For further analysis it will be useful to have a distribution function for margins of success.
To obtain such a distribution we start by observing that each roll can be divided into four groups: critical successes, successes, failures and critical failures, as shown in \cref{equ:dice_groups}. Further, this tuple has a multinomial distribution
p = \left[1/12,\frac{12-\stat{dt}-1}{12},\frac{\stat{dt}-1}{12},\frac{1}{12}\right]
\end{align}
We can easily map a tuple to a success margin:
\begin{equation}
M = 2C_s + S - F - 2C_f
\end{equation}
Therefore, our distribution for $M$ is a sum over all tuples that produce the correct $m$
\begin{equation}
Pr(M = m) = \sum_{C_s,S,F,C_f; M = m}\text{MN}\left(d, 4, p\right)
\end{equation}
Instead of trying all combinations of variables and summing the correct ones, we can use a system of Diophantine equations to reduce the search space.
We start with two equations
\begin{align}
2c_s + s - f - 2c_f & = m \\
c_s + s + f + c_f & = d
\end{align}
which resolve into a two-dimensional system
\begin{align}
c_s & = m - d + 2f + 3c_f \\
s & = -m + 2d - 3f - 4c_f
\end{align}
Next, we use inequalities $c_s\geq0, s\geq0$ to constrain the space for $c_f$, based on $f$. An example plot in \cref{fig:candidates} graphically shows the area defined by these equations and inequalities for one combination of $d$ and $m$.
\begin{align}
\frac{m-d+2f}{-3}\leq c_f \leq\frac{-m+2d-3f}{4}
\end{align}
Because the lower bound is not necessarily greater than 0, we will also use $c_f\geq0$.
We can now also constrain $f$ by observing that the lower bound for $c_f$ can cross $c_f=0$ before $f=d$.
\begin{equation}
0 \leq f \leq\frac{2d-m}{3}
\end{equation}
Putting the sum together, and adding floors and ceilings to limit the sum to integers, we get:
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.
\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}}
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.
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.
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.
The variance on the upper half of the \stat{DT} range is very low. Even large dice pools can only expect 1 or 2 successes. With a drop in \stat{DT} the dice pool really comes into its own and can be expected to produce more successes.
This means that building a large dice pool is not enough to ensure a large number of successes. The player must actually reduce the \stat{DT} to get a high likelihood of more successes.
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.
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 $\text E[d]$.
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.
