Can someone please explain to me how these two absorbing matrices,
$ M_1 = \left[ \begin{array}{ccccc} \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\ \frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 \\ \frac{2}{3} & 0 & 0 & \frac{1}{3} & 0 \\ \frac{1}{3} & 0 & 0 & 0 & \frac{2}{3} \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right]^n $
$ M_2 = \left[ \begin{array}{ccccc} \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 & 0 & \frac{1}{3} \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right]^n $
when enumerated, that the element in the 1st row and 5th column is seemingly always equal to one another?
As for some background, these matrices can be used to calculate the odds of getting 4 consecutive 1's or four 2's in a row when rolling a 3-sided die. So a die labelled $\{1, 2, 3\}$, you care if you get $\{..., 1, 1, 1, 1, ...\}$ or $\{..., 2, 2, 2, 2, ...\}$. $M_2$ is more intuitive in state-model design, whereas I accidentally stumbled upon $M_1$ as a valid solution.
Also, I could be wrong that the aforementioned two elements are equal $\forall \mathbb{Z^*}$, which I have tested exhaustively, but unable to generate a closed form for.
Suppose the die faces are A, B, C.
(1,5) of the second matrix is as you said, the chance that within n steps you have seen four As in a row or four Bs in a row.
(1,5) of the first matrix is the chance that within n steps that you have seen three As in a row, then something not an A.
In the second matrix, you have a 2/3 chance of starting such a subsequence (from scratch), a 1/3 chance of continuing it (twice), and a 1/3 chance of finishing it.
In the first matrix, you have a 1/3 chance of starting such a subsequence (from scratch), a 1/3 chance of continuing it (twice), and a 2/3 chance of finishing it.
There some more complications because the chances of not being in a subsequence vs the chance of restarting a subsequence in the middle also has to balance, but that's the rough gist of it.