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长列表的格式化

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长列表的格式化

自从几周前我上次发帖以来，我已经写了大约 10 个小时的 LaTeX。我想我已经掌握了窍门。我正在完成家庭作业，我的文档看起来有点混乱。有没有更好的方法来缩进长列表？

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}


\begin{document}

\textbf{6)} Suppose we roll a fair 6 sided die with the number [1,6] written on them. After the first die roll we roll the die $k$ times where $k$ is the number on the first die roll. The number of points you score is the sum of the face-values on all die rolls (including the first). What is the expected number of points you will score?

There are six different cases for the first roll 

\textbf{Case 1: Roll 1}
\begin{itemize}
    \item $P(D_1=1) =\frac{1}{6}$ 
    \item The expected number on the next roll is $E(D_2)=\frac{1}{6}\times(1+2+3+4+5+6)=3.5$
    \item $E(S_1) = 3.5+1=4.5$ points
\end{itemize}


\textbf{Case 2: Roll 2}

\begin{itemize}
    \item We know what the expected value of one roll is, and since rolling the die is independent we can use the previous expected value for the next roll.  
    \item $E(S_2) = (3.5\times 2)+2=9$ points
\end{itemize}

\textbf{Case 3: Roll 3}
\begin{itemize}
\item$E(S_3) = (3.5\times 3)+3=13.5$ points
\end{itemize}

\textbf{Case 4: Roll 4}

\begin{itemize}
    \item $E(S_4) = (3.5\times 4)+4=18$ points
\end{itemize}

\textbf{Case 5: Roll 5}

\begin{itemize}
    \item $E(S_5) = (3.5\times 5)+5=22.5$ points
\end{itemize}

\textbf{Case 6: Roll 6}

\begin{itemize}
    \item $E(S_6) = (3.5\times 6)+6=27$ points
\end{itemize}
Therefore, the expected number of points scored is
\begin{align}
    =\frac{1}{6}\times(4.5+9+13.5+18+22.5+27)=15.75 \notag
\end{align}
\end{document}

答案1

这个怎么样？里面有enumerate一张单曲。itemize

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}
\usepackage{enumitem}
\usepackage{showframe}
\begin{document}

\textbf{6)} Suppose we roll a fair 6 sided die with the number [1,6] written on them. After the first die roll we roll the die $k$ times where $k$ is the number on the first die roll. The number of points you score is the sum of the face-values on all die rolls (including the first). What is the expected number of points you will score?

There are six different cases for the first roll 

\begin{enumerate}[label={\bfseries Case \arabic*:},leftmargin=1.5cm]
\item Roll 1
\begin{itemize}
    \item $P(D_1=1) =\frac{1}{6}$ 
    \item The expected number on the next roll is $E(D_2)=\frac{1}{6}\times(1+2+3+4+5+6)=3.5$
    \item $E(S_1) = 3.5+1=4.5$ points
\end{itemize}


\item Roll 2

\begin{itemize}
    \item We know what the expected value of one roll is, and since rolling the die is independent we can use the previous expected value for the next roll.  
    \item $E(S_2) = (3.5\times 2)+2=9$ points
\end{itemize}

\item Roll 3
\begin{itemize}
\item$E(S_3) = (3.5\times 3)+3=13.5$ points
\end{itemize}

\item Roll 4
\begin{itemize}
    \item $E(S_4) = (3.5\times 4)+4=18$ points
\end{itemize}

\item Roll 5
\begin{itemize}
    \item $E(S_5) = (3.5\times 5)+5=22.5$ points
\end{itemize}

\item Roll 6
\begin{itemize}
    \item $E(S_6) = (3.5\times 6)+6=27$ points
\end{itemize}

\end{enumerate}

Therefore, the expected number of points scored is
\begin{align}
    =\frac{1}{6}\times(4.5+9+13.5+18+22.5+27)=15.75 \notag
\end{align}

\end{document}

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