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HeNine 2 years ago
parent acdeb9c490
commit b93c1cc72e

@ -107,8 +107,8 @@
\subsection{Expected Margin of Success} \subsection{Expected 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.
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. 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.
\end{multicols} \end{multicols}
@ -136,25 +136,25 @@
\section{Combat Pool Steady State} \section{Combat Pool Steady State}
\begin{multicols}{2} \begin{multicols}{2}
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 $E[d]$. 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 $\text E[d]$.
We start by assuming that $E[d]$ is a number such that We start by assuming that $\text E[d]$ is a number such that
\begin{equation} \begin{equation}
E[d_{n+1}] = E[d_{n}] \label{equ:dice_ss_a1} \text E[d_{n+1}] = \text E[d_{n}] \label{equ:dice_ss_a1}
\end{equation} \end{equation}
where $n$ is the turn after which the steady state is achieved. where $n$ is the turn after which the steady state is achieved.
To compute $E[d_n]$ for any $n$ we can use To compute $\text E[d_n]$ for any $n$ we can use
\begin{equation} \begin{equation}
E[d_n] = E[d_{n-1}] - p_f \cdot E[d_{n-1}] + \stat{con} \label{equ:dice_ss_a2} \text E[d_n] = \text E[d_{n-1}] - p_f \cdot \text E[d_{n-1}] + \stat{con} \label{equ:dice_ss_a2}
\end{equation} \end{equation}
where $p_f = \stat{dt}/12$ is the probability of failure. where $p_f = \stat{dt}/12$ is the probability of failure.
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}
E[d] & = E[d] - p_f \cdot E[d] + \stat{con} \\ \text E[d] & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\
p_f\cdot E[d] & = \stat{con} \\ p_f\cdot \text E[d] & = \stat{con} \\
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 toward it on every turn.

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