An urn contains two two coins with probabilities of heads $p_1$ and $p_2$, respectively, such that $p_1+p_2>1$; that is, the coins are on average biased in favor of heads. Two people each have initial wealth of $1$. Each takes a coin from the urn, and each flips their coin. Each gets $+1$ for a Head and $-1$ for a Tail. They put their coins back in the urn, shake it up, and draw again. They repeat this process until someone exhausts their wealth. This describes a random walk in the plane that begins in $\mathbb Z^2_{++}$ and is absorbed at the boundary. The walk has a strictly positive drift. Does this random walk terminate, that is, hit the boundary of $\mathbb Z^2_+$, with probability $1$? Any one person has a positive probability of surviving forever because her marginal process is a one-dimensional simple walk with positive drift. But the two walks are dependent because if one person gets the good coin, the other gets the bad coin. I want to know if, with positive probability, both survive forever.
Suppose I have $k$ coins like $p_1$ and $l$ coins like $p_2$ with $kp_1+lp_2>(k+l)/2$, and $k+l$ people, so the walk is now in $\mathbb{Z}^{k+l}_+$. Does this random walk hit the boundary of $\mathbb Z^{k+1}_+$ with probability $1$?
The termination probability is always less than $1$. This follows from using the law of large numbers twice. First, for any $\varepsilon>0$ and for sufficiently large $n$, there is a positive probability that each player chooses the better coin at least $(\frac12-\varepsilon)t$ times out of the first $t$ for every $t>n$. We can choose $\varepsilon$ sufficiently small that $(\frac12-\varepsilon)p_1+(\frac12+\varepsilon)p_2>\frac12$. We also have positive probability that all the tosses before this $n$ will be heads.
Now for any sequence of coin choices with this property, there is a positive probability that for all $t>n$ each player has got a proportion at least $p_1-\delta$ heads on the times he has tossed coin $1$ up to time $t$, and a proportion of at least $p_2-\delta$ heads on the times he has tossed coin $2$. Choosing $\delta$ sufficiently small, this implies that neither player has gone bankrupt.