There is a game with 5 positions starting in position 3. two teams have to compete against each other with tasks in math. There cannot be a draw and every round one of the teams wins. One team is stronger than the other team. They start in position three and the goal of team 1 is to reach Position five and the goal of team 2 is to reach position 1. Every round they move one position. This is the matrix.
$$A=\begin{bmatrix} 0 & 0.25 & 0 & 0 & 0 \\ 1 & 0 & 0.75 & 0 & 0 \\ 0 & 0.75 & 0 & 0.75 & 0 \\ 0 & 0 & 0.25 & 0 & 0 \\ 0 & 0 & 0 & 0.25 & 1 \\ \end{bmatrix}$$
and this is the vector with a 1 on $a_{31}$ as it starts in position 3. position 1 is $a_{11}$, position 2 is $a_{21}$ and so on.
$$b=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}$$
The game ends when one of the teams reach position 1 or 5.
Now my question:
$A*b$ gives us
$$\begin{bmatrix} 0 \\ 0.75 \\ 0 \\ 0.25 \\ 0 \\ \end{bmatrix}$$
Perfectly fine, the probability of position 1,3 and 5 is 0 and the probabilities of the other positions vary.
Now lets focus on $A^3*b$
$$\begin{bmatrix} 0.5625 \\ 0.28125 \\ 0 \\ 0.09375 \\ 0.0625 \\ \end{bmatrix}$$
How is it possible that there is a probability to win (land on 1 or 5)? This is not possible after three moves is it? Starting from 3 the shortest way to win is either from 3 to 2 to 1 for Team 2 or from 3 to 4 to 5 for Team 1.
I do not understand why there are probabilities for the first and fifth position.
In my point of view the matrix has to look like this: $$\begin{bmatrix} 0 \\ k \\ 0 \\ h \\ 0 \\ \end{bmatrix}$$ whereas k and h represent probabilities between 0 and 1
As the matrix shows the cumulative probability, $a_{11}$ and $a_{51}$ show the probabilities of $U^2\cdot v$.
Hence, $P(X=3)$ would be $U^3\cdot v-U^2\cdot v$.