Even more so, seems like an actual uniform generator would be simpler…

Randy has an updated suggestion for how the button should behave at door 2. What hasn’t changed is that if a contestant at door 2 and moves to an adjacent door, that new door will be 1 or 3 with equal probability.

But this time, on the first, third, fifth, and other odd button presses that happen to be at door 2, there’s a 20 percent the contestant remains at door 2. On the second, fourth, sixth, and other even button presses that happen to be at door 2, there’s a 50 percent chance the contest remains at door 2.

Meanwhile, the button’s behavior at doors 1 and 3 should in no way depend on the number of times the button has been pressed.

As the producer, you want the chances of winding up at each of the three doors—after a large even number of button presses— to be nearly equal. If a contestant presses the button while at door 1 (or door 3), what should the probability be that they remain at that door?

In this case, let $q$ be the probability of remaining at door 1 (or at door 3), then we can treat the two different behaviors at door 2 sequentially in order to come with the two-step transition matrix \begin{align*}Q & = \begin{pmatrix} q & 1-q & 0 \\ 0.4 & 0.2 & 0.4 \\ 0 & 1-q & q \end{pmatrix} \begin{pmatrix} q & 1-q & 0 \\ 0.25 & 0.5 & 0.25 \\ 0 & 1-q & q \end{pmatrix} \\ & = \begin{pmatrix} q^2 -\frac{1}{4}q +\frac{1}{4} & -q^2 + \frac{1}{2} q +\frac{1}{2} & -\frac{1}{4}q + \frac{1}{4}\\ \frac{2}{5}q + \frac{1}{20} & -\frac{4}{5}q + \frac{9}{10} & \frac{2}{5}q + \frac{1}{20}\\ -\frac{1}{4}q + \frac{1}{4} & -q^2 + \frac{1}{2} q + \frac{1}{2} & q^2 -\frac{1}{4}q + \frac{1}{4}\end{pmatrix}.\end{align*}

We will lean upon our own great (?) shoulders from the Classic problem to show that we need to solve for $q$ that makes the transition matrix symmetric. In this case, that requirement yields $$\frac{2}{5}q + \frac{1}{20} = -q^2 + \frac{1}{2} q + \frac{1}{2},$$ or equivalently, $$q^2 -\frac{1}{10} q - \frac{9}{20} = 0.$$ Solving this quadratic for the positive root (since after all we need $q\in[0,1]$ as a probability), gives that the appropriate probability to remain at door 1 in this even more complicated Markov scheme is $$q=\frac{ \frac{1}{10} + \sqrt{ \frac{1}{100} + 4 \frac{9}{20} } }{2} = \frac{1+\sqrt{181}}{20} \approx 0.722681202354\dots$$