diff --git a/dice.tex b/dice.tex index 9cf1e2c..df5ce53 100644 --- a/dice.tex +++ b/dice.tex @@ -65,12 +65,12 @@ \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} -\subsection{Probability of Success} +\subsection{Probability of Success}\label{sec:pos} \begin{multicols}{2} The first question to ask is: ``What is the probability of success, given a fixed \stat{DT}?'' - 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. + 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. \Cref{tab:cos} shows the probability of success for a given \stat{DT} as well as number of successes in a number of attempts. \end{multicols} @@ -78,7 +78,7 @@ \begin{table}[htb] \centering \begin{tabular}{r | r r c l} - \stat{DT} & $p$ & \multicolumn{3}{c}{Succeeds} \\ + \stat{DT} & $p$ & \multicolumn{3}{c}{Succeeds} \\ \cline{2-5} 1 & 98\% & 51 & in & 52 \\ 2 & 95\% & 20 & in & 21 \\ 3 & 90\% & 8 & in & 9 \\ @@ -104,7 +104,7 @@ % \caption{Expected chance of success.\label{tab:cos2}} % \end{figure} -\subsection{Expected Margin of Success} +\subsection{Margin of Success} \begin{multicols}{2} 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. @@ -133,6 +133,350 @@ \caption{Expected number of successes.\label{tab:exp_mos}} \end{table} +\begin{multicols}{2} + 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. + + 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. +\end{multicols} + +\begin{table}[ht] + \centering + \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]} + \stat{DT} & $q_{10}$ & $q_{25}$ & $q_{50}$ & $q_{75}$ & $q_{90}$ \\ \cline{2-6} + 1 & 3.1 & 4.3 & 6.6 & 7.9 & 8.1 \\ + 2 & 1.7 & 3.3 & 5.1 & 6.8 & 8.0 \\ + 3 & 0.8 & 2.4 & 4.2 & 5.8 & 7.4 \\ + 4 & 0.4 & 1.6 & 3.3 & 5.0 & 6.6 \\ + 5 & 0.0 & 0.9 & 2.6 & 4.2 & 5.7 \\ + 6 & 0.0 & 0.7 & 1.9 & 3.6 & 5.0 \\ + 7 & 0.0 & 0.4 & 1.6 & 2.9 & 4.5 \\ + 8 & 0.0 & 0.0 & 1.0 & 2.4 & 3.9 \\ + 9 & 0.0 & 0.0 & 1.0 & 2.0 & 3.4 \\ + 10 & 0.0 & 0.0 & 0.8 & 1.8 & 2.9 \\ + 11 & 0.3 & 0.5 & 1.0 & 1.0 & 2.4 \\ + \end{tabular} + + \caption{Quantiles of the number of successes for a given \stat{DT}.\label{tab:quants}} +\end{table} + +\subsection{Threshold--Margin Combinations} + +\begin{multicols}{2} + 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}. + + 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. +\end{multicols} + + +% \begin{table}[htb] +% \centering +% \begin{tabular}[t]{c | c c } +% $p$ & \stat{DT} & $m$ \\ +% \multirow[t]{3}{*}{100\%} & 1 & 0 \\ +% & 1 & 1 \\ +% & 2 & 0 \\ +% \multirow[t]{5}{*}{90\%} & 1 & 2 \\ +% & 1 & 3 \\ +% & 2 & 1 \\ +% & 2 & 2 \\ +% & 3 & 0 \\ +% \multirow[t]{4}{*}{80\%} & 1 & 4 \\ +% & 2 & 3 \\ +% & 3 & 1 \\ +% & 4 & 0 \\ +% \multirow[t]{5}{*}{70\%} & 1 & 5 \\ +% & 2 & 4 \\ +% & 3 & 2 \\ +% & 4 & 1 \\ +% & 5 & 0 +% \end{tabular} +% \hfill +% \begin{tabular}[t]{c | c c} +% $p$ & \stat{DT} & $m$ \\ +% \multirow[t]{5}{*}{60\%} & 1 & 6 \\ +% & 3 & 3 \\ +% & 4 & 2 \\ +% & 5 & 1 \\ +% & 6 & 0 \\ +% \multirow[t]{4}{*}{50\%} & 2 & 5 \\ +% & 3 & 4 \\ +% & 4 & 3 \\ +% & 5 & 2 \\ +% \multirow[t]{6}{*}{40\%} & 1 & 7 \\ +% & 2 & 6 \\ +% & 3 & 5 \\ +% & 4 & 4 \\ +% & 6 & 1 \\ +% & 7 & 0 \\ +% \end{tabular} +% \hfill +% \begin{tabular}[t]{c | c c} +% $p$ & \stat{DT} & $m$ \\ +% \multirow[t]{7}{*}{30\%} & 1 & 8 \\ +% & 2 & 7 \\ +% & 3 & 6 \\ +% & 5 & 3 \\ +% & 6 & 2 \\ +% & 7 & 1 \\ +% & 8 & 0 \\ +% \multirow[t]{10}{*}{20\%} & 1 & 9 \\ +% & 2 & 8 \\ +% & 3 & 7 \\ +% & 4 & 5 \\ +% & 4 & 6 \\ +% & 5 & 4 \\ +% & 6 & 3 \\ +% & 7 & 2 \\ +% & 8 & 1 \\ +% & 9 & 0 +% \end{tabular} +% \hfill +% \begin{tabular}[t]{c | c c} +% $p$ & \stat{DT} & $m$ \\ +% \multirow[t]{19}{*}{10\%} & 1 & 10 \\ +% & 1 & 11 \\ +% & 2 & 9 \\ +% & 2 & 10 \\ +% & 3 & 8 \\ +% & 3 & 9 \\ +% & 4 & 7 \\ +% & 4 & 8 \\ +% & 5 & 5 \\ +% & 5 & 6 \\ +% & 6 & 4 \\ +% & 6 & 5 \\ +% & 7 & 3 \\ +% & 7 & 4 \\ +% & 8 & 2 \\ +% & 8 & 3 \\ +% & 9 & 1 \\ +% & 9 & 2 \\ +% & 10 & 0 +% \end{tabular} + +% \caption{Combinations of \stat{DT} and margin that give a certain chance of success.\label{tab:dtm}} +% \end{table} + +\begin{table}[ht] + \centering + % \begin{tabular}[t]{r|cc} + % Succeeds & \stat{DT} & $m$ \\ \hline + % \multirow[t]{2}{*}{49 in 50} & 1 & 0 \\ + % & 1 & 1 \\ + % \multirow[t]{2}{*}{9 in 10} & 2 & 0 \\ + % & 1 & 2 \\ + % & 2 & 1 \\ + % & 3 & 0 \\ + % \multirow[t]{2}{*}{4 in 5} & 1 & 3 \\ + % & 3 & 1 \\ + % & 2 & 2 \\ + % & 1 & 4 \\ + % & 4 & 0 \\ + % \multirow[t]{2}{*}{2 in 3} & 2 & 3 \\ + % & 3 & 2 \\ + % & 4 & 1 \\ + % & 5 & 0 \\ + % & 1 & 5 \\ + % & 2 & 4 \\ + % & 3 & 3 \\ + % & 4 & 2 \\ + % \multirow[t]{2}{*}{1 in 2} & 5 & 1 \\ + % & 6 & 0 \\ + % & 1 & 6 \\ + % & 2 & 5 \\ + % & 3 & 4 \\ + % & 4 & 3 \\ + % & 5 & 2 \\ + % & 6 & 1 \\ + % & 7 & 0 \\ + % & 1 & 7 + % \end{tabular} + % \hfill + % \begin{tabular}[t]{r|cc} + % Succeeds & \stat{DT} & $m$ \\ \hline + + % \multirow[t]{2}{*}{1 in 3} & 2 & 6 \\ + % & 3 & 5 \\ + % & 4 & 4 \\ + % & 5 & 3 \\ + % & 6 & 2 \\ + % & 7 & 1 \\ + % & 1 & 8 \\ + % & 8 & 0 \\ + % & 2 & 7 \\ + % & 3 & 6 \\ + % & 4 & 5 \\ + % & 5 & 4 \\ + % \multirow[t]{2}{*}{1 in 5} & 6 & 3 \\ + % & 7 & 2 \\ + % & 8 & 1 \\ + % & 1 & 9 \\ + % & 2 & 8 \\ + % & 3 & 7 \\ + % & 9 & 0 \\ + % & 4 & 6 \\ + % & 5 & 5 \\ + % & 6 & 4 \\ + % & 7 & 3 \\ + % & 8 & 2 + % \end{tabular} + % \hfill + % \begin{tabular}[t]{rrr|cc} + % Succeeds & \stat{DT} & $m$ \\ \hline + + % \multirow[t]{2}{*}{1 in 20} & 1 & 10 \\ + % & 2 & 9 \\ + % & 9 & 1 \\ + % & 3 & 8 \\ + % & 4 & 7 \\ + % & 5 & 6 \\ + % & 10 & 0 \\ + % & 6 & 5 \\ + % & 7 & 4 \\ + % & 8 & 3 \\ + % & 9 & 2 \\ + % & 2 & 10 \\ + % & 1 & 11 \\ + % & 3 & 9 \\ + % & 4 & 8 \\ + % & 5 & 7 \\ + % & 10 & 1 \\ + % & 6 & 6 + % \end{tabular} + % \hfill + % \begin{tabular}[t]{rrr|cc} + % Succeeds & \stat{DT} & $m$ \\ \hline + % \multirow[t]{2}{*}{1 in 100} & 7 & 5 \\ + % & 11 & 0 \\ + % & 8 & 4 \\ + % & 9 & 3 \\ + % & 3 & 10 \\ + % & 2 & 11 \\ + % & 4 & 9 \\ + % & 1 & 12 \\ + % & 5 & 8 \\ + % & 10 & 2 \\ + % & 6 & 7 \\ + % & 7 & 6 \\ + % & 11 & 1 \\ + % & 8 & 5 \\ + % & 9 & 4 \\ + % & 3 & 11 \\ + % & 4 & 10 \\ + % & 2 & 12 \\ + % & 10 & 3 \\ + % & 5 & 9 \\ + % & 1 & 13 \\ + % & 6 & 8 + % \end{tabular} + + \begin{tabular}[t]{r|cc} + Succeeds & \stat{DT} & $m$ \\ \hline + 19 in 20 & 1 & 0--1 \\ + % & & 1 \\ + & 2 & 0 \\ + 9 in 10 & 1 & 2--3 \\ + % & & 3 \\ + & 2 & 1--2 \\ + % & & 2 \\ + & 3 & 0--1 \\ + % & & 1 \\ + 4 in 5 & 1 & 4--5 \\ + % & & 5 \\ + & 2 & 3--4 \\ + % & & 4 \\ + & 3 & 2 \\ + & 4 & 0--1 \\ + % & & 1 \\ + & 5 & 0 \\ + 1 in 2 & 1 & 6--7 \\ + % & & 7 \\ + & 2 & 5 \\ + & 3 & 3--4 \\ + % & & 4 \\ + & 4 & 2--3 \\ + % & & 3 \\ + & 5 & 1--2 \\ + % & & 2 \\ + & 6 & 0--1 \\ + % & & 1 \\ + & 7 & 0 + \end{tabular} + \hfill + \begin{tabular}[t]{r|cc} + Succeeds & \stat{DT} & $m$ \\ \hline + 1 in 3 & 1 & 8 \\ + & 2 & 6--7 \\ + % & & 7 \\ + & 3 & 5--6 \\ + % & & 6 \\ + & 4 & 4--5 \\ + % & & 5 \\ + & 5 & 3--4 \\ + % & & 4 \\ + & 6 & 2--3 \\ + % & & 3 \\ + & 7 & 1--2 \\ + % & & 2 \\ + & 8 & 0 \\ + 1 in 10 & 1 & 9--10 \\ + % & 1 & 10 \\ + & 2 & 8--9 \\ + % & & 9 \\ + & 3 & 7--8 \\ + % & & 8 \\ + & 4 & 6--7 \\ + % & & 7 \\ + & 5 & 5--6 \\ + % & & 6 \\ + & 6 & 4--5 \\ + % & & 5 \\ + & 7 & 3 \\ + & 8 & 1--2 \\ + % & & 2 \\ + & 9 & 0--1 \\ + % & & 1 \\ + & 10 & 0 + \end{tabular} + \hfill + \begin{tabular}[t]{r|cc} + Succeeds & \stat{DT} & $m$ \\ \hline + 1 in 20 & 1 & 11--12 \\ + % & & 12 \\ + & 2 & 10--11 \\ + % & & 11 \\ + & 3 & 9 --10 \\ + % & & 10 \\ + & 4 & 8--9 \\ + % & & 9 \\ + & 5 & 7--8 \\ + % & & 8 \\ + & 6 & 6--7 \\ + % & & 7 \\ + & 7 & 4--5 \\ + % & & 5 \\ + & 8 & 3--4 \\ + % & & 4 \\ + & 9 & 2--3 \\ + % & & 3 \\ + & 10 & 1--2 \\ + % & & 2 \\ + & 11 & 0 \\ + 1 in 100 & 1 & 13 \\ + & 2 & 12 \\ + & 3 & 11 \\ + & 4 & 10 \\ + & 5 & 9 \\ + & 6 & 8 \\ + & 7 & 6 \\ + & 8 & 5 \\ + & 9 & 4 \\ + & 10 & 3 \\ + & 11 & 1 \\ + \end{tabular} + \caption{Combinations of \stat{DT} and margin that give a certain chance of success.\label{tab:dtm}} +\end{table} + \section{Combat Pool Steady State} \begin{multicols}{2} @@ -152,10 +496,10 @@ Using assumptions in \cref{equ:dice_ss_a1,equ:dice_ss_a2} we write the following equations: \begin{align} - \text E[d] & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\ - p_f\cdot \text E[d] & = \stat{con} \\ - \text E[d] & = \frac{\stat{con}}{p_f} + \text E[d] & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\ + p_f \cdot \text E[d] & = \stat{con} \\ + \text 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. + 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. \end{multicols}