先给出结论，在 \(n\) 足够大时，期望近似于 \(\frac{n}{3}\)
纯数学推导
\[2*\frac{\sum_{l=1}^n\sum_{r=l}^n(r-l+1)}{n*n}
\]
先抛开2和分母
\[\sum_{l=1}^n\sum_{r=l}^n(r-l+1)
\]
\[\sum_{l=1}^n\sum_{r=l}^n r-\sum_{l=1}^n\sum_{r=l}^n l+\sum_{l=1}^n\sum_{r=l}^n1
\]
\[\sum_{i=1}^n i^2-\sum_{i=1}^ni*(n-i+1)+\frac{n(n+1)}{2}
\]
\[2*\sum_{i=1}^ni^2-\sum_{i=1}^ni*n
\]
\[\frac{n(n+1)(2n+1)}{3}-\frac{n^2(n+1)}{2}
\]
整体乘6，拆开
\[4n^3+6n^2+2n-2n^3-3n^2\rightarrow n^3+3n^2+2n\rightarrow n(n+2)(n+1)
\]
代回原柿
\[2*\frac{n(n+2)(n+1)}{6n^2}
\]
最终得到
\[\frac{(n+2)(n+1)}{3n}
\]