metaphysics/dice.tex

\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}