the warrior group. Some classes, such as rangers, add mental stats, to support other classes.
+
+\paragraph{Standard Weapons}
+The standard weapons are purely for flavor. They act as a guide to how that class should feel.
+
+\paragraph{Weapon Proficiencies}
+Weapon proficiencies are chosen such that they include the standard weapons.
+
+\paragraph{Armor proficiencies}
+Armor proficiencies are also selected to support the flavor of the class. It should reflect how vulnerable the class should feel, taking into account how the stats, i.e., \stat{STB} and \stat{DEX} compensate for any missing armor.
+
+Note that every class can wear no armor, e.g., normal clothes.
+
+\paragraph{Skills}
+The number of skills should roughly reflect the mental stats a class grants. More support-focused classes should grant more skills.
+
+\paragraph{Abilities}
+Each class group has one ability that all classes in that group share. Other than that, each class should have at least 4 abilities.
+
+The number of passive and reactive abilities affects how passive the class feels. Something like a guard should feel more like guarding -- standing still and reacting to what other people do.
\ No newline at end of file
diff --git a/dice.tex b/dice.tex
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+\chapter{Dice}
+
+\section{Distribution of Successes}
+
+\begin{multicols}{2}
+    For further analysis it will be useful to have a distribution function for margins of success.
+
+    To obtain such a distribution we start by observing that each roll can be divided into four groups: critical successes, successes, failures and critical failures, as shown in \cref{equ:dice_groups}. Further, this tuple has a multinomial distribution
+    \begin{align}
+        \left(C_s, S, F, C_f\right) \sim \text{MN}\left(d, 4, p\right)\label{equ:dice_groups} \\
+        p = \left[1/12,\frac{12-\stat{dt}-1}{12},\frac{\stat{dt}-1}{12},\frac{1}{12}\right]
+    \end{align}
+
+    We can easily map a tuple to a success margin:
+    \begin{equation}
+        M = 2C_s + S - F - 2C_f
+    \end{equation}
+
+    Therefore, our distribution for $M$ is a sum over all tuples that produce the correct $m$
+    \begin{equation}
+        Pr(M = m) = \sum_{C_s,S,F,C_f; M = m}  \text{MN}\left(d, 4, p\right)
+    \end{equation}
+
+    Instead of trying all combinations of variables and summing the correct ones, we can use a system of Diophantine equations to reduce the search space.
+
+    We start with two equations
+    \begin{align}
+        2c_s + s - f - 2c_f & = m \\
+        c_s + s + f + c_f   & = d
+    \end{align}
+    which resolve into a two-dimensional system
+    \begin{align}
+        c_s & = m - d + 2f + 3c_f   \\
+        s   & = -m + 2d - 3f - 4c_f
+    \end{align}
+
+    Next, we use inequalities $c_s\geq0, s\geq0$ to constrain the space for $c_f$, based on $f$. An example plot in \cref{fig:candidates} graphically shows the area defined by these equations and inequalities for one combination of $d$ and $m$.
+    \begin{align}
+        \frac{m-d+2f}{-3} \leq c_f \leq \frac{-m+2d-3f}{4}
+    \end{align}
+    Because the lower bound is not necessarily greater than 0, we will also use $c_f\geq0$.
+
+    We can now also constrain $f$ by observing that the lower bound for $c_f$ can cross $c_f=0$ before $f=d$.
+    \begin{equation}
+        0 \leq f \leq \frac{2d-m}{3}
+    \end{equation}
+
+    Putting the sum together, and adding floors and ceilings to limit the sum to integers, we get:
+    \begin{align}
+        Pr(M=m) & = \sum_{f=0}^{\left\lfloor \frac{2d-m}{3} \right\rfloor}\sum_{c_f=\max(0,\left\lceil\frac{m-d+2f}{-3}\right\rceil)}^{\left\lfloor\frac{-m+2d-3f}{4}\right\rfloor} \\
+                &\quad\; \frac{d!}{c_s!s!f!c_f!}
+        \frac{n_s^s n_f^f}{12^{d}} \\
+        n_f     & = \stat{dt}-1 \\
+        n_s     & = 12-n_s
+    \end{align}
+    While this equation is not very illuminating, it does allow us to efficiently compute the distribution and thus answer various questions regarding the results of dice rolls.
