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margin of success

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HeNine 2 years ago
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dice.tex
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@ -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}} \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} \end{figure}
\subsection{Probability of Success} \subsection{Probability of Success}\label{sec:pos}
\begin{multicols}{2} \begin{multicols}{2}
The first question to ask is: ``What is the probability of success, given a fixed \stat{DT}?'' 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. \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} \end{multicols}
@ -78,7 +78,7 @@
\begin{table}[htb] \begin{table}[htb]
\centering \centering
\begin{tabular}{r | r r c l} \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 \\ 1 & 98\% & 51 & in & 52 \\
2 & 95\% & 20 & in & 21 \\ 2 & 95\% & 20 & in & 21 \\
3 & 90\% & 8 & in & 9 \\ 3 & 90\% & 8 & in & 9 \\
@ -104,7 +104,7 @@
% \caption{Expected chance of success.\label{tab:cos2}} % \caption{Expected chance of success.\label{tab:cos2}}
% \end{figure} % \end{figure}
\subsection{Expected Margin of Success} \subsection{Margin of Success}
\begin{multicols}{2} \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. 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}} \caption{Expected number of successes.\label{tab:exp_mos}}
\end{table} \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} \section{Combat Pool Steady State}
\begin{multicols}{2} \begin{multicols}{2}
@ -152,10 +496,10 @@
Using assumptions in \cref{equ:dice_ss_a1,equ:dice_ss_a2} we write the following equations: Using assumptions in \cref{equ:dice_ss_a1,equ:dice_ss_a2} we write the following equations:
\begin{align} \begin{align}
\text E[d] & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\ \text E[d] & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\
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] & = \frac{\stat{con}}{p_f}
\end{align} \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} \end{multicols}

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