diff --git a/dice.tex b/dice.tex
index 11a8bec..9cf1e2c 100644
--- a/dice.tex
+++ b/dice.tex
@@ -107,8 +107,8 @@
 \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. 
-    
+    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.
 \end{multicols}
 
@@ -136,25 +136,25 @@
 \section{Combat Pool Steady State}
 
 \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}
-        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}
     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}
-        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}
     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:
     \begin{align}
-        E[d]          & = E[d] - p_f \cdot E[d] + \stat{con} \\
-        p_f\cdot E[d] & = \stat{con}                         \\
-        E[d]          & = \frac{\stat{con}}{p_f}
+        \text E[d]          & = \text E[d] - p_f \cdot \text E[d] + \stat{con} \\
+        p_f\cdot \text E[d] & = \stat{con}                         \\
+        \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.