probability of success

main
HeNine 2 years ago
parent fc33db676a
commit e22154044d

@ -1,6 +1,8 @@
\chapter{Dice} \chapter{Dice}
\section{Distribution of Successes} \section{Basics of Dice}
\subsection{Distribution of Successes}
\begin{multicols}{2} \begin{multicols}{2}
For further analysis it will be useful to have a distribution function for margins of success. For further analysis it will be useful to have a distribution function for margins of success.
@ -48,9 +50,9 @@
Putting the sum together, and adding floors and ceilings to limit the sum to integers, we get: Putting the sum together, and adding floors and ceilings to limit the sum to integers, we get:
\begin{align} \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} \\ 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!} & \quad\; \frac{d!}{c_s!s!f!c_f!}
\frac{n_s^s n_f^f}{12^{d}} \\ \frac{n_s^s n_f^f}{12^{d}} \\
n_f & = \stat{dt}-1 \\ n_f & = \stat{dt}-1 \\
n_s & = 12-n_s n_s & = 12-n_s
\end{align} \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. 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.
@ -63,6 +65,53 @@
\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}
\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.
\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}
\begin{table}[htb]
\centering
\begin{tabular}{r | r r c l}
\stat{DT} & $p$ & \multicolumn{3}{c}{Succeeds} \\
1 & 98\% & 51 & in & 52 \\
2 & 95\% & 20 & in & 21 \\
3 & 90\% & 8 & in & 9 \\
4 & 82\% & 4 & in & 5 \\
5 & 70\% & 2 & in & 3 \\
6 & 56\% & 1 & in & 2 \\
7 & 42\% & 2 & in & 5 \\
8 & 28\% & 1 & in & 4 \\
9 & 17\% & 1 & in & 6 \\
10 & 9\% & 1 & in & 12 \\
11 & 4\% & 1 & in & 26 \\
\end{tabular}
\caption{Expected chance of success.\label{tab:cos}}
\end{table}
% \begin{figure}[htb]
% \centering
% \begin{tabular}{r c c c c c c c c c c c}
% \stat{DT} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline
% $p$ & 98\% & 95\% & 90\% & 82\% & 70\% & 56\% & 42\% & 28\% & 17\% & 9\% & 4\% \\
% Succeeds & 51 in 52 & 20 in 21 & 8 in 9 & 4 in 5 & 2 in 3 & 1 in 2 & 2 in 5 & 1 in 4 & 1 in 6 & 1 in 12 & 1 in 26
% \end{tabular}
% \caption{Expected chance of success.\label{tab:cos2}}
% \end{figure}
\subsection{Expected 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.
Of note are the expected numbers of successes for \stat{DT} 11. The requirement for critical successes there causes a large variance in results.
\end{multicols}
\begin{figure}[htb] \begin{figure}[htb]
\centering \centering
\begin{tabular}{r | c c c c c c c c} \begin{tabular}{r | c c c c c c c c}

Loading…
Cancel
Save