Soccer Passing Game Probability Matrix Analysis
<p>Let the transition matrix be \( P \) and the steady-state vector be \( \vec{s} \), where</p>
<p>\[ P = \begin{bmatrix} 0 & 0.5 & 0.25 & q & r & 0 \\ 0.5 & 0 & 0.25 & 0 & 0 & p \\ 0.25 & 0.5 & 0 & 0 & 0 & 0 \\ p & 0 & 0 & 0 & 0.5 & 0.25 \\ 0 & 0 & 0 & 0.5 & 0 & 0.25 \\ 0.25 & 0 & r & 0 & 0.25 & 0 \end{bmatrix} \]</p>
<p>\[ \vec{s} = \begin{bmatrix} s \\ s \\ s \\ s \\ s \\ s \end{bmatrix} \]</p>
<p>For a steady-state vector \( \vec{s} \), the equation \( \vec{s} = \vec{s}P \) must be satisfied.</p>
<p>\( s = 0.5s + 0.25s + ps \)</p>
<p>\( s = 0.75s + ps \)</p>
<p>\( s(1 - 0.75 - p) = 0 \)</p>
<p>Since \( s \neq 0 \),</p>
<p>\( 1 - 0.75 - p = 0 \)</p>
<p>\( p = 0.25 \)</p>
<p>Similarly, for \( r \):</p>
<p>\( s = 0.25s + 0.25s + rs \)</p>
<p>\( s = 0.5s + rs \)</p>
<p>\( s(1 - 0.5 - r) = 0 \)</p>
<p>Since \( s \neq 0 \),</p>
<p>\( 1 - 0.5 - r = 0 \)</p>
<p>\( r = 0.5 \)</p>
<p>And also for \( q \):</p>
<p>\( s = 0.5s + 0.25s + qs \)</p>
<p>\( s = 0.75s + qs \)</p>
<p>\( s(1 - 0.75 - q) = 0 \)</p>
<p>Since \( s \neq 0 \),</p>
<p>\( 1 - 0.75 - q = 0 \)</p>
<p>\( q = 0.25 \)</p>
<p>Therefore, \( s = s \), \( p = 0.25 \), \( q = 0.25 \), and \( r = 0.5 \).</p>
