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286 lines
22 KiB
TeX
286 lines
22 KiB
TeX
\chapter{Dice}
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\section{Basics of Dice}
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\subsection{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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\subsection{Probability of Success}\label{sec:pos}
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\begin{multicols}{2}
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The first question to ask is: ``What is the probability of success, given a fixed \stat{DT}?''
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We will use the binomial distribution $\text{B}(20,7/20)$ for the size of the dice pool, to compute the expectation. This is a fairly arbitrary choice that assumes 7 is the average pool size. However, it allows us to analyze the probabilities regardless of what the actual current size of the pool is.
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\Cref{tab:cos} shows the probability of success for a given \stat{DT} as well as number of successes in a number of attempts.
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\end{multicols}
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\begin{table}[htb]
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\centering
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\begin{tabular}{r | r r c l}
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\stat{DT} & $p$ & \multicolumn{3}{c}{Succeeds} \\ \cline{2-5}
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1 & 98\% & 51 & in & 52 \\
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2 & 95\% & 20 & in & 21 \\
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3 & 90\% & 8 & in & 9 \\
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4 & 82\% & 4 & in & 5 \\
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5 & 70\% & 2 & in & 3 \\
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6 & 56\% & 1 & in & 2 \\
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7 & 42\% & 2 & in & 5 \\
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8 & 28\% & 1 & in & 4 \\
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9 & 17\% & 1 & in & 6 \\
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10 & 9\% & 1 & in & 12 \\
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11 & 4\% & 1 & in & 26 \\
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\end{tabular}
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\caption{Expected chance of success.\label{tab:cos}}
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\end{table}
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\subsection{Margin of Success}
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\begin{multicols}{2}
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We also want to know what the expected margin of success is, given that the player succeeded in the roll, $\text{E}\left[M | M \geq 0\right]$. \Cref{tab:exp_mos} shows the expectations.
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Of note are the expected numbers of successes for \stat{DT} 11. The requirement for critical successes there causes a large variance in results. It is also important to remember that this does not show the full range of results. The highest possible result is $2d$, which for one die is the only possible (successful) result, while other numbers of dice have wider ranges.
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The variance on the upper half of the \stat{DT} range is very low. Even large dice pools can only expect 1 or 2 successes. With a drop in \stat{DT} the dice pool really comes into its own and can be expected to produce more successes.
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This means that building a large dice pool is not enough to ensure a large number of successes. The player must actually reduce the \stat{DT} to get a high likelihood of more successes.
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\end{multicols}
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\begin{table}[htb]
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\centering
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\begin{tabular}{r | S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1]}
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& \multicolumn{10}{c}{Difficulty Threshold} \\
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$d$ & \multicolumn{1}{c}{1} & \multicolumn{1}{c}{2} & \multicolumn{1}{c}{3} & \multicolumn{1}{c}{4} & \multicolumn{1}{c}{5} & \multicolumn{1}{c}{6} & \multicolumn{1}{c}{7} & \multicolumn{1}{c}{8} & \multicolumn{1}{c}{9} & \multicolumn{1}{c}{10} & \multicolumn{1}{c}{11} \\ \cline{2-12}
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1 & 1.1 & 1.1 & 1.1 & 1.1 & 1.1 & 1.2 & 1.2 & 1.2 & 1.3 & 1.5 & 2.0 \\
