先看完整的一輪之後，各水量的流動
亦即，依照 \(A\rightarrow B\rightarrow C\rightarrow A\) 每次都倒一半給下一桶，
水量流動如下：
\((a,b,c) \rightarrow (\frac{1}{2} a, \frac{1}{2}a+b, c) \rightarrow (\frac{1}{2} a, \frac
{1}{4} a+\frac{1}{2}b, \frac{1}{4} a+\frac{1}{2}b +c) \rightarrow (\frac{5}{8}a+\frac{1}{4}b+\frac{1}{2}c,\frac
{1}{4} a+\frac{1}{2}b , \frac{1}{8}a+ \frac{1}{4}b+\frac{1}{2}c)\)
所以，
轉移矩陣為
\(\displaystyle \left[ \begin{array}{*{20}{c}}
\frac{5}{8} & \frac{1}{4} & \frac{1}{2} \\
\frac{1}{4} & \frac{1}{2} & 0 \\
\frac{1}{8} & \frac{1}{4} & \frac{1}{2} \\
\end{array}\right]\)
達穩定狀態時，
\(\displaystyle \left[\begin{array}{*{20}{c}}
\frac{5}{8} & \frac{1}{4} & \frac{1}{2} \\
\frac{1}{4} & \frac{1}{2} & 0 \\
\frac{1}{8} & \frac{1}{4} & \frac{1}{2} \\
\end{array} \right]\left[ \begin{array}{*{20}{c}}
a \\
b \\
c \\
\end{array} \right] = \left[ \begin{array}{*{20}{c}}
a \\
b \\
c \\
\end{array} \right]\)
可以解得
\(a:b:c=2:1:1.\)
或是，另解
轉移矩陣為
\(\displaystyle \left[ \begin{array}{*{20}{c}}
1 & 0 & \frac{1}{2} \\
0 & 1 & 0 \\
0 & 0 & \frac{1}{2} \\
\end{array} \right]\left[ \begin{array}{*{20}{c}}
1 & 0 & 0 \\
0 & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & 1 \\
\end{array} \right]\left[ \begin{array}{*{20}{c}}
\frac{1}{2} & 0 & 0 \\
\frac{1}{2} & 1 & 0 \\
0 & 0 & 1 \\
\end{array} \right] = \left[ \begin{array}{*{20}{c}}
\frac{5}{8} & \frac{1}{4} & \frac{1}{2} \\
\frac{1}{4} & \frac{1}{2} & 0 \\
\frac{1}{8} & \frac{1}{4} & \frac{1}{2} \\
\end{array} \right].\)
^__^
