\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.
%\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, $\text{E}\left[M | M \geq0\right]$. \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.
\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}
@ -157,5 +501,5 @@
\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.