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2 & 2.1 & 1.8 & 1.5 & 1.3 & 1.1 & 1.0 & 0.8 & 0.7 & 0.7 & 0.7 & 1.0 \\
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3 & 2.6 & 2.3 & 2.1 & 1.9 & 1.8 & 1.6 & 1.5 & 1.3 & 1.1 & 0.7 & 0.3 \\
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4 & 3.5 & 3.1 & 2.7 & 2.3 & 1.9 & 1.6 & 1.4 & 1.2 & 1.1 & 1.2 & 2.0 \\
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5 & 4.4 & 3.7 & 3.1 & 2.7 & 2.3 & 2.0 & 1.7 & 1.4 & 1.1 & 0.9 & 1.1 \\
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6 & 5.1 & 4.3 & 3.6 & 3.0 & 2.5 & 2.1 & 1.7 & 1.5 & 1.3 & 1.1 & 0.5 \\
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7 & 6.0 & 5.0 & 4.1 & 3.4 & 2.8 & 2.3 & 2.0 & 1.6 & 1.3 & 1.2 & 1.9 \\
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8 & 6.8 & 5.6 & 4.6 & 3.7 & 3.0 & 2.5 & 2.0 & 1.7 & 1.4 & 1.1 & 1.0 \\
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9 & 7.5 & 6.2 & 5.1 & 4.1 & 3.3 & 2.7 & 2.2 & 1.8 & 1.4 & 1.2 & 0.6 \\
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10 & 8.4 & 6.9 & 5.5 & 4.4 & 3.5 & 2.8 & 2.2 & 1.8 & 1.5 & 1.2 & 1.8 \\
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11 & 9.2 & 7.5 & 6.0 & 4.8 & 3.7 & 3.0 & 2.4 & 1.9 & 1.5 & 1.2 & 1.0 \\
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12 & 10.0 & 8.2 & 6.5 & 5.1 & 4.0 & 3.1 & 2.4 & 1.9 & 1.6 & 1.3 & 0.7
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\end{tabular}
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\caption{Expected number of successes, given the test was successful.\label{tab:exp_mos}}
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\end{table}
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\begin{table}[htb]
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\centering
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\begin{tabular}{r | S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1]}
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& \multicolumn{10}{c}{Difficulty Threshold} \\
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$d$ & \multicolumn{1}{c}{1} & \multicolumn{1}{c}{2} & \multicolumn{1}{c}{3} & \multicolumn{1}{c}{4} & \multicolumn{1}{c}{5} & \multicolumn{1}{c}{6} & \multicolumn{1}{c}{7} & \multicolumn{1}{c}{8} & \multicolumn{1}{c}{9} & \multicolumn{1}{c}{10} & \multicolumn{1}{c}{11} \\ \cline{2-12}
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1 & 1.0 & 0.9 & 0.8 & 0.8 & 0.7 & 0.6 & 0.5 & 0.4 & 0.3 & 0.2 & 0.2 \\
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2 & 1.8 & 1.5 & 1.3 & 1.0 & 0.8 & 0.7 & 0.5 & 0.4 & 0.3 & 0.2 & 0.2 \\
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3 & 2.6 & 2.2 & 1.8 & 1.4 & 1.1 & 0.9 & 0.6 & 0.4 & 0.3 & 0.2 & 0.1 \\
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4 & 3.4 & 2.8 & 2.3 & 1.8 & 1.3 & 1.0 & 0.7 & 0.4 & 0.3 & 0.2 & 0.1 \\
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5 & 4.2 & 3.5 & 2.7 & 2.1 & 1.6 & 1.1 & 0.7 & 0.4 & 0.2 & 0.1 & 0.1 \\
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6 & 5.0 & 4.1 & 3.2 & 2.4 & 1.8 & 1.2 & 0.8 & 0.4 & 0.2 & 0.1 & 0.0 \\
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7 & 5.9 & 4.8 & 3.7 & 2.8 & 2.0 & 1.3 & 0.8 & 0.4 & 0.2 & 0.1 & 0.0 \\
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8 & 6.7 & 5.4 & 4.2 & 3.1 & 2.1 & 1.4 & 0.8 & 0.4 & 0.2 & 0.1 & 0.0 \\
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9 & 7.5 & 6.1 & 4.7 & 3.4 & 2.3 & 1.5 & 0.8 & 0.4 & 0.2 & 0.1 & 0.0 \\
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10 & 8.3 & 6.7 & 5.2 & 3.7 & 2.5 & 1.5 & 0.8 & 0.4 & 0.2 & 0.1 & 0.0 \\
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11 & 9.2 & 7.4 & 5.6 & 4.1 & 2.7 & 1.6 & 0.9 & 0.4 & 0.1 & 0.0 & 0.0 \\
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12 & 10.0 & 8.0 & 6.1 & 4.4 & 2.9 & 1.7 & 0.9 & 0.4 & 0.1 & 0.0 & 0.0 \\
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\end{tabular}
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\caption{Expected number of successes overall.\label{tab:exp_mosp}}
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\end{table}
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\begin{multicols}{2}
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Further, we may want to know more about the distribution of successes. Once again employing the distribution of dice pools from \cref{sec:pos} we can compute the quantiles (for $[0.1,0.25,0.5,0.75,0.9]$) in \cref{tab:quants}. Note that this is still conditioned on the whole roll being successful, i.e., we are considering only non-negative margins of success.
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Like with expectations, there is an anomaly in \stat{DT} 11, because only critical successes pass that \stat{DT}. Another thing worth noting is that for \stat{DT} 8--11, the median number of successes is 1. This means that even if the roll is successful, the margin is likely as not at most 1.