+
+\end{multicols}
+
+\begin{figure}[htb]
+    \centering
+    \includegraphics{images/dice_candidates_example.pdf}
+    \caption{Example of candidate combinations of $f$ and $c_f$ with upper and lower bounds for $c_f$ shown. In this plot, $d=12$ and $m=8$.\label{fig:candidates}}
+\end{figure}
+
+\begin{figure}[htb]
+    \centering
+    \begin{tabular}{r | c c c c c c c c}
+           & \multicolumn{8}{c}{Difficulty Threshold}                             \\
+        d  & 0                                        & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \cline{2-9}
+        1  & 1                                        & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\
+        2  & 2                                        & 2 & 1 & 1 & 1 & 1 & 0 & 0 \\
+        3  & 3                                        & 3 & 2 & 2 & 1 & 1 & 1 & 1 \\
+        4  & 4                                        & 3 & 3 & 2 & 2 & 1 & 1 & 0 \\
+        5  & 5                                        & 4 & 3 & 3 & 2 & 1 & 1 & 1 \\
+        6  & 6                                        & 5 & 4 & 3 & 2 & 2 & 1 & 1 \\
+        7  & 7                                        & 6 & 5 & 4 & 3 & 2 & 1 & 1 \\
+        8  & 8                                        & 7 & 5 & 4 & 3 & 2 & 1 & 1 \\
+        9  & 9                                        & 8 & 6 & 5 & 3 & 2 & 1 & 1 \\
+        10 & 10                                       & 8 & 7 & 5 & 4 & 2 & 1 & 1 \\
+    \end{tabular}
+    \caption{Expected number of successes.\label{tab:exp_mos}}
+\end{figure}
+
+\section{Combat Pool Steady State}
+
+\begin{multicols}{2}
+    The combat pool starts with the proficiency state, i.e., \stat{STR} or \stat{DEX}. However, after rolling, the change in pool size is determined by \stat{CON}. After a number of turns, the expected size of the pool should converge on a number $E[d]$.
+
+    We start by assuming that $E[d]$ is a number such that
+    \begin{equation}
+        E[d_{n+1}] = E[d_{n}] \label{equ:dice_ss_a1}
+    \end{equation}
+    where $n$ is the turn after which the steady state is achieved.
+
+    To compute $E[d_n]$ for any $n$ we can use
+    \begin{equation}
+        E[d_n] = E[d_{n-1}] - p_f \cdot E[d_{n-1}] + \stat{con} \label{equ:dice_ss_a2}
+    \end{equation}
+    where $p_f = \stat{dt}/12$ is the probability of failure.
+
+    Using assumptions in \cref{equ:dice_ss_a1,equ:dice_ss_a2} we write the following equations:
+    \begin{align}
+        E[d]          & = E[d] - p_f \cdot E[d] + \stat{con} \\
+        p_f\cdot E[d] & = \stat{con}                         \\
+        E[d]          & = \frac{\stat{con}}{p_f}
+    \end{align}
+
+    As $p_f$ approaches 1, the steady state becomes just \stat{CON}, meaning the dice pool is entirely discarded and refreshed by \stat{con} on every turn. On the other hand, as $p_f$ approaches 0, the expected dice pool size increases. Specifically at 0, it shoots off toward infinity, as  the dice pool loses no dice and \stat{con} dice get added toward it on every turn.
+\end{multicols}
+
diff --git a/main.tex b/main.tex
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+\documentclass{kartsstyle}
+
+\usepackage{multirow}
+
+\title{On The Metaphysics Of Savits}
+\author{\texttt{HeNine}}
+\date{2023}
+
+\newcommand{\tochange}[1]{\textcolor{red}{#1}}
+
+\begin{document}
+
+\frontmatter
+
+\maketitle
+
+\tableofcontents
+
+\mainmatter
+
+\include{dice}
+
+\include{classes}
+
+\end{document}