Question - Soccer Passing Game Probability Matrix Analysis

Let the transition matrix be \( P \) and the steady-state vector be \( \vec{s} \), where

\[ P = \begin{bmatrix} 0 & 0.5 & 0.25 & q & r & 0 \\ 0.5 & 0 & 0.25 & 0 & 0 & p \\ 0.25 & 0.5 & 0 & 0 & 0 & 0 \\ p & 0 & 0 & 0 & 0.5 & 0.25 \\ 0 & 0 & 0 & 0.5 & 0 & 0.25 \\ 0.25 & 0 & r & 0 & 0.25 & 0 \end{bmatrix} \]

\[ \vec{s} = \begin{bmatrix} s \\ s \\ s \\ s \\ s \\ s \end{bmatrix} \]

For a steady-state vector \( \vec{s} \), the equation \( \vec{s} = \vec{s}P \) must be satisfied.

\( s = 0.5s + 0.25s + ps \)

\( s = 0.75s + ps \)

\( s(1 - 0.75 - p) = 0 \)

Since \( s \neq 0 \),

\( 1 - 0.75 - p = 0 \)

\( p = 0.25 \)

Similarly, for \( r \):

\( s = 0.25s + 0.25s + rs \)

\( s = 0.5s + rs \)

\( s(1 - 0.5 - r) = 0 \)

Since \( s \neq 0 \),

\( 1 - 0.5 - r = 0 \)

\( r = 0.5 \)

And also for \( q \):

\( s = 0.5s + 0.25s + qs \)

\( s = 0.75s + qs \)

\( s(1 - 0.75 - q) = 0 \)

Since \( s \neq 0 \),

\( 1 - 0.75 - q = 0 \)

\( q = 0.25 \)

Therefore, \( s = s \), \( p = 0.25 \), \( q = 0.25 \), and \( r = 0.5 \).