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\end{multicols}
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\begin{table}[ht]
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\centering
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\begin{tabular}{r | S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1] S[table-format=2.1]}
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\stat{DT} & $q_{10}$ & $q_{25}$ & $q_{50}$ & $q_{75}$ & $q_{90}$ \\ \cline{2-6}
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1 & 3.1 & 4.3 & 6.6 & 7.9 & 8.1 \\
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2 & 1.7 & 3.3 & 5.1 & 6.8 & 8.0 \\
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3 & 0.8 & 2.4 & 4.2 & 5.8 & 7.4 \\
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4 & 0.4 & 1.6 & 3.3 & 5.0 & 6.6 \\
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5 & 0.0 & 0.9 & 2.6 & 4.2 & 5.7 \\
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6 & 0.0 & 0.7 & 1.9 & 3.6 & 5.0 \\
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7 & 0.0 & 0.4 & 1.6 & 2.9 & 4.5 \\
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8 & 0.0 & 0.0 & 1.0 & 2.4 & 3.9 \\
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9 & 0.0 & 0.0 & 1.0 & 2.0 & 3.4 \\
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10 & 0.0 & 0.0 & 0.8 & 1.8 & 2.9 \\
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11 & 0.3 & 0.5 & 1.0 & 1.0 & 2.4 \\
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\end{tabular}
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\caption{Quantiles of the number of successes for a given \stat{DT}.\label{tab:quants}}
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\end{table}
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\subsection{Threshold--Margin Combinations}
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\begin{multicols}{2}
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It is useful to know how to give players a challenge with a certain chance of success. There is no simple formula for this, but in \cref{tab:dtm} we show, for 10\% increments, what combinations of \stat{DT} and margin give a given chance of passing the test. We are using the same distribution for dice pool sizes as in \cref{sec:pos}.
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We separated \cref{tab:dtm} into what we might call ``difficulty classes'', denoted by an intuitive chance of success. A quick glance at the numbers reveals that moving from once class to the next requires adding 1 to the \stat{DT} or adding 1--2 to the required margin.
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\end{multicols}
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\begin{table}[ht]
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\centering
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\begin{tabular}[t]{r|cc}
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Succeeds & \stat{DT} & $m$ \\ \hline
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19 in 20 & 1 & 0--1 \\
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& 2 & 0 \\
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9 in 10 & 1 & 2--3 \\
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& 2 & 1--2 \\
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& 3 & 0--1 \\
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4 in 5 & 1 & 4--5 \\
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& 2 & 3--4 \\
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& 3 & 2 \\
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& 4 & 0--1 \\
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& 5 & 0 \\
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1 in 2 & 1 & 6--7 \\
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& 2 & 5 \\
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& 3 & 3--4 \\
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& 4 & 2--3 \\
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& 5 & 1--2 \\
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& 6 & 0--1 \\
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& 7 & 0
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\end{tabular}
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\hfill
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\begin{tabular}[t]{r|cc}
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Succeeds & \stat{DT} & $m$ \\ \hline
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1 in 3 & 1 & 8 \\
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& 2 & 6--7 \\
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& 3 & 5--6 \\
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& 4 & 4--5 \\
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& 5 & 3--4 \\
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& 6 & 2--3 \\
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& 7 & 1--2 \\
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& 8 & 0 \\
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1 in 10 & 1 & 9--10 \\
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& 2 & 8--9 \\
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& 3 & 7--8 \\
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& 4 & 6--7 \\
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& 5 & 5--6 \\
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& 6 & 4--5 \\
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& 7 & 3 \\
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& 8 & 1--2 \\
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& 9 & 0--1 \\
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& 10 & 0
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\end{tabular}
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\hfill
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\begin{tabular}[t]{r|cc}
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Succeeds & \stat{DT} & $m$ \\ \hline
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1 in 20 & 1 & 11--12 \\
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& 2 & 10--11 \\
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& 3 & 9 --10 \\
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& 4 & 8--9 \\
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& 5 & 7--8 \\
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& 6 & 6--7 \\
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& 7 & 4--5 \\
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& 8 & 3--4 \\
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& 9 & 2--3 \\
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& 10 & 1--2 \\
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& 11 & 0 \\
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1 in 100 & 1 & 13 \\
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& 2 & 12 \\
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& 3 & 11 \\
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& 4 & 10 \\
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& 5 & 9 \\
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& 6 & 8 \\
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& 7 & 6 \\
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& 8 & 5 \\
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& 9 & 4 \\
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& 10 & 3 \\
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& 11 & 1 \\
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\end{tabular}
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\caption{Combinations of \stat{DT} and margin that give a certain chance of success.\label{tab:dtm}}
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\end{table}
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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 stat, 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 $\text E[d]$.
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We start by assuming that $\text E[d]$ is a number such that
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\begin{equation}
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\text E[d_{n+1}] = \text 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 $\text E[d_n]$ for any $n$ we can use
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\begin{equation}
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\text E[d_n] = \text E[d_{n-1}] - p_f \cdot \text 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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\text E[d] & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\
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p_f \cdot \text E[d] & = \stat{con} \\
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\text 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 to it on every turn.
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\end{multicols}
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