Table 3 State transitions for the two users case: uncoded transmission ( p _a = p _b = p )

$\left(\begin{array}{cccc}{p}^2+(1-p)& \frac{1}{2}p(1-p)& \frac{1}{2}p(1-p)& 0\\ \frac{1}{2}(1-p)& p+\frac{1}{2}{(1-p)}^2& 0& \frac{1}{2}p(1-p)\\ \frac{1}{2}(1-p)& 0& p+\frac{1}{2}{(1-p)}^2& \frac{1}{2}p(1-p)\\ 0& \frac{1}{2}(1-p)& \frac{1}{2}(1-p)& p\end{array}\right)$

$\left(\begin{array}{l}1-p\\ 1-p\\ 1-p\\ 1-p\end{array}\right)$