dice and classes
HeNine
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\chapter{Classes}
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\section*{Basic Class Structure}
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\subsection*{Combat}
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\paragraph{Stats}
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Each class grants +1 in three separate stats; no two classes grant the exact same stats.
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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.
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\paragraph{Standard Weapons}
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The standard weapons are purely for flavor. They act as a guide to how that class should feel.
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\paragraph{Weapon Proficiencies}
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Weapon proficiencies are chosen such that they include the standard weapons.
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\paragraph{Armor proficiencies}
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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.
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Note that every class can wear no armor, e.g., normal clothes.
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\paragraph{Skills}
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The number of skills should roughly reflect the mental stats a class grants. More support-focused classes should grant more skills.
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\paragraph{Abilities}
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Each class group has one ability that all classes in that group share. Other than that, each class should have at least 4 abilities.
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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.
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\chapter{Dice}
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\section{Distribution of Successes}
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\begin{multicols}{2}
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For further analysis it will be useful to have a distribution function for margins of success.
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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
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\begin{align}
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\left(C_s, S, F, C_f\right) \sim \text{MN}\left(d, 4, p\right)\label{equ:dice_groups} \\
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p = \left[1/12,\frac{12-\stat{dt}-1}{12},\frac{\stat{dt}-1}{12},\frac{1}{12}\right]
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\end{align}
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We can easily map a tuple to a success margin:
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\begin{equation}
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M = 2C_s + S - F - 2C_f
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\end{equation}
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Therefore, our distribution for $M$ is a sum over all tuples that produce the correct $m$
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\begin{equation}
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Pr(M = m) = \sum_{C_s,S,F,C_f; M = m} \text{MN}\left(d, 4, p\right)
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\end{equation}
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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.
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We start with two equations
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\begin{align}
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2c_s + s - f - 2c_f & = m \\
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c_s + s + f + c_f & = d
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\end{align}
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which resolve into a two-dimensional system
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\begin{align}
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c_s & = m - d + 2f + 3c_f \\
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s & = -m + 2d - 3f - 4c_f
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\end{align}
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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 $m$.
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\begin{align}
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\frac{m-d+2f}{-3} \leq c_f \leq \frac{-m+2d-3f}{4}
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\end{align}
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Because the lower bound is not necessarily greater than 0, we will also use $c_f\geq0$.
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We can now also constrain $f$ by observing that the lower bound for $c_f$ can cross $c_f=0$ before $f=d$.
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\begin{equation}
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0 \leq f \leq \frac{2d-m}{3}
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\end{equation}
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Putting the sum together, and adding floors and ceilings to limit the sum to integers, we get:
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\begin{align}
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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} \\
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&\quad\; \frac{d!}{c_s!s!f!c_f!}
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\frac{n_s^s n_f^f}{12^{d}} \\
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n_f & = \stat{dt}-1 \\
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n_s & = 12-n_s
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\end{align}
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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.
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\end{multicols}
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\begin{figure}[htb]
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\centering
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\includegraphics{images/dice_candidates_example.pdf}
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\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}}
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\end{figure}
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\begin{figure}[htb]
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\centering
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\begin{tabular}{r | c c c c c c c c}
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& \multicolumn{8}{c}{Difficulty Threshold} \\
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d & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \cline{2-9}
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1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\
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2 & 2 & 2 & 1 & 1 & 1 & 1 & 0 & 0 \\
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3 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 \\
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4 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 0 \\
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5 & 5 & 4 & 3 & 3 & 2 & 1 & 1 & 1 \\
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6 & 6 & 5 & 4 & 3 & 2 & 2 & 1 & 1 \\
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7 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 1 \\
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8 & 8 & 7 & 5 & 4 & 3 & 2 & 1 & 1 \\
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9 & 9 & 8 & 6 & 5 & 3 & 2 & 1 & 1 \\
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10 & 10 & 8 & 7 & 5 & 4 & 2 & 1 & 1 \\
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\end{tabular}
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\caption{Expected number of successes.\label{tab:exp_mos}}
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\end{figure}
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\section{Combat Pool Steady State}
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\begin{multicols}{2}
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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 $E[d]$.
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We start by assuming that $E[d]$ is a number such that
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\begin{equation}
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E[d_{n+1}] = E[d_{n}] \label{equ:dice_ss_a1}
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\end{equation}
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where $n$ is the turn after which the steady state is achieved.
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To compute $E[d_n]$ for any $n$ we can use
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\begin{equation}
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E[d_n] = E[d_{n-1}] - p_f \cdot E[d_{n-1}] + \stat{con} \label{equ:dice_ss_a2}
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\end{equation}
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where $p_f = \stat{dt}/12$ is the probability of failure.
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Using assumptions in \cref{equ:dice_ss_a1,equ:dice_ss_a2} we write the following equations:
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\begin{align}
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E[d] & = E[d] - p_f \cdot E[d] + \stat{con} \\
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p_f\cdot E[d] & = \stat{con} \\
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E[d] & = \frac{\stat{con}}{p_f}
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\end{align}
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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.
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\end{multicols}
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@ -0,0 +1,25 @@
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\documentclass{kartsstyle}
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\usepackage{multirow}
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\title{On The Metaphysics Of Savits}
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\author{\texttt{HeNine}}
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\date{2023}
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\newcommand{\tochange}[1]{\textcolor{red}{#1}}
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\begin{document}
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\frontmatter
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\maketitle
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\tableofcontents
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\mainmatter
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\include{dice}
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\include{classes}
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\end{document}
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