Suppose that we have a state space $\mathcal{X}=\left\{A,B,C,D,E\right\}$ and a markov chain $\left\{X_{n}\right\}$ that modelize the transition of a mice with probability matrix
$P:= \begin{bmatrix}0 & 1/2 &1/2 &0 &0\\1/2 &0&1/2&0&0 \\ 1/4&1/4&0&1/4&1/4 \\0& 0&1&0&0 \\ 0&0&1&0&0 \end{bmatrix}$
Assume that our mice starts at state $E$ and there is also a second mice which starts at state $B$ and I want to find the probability of the two mices to meet each other.
So I define a new markov chain $\left\{Y_{n}\right\}$ which will have the same probability matrix as $X_{n}$ and there initial distributions will differ.
$\pi^{Y_{n}}_{0}=(0,1,0,0,0)$ and $\pi_{0}^{X_{n}}=(0,0,0,0,1)$
So I suppose that we have to caclculate the probability $\mathbb{P}(Y_{n}=x,X_{n}=x)$ for $x\in \mathcal{X}$
But my question here is , can we assume that $Y_{n}$ and $X_{n}$ are independent, in order to rewrite $\mathbb{P}(Y_{n}=x,X_{n}=x)= \mathbb{P}(Y_{n}=x)\mathbb{P}(X_{n}=x)$
I am not the greatest expert on Markov chains, but I would say your problem admits a simple solution through Kronecker products. You may define the double mice $(M_1,M_2)$ and its random walk over the states $\{A,B,C,D,E\}\times\{A,B,C,D,E\}$. The associated transition matrix is just $Q=P\otimes P$ and the wanted probability is the probability of lying in $\{(A,A),(B,B),(C,C),(D,D),(E,E)\}$ after $n$ steps, which can be computed from $Q^n$.