This question relates to: How to pick the winner in the "Play the Winner" treatment assignment scheme (Urn model) which is like a Pólya urn model where the additions of balls into the urn relate to an additional randon variable.
Let's start with an urn that contains $x=2,y=1$ red and green balls.*
Pólya urn: We draw randomly balls from the urn. If the draw is green, then we put that drawn green ball back into the urn plus another green ball. If the draw is red, then we put that drawn red ball back into the urn plus another red ball.
In the Pólya urn the transition probabilities are either an increase in $x$ or an increase in $y$
$$P(\text{$x$ increase}) = \frac{x}{n}$$ $$P(\text{$y$ increase}) = \frac{y}{n}$$
where $n$ is the total of balls in the urn.
Alternative: after the draw an experiment is performed with success probabilities $p_{y}$ or $p_{x}$ depending on which ball had been drawn. If the experiment is a success then we add an additional ball of the same colour otherwise we add a ball of the opposite colour. Now we have transition probabilities $$P(\text{$x$ increase}) = \frac{x}{n} p_{x} + \frac{y}{n} (1-p_{y}) $$ $$P(\text{$y$ increase}) = \frac{x}{n} (1-p_{x}) + \frac{y}{n} p_{y} $$
I wonder what happens when we consider an absorbing boundary at $x=y$ and the probability of ending at the boundary.
In the special case of $p_x=p_y = 1$, then we have the typical Pólya urn and we know that $P(\textrm{boundary}) < 1$. More specifically, when we start with $(x,y) = (2,1)$ then half the cases will end up at the boundary (we can use the betabinomial distribution along with the reflection principle explained in: What is the distribution of time's to ruin in the gambler's ruin problem (random walk)?) and $P(\text{boundary}) = 0.5$ for $n \to \infty$.
So, when $p_x = 1$ and $p_y = 1$ we have a non-zero probability that the process never reaches the boundary. Is this also true for any $p_x < 1$ and $p_y = 1$?
A schematic view of the process is in the image below which displays 200 simulated paths among which 151 ended at the boundary $x=y$. What fraction ends at the boundary when we simulate an infinitely long time?
Contrasting perspective.
When we have a discrete random walk with steps in x or y direction depending on a Bernoulli variable with probability $p_x$ and $p_y$, then there is a balance for the probability $p_x=p_y = 0.5$ in which case the random walk almost certainly ends.
That point is a boundary case. If we have $p_x \leq p_y$ then the process almost certainly ends at the boundary. If we have $p_x > p_y$ then the process has a non-zero probability to never end at the boundary.
For this Pólya urn we know that at least in the case of $p_x = p_y = 1$ we do not end up almost certainly at the boundary (or not even close since only halve the cases end up at the boundary). The question is whether this situation is still a balance and whether we have for $p_x \in [0,1)$ almost certain absorption at the boundary, or whether the point between almost certain absorption and non-zero probability of non-absorption is at some other place.
* Motivation for relationship with the linked question: I wonder if there is a non-zero possibility that the process will never be in favour of y even when $p_y > p_x$. For this purpose, I consider what happens when there is an initial offset $(x,y) = (2,1)$ in favour of $x$, which can happen for a small part of the cases. Is there, from that point onwards, a sufficient drift away from the line $x=y$ such that a fraction of the paths will never get into the region $y>x$?
