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I have the matrix:

  A     B     C
A 0.80  0.10  0.10
B 0.2   0.75  0.05
C 0.10  0.10  0.80

They ask me: if $ A $ is 40% right now, what's the probability of $A$ after two passes?

I proposed solving as

$$ \left( \begin{array}{ccc}0.4 & b & (0.6-b)\end{array} \right) \left( \begin{array}{ccc} 0.8 & 0.1 & 0.1 \\ 0.2 & 0.75 & 0.05 \\ 0.1 & 0.1 & 0.8 \end{array} \right)^2 $$

But I can't find how to get $ P_0(B)=b $.

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Suppose that the matrix squaring is done. So, $$ \left(\begin{array} \ P_2(A)&P_2(B)&P_2(C) \end{array}\right)=$$ $$=\left( \begin{array}{ccc}0.4 & b & (0.6-b)\end{array} \right) \left( \begin{array}{ccc} 0.8 & 0.1 & 0.1 \\ 0.2 & 0.75 & 0.05 \\ 0.1 & 0.1 & 0.8 \end{array} \right)^2=$$ $$=\left( \begin{array}{ccc}0.4 & b & (0.6-b)\end{array} \right) \left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)=$$ $$=\left( \begin{array}{ccc} 0.4\cdot a_{1,1} + b\cdot a_{2,1} + (0.6-b)\cdot a_{3,1} \\ 0.4\cdot a_{1,2} + b\cdot a_{2,2} + (0.6-b)\cdot a_{3,2} \\ 0.4\cdot a_{1,3} + b\cdot a_{2,3} + (0.6-b)\cdot a_{3,3} \end{array} \right)^T.$$ From here

$$P_2(A)=0.4\cdot a_{1,1} + b\cdot a_{2,1} + (0.6-b)\cdot a_{3,1}$$ and indeed, $0\le b \le 0.6$ is unknown. But this means only that $P_2(A)$ depends on $P_0(B)=b$. This is not counter intuitive.

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