Lattice random walk under gravity Suppose a random walk on $\mathbb{Z}^2$ takes a step left or right with
probability $\frac{1}{4}$, but up with probability $\frac{1}{2} p$
and down with probability $\frac{1}{2} (1-p)$, where $p \in [0,1]$ indicates
a constant bias against north when $p < \frac{1}{2}$.
(So we have $\frac{1}{4}+\frac{1}{4}+ p \frac{1}{2}+ (1-p)\frac{1}{2}=1$.)
For example, here are two $10^5$-step random walks with $p=0.49$:

          


          


          

The red dot marks the starting origin.



What is the probability of returning to the origin for $p < \frac{1}{2}$?

It is not $0$, and I don't think Polya's recurrence theorem
holds unless $p = \frac{1}{2}$.
Added. Here is a distribution of path endpoints,
again using $p=0.49$, and paths of $10^5$ steps:

          


          

$500$ random walk endpoints, with $p=0.49$ bias against north.


 A: Let $E_n$ denote the probability of the first return after $n$ steps, and let $P_n$ denote the probability of return after $n$ steps (but not necessarily the first return).  Note that $E_n=P_n=0$ if $n$ is odd.
 We are interested in $E= E(p)= \sum_{n=1}^{\infty} E_n$.   Now note that for $n\ge 1$
 $$
 P_n = E_n + \sum_{k=1}^{n-1}  E_k P_{n-k},
 $$ 
 from which it follows that 
 $$ 
 P = \sum_{n=1}^{\infty} P_n = \sum_{n=1}^{\infty} E_n + \sum_{k=1}^{\infty} E_k \sum_{j=1}^{\infty} P_j,
 $$ 
 or in other words $P= E(1+P)$, or 
 $$ 
 E= \frac{P}{1+P}. 
 $$ 
 All this is due to Polya, of course. 
Now $P_n$ is quite easy to calculate:  for odd $n$ it is zero, and for even $n$ it is simply (counting $a$ steps to the right and left, and $b$ steps up and down)
 $$ 
 P_n  = \sum_{a+b=n/2} \frac{n!}{a!^2 b!^2} \Big(\frac 14\Big)^a \Big(\frac 14\Big)^a \Big(\frac 12p \Big)^b \Big(\frac 12 (1-p)\Big)^b.
 $$ 
 We can also express $P_n$ as a double integral: 
 $$ 
\frac{1}{(2\pi )^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \Big( \frac 14 e^{i\alpha} + \frac 14 e^{-i\alpha} + \frac 12p e^{i\beta} + \frac 12 (1-p) e^{-i\beta} \Big)^n d\alpha d\beta.
$$
From this we can see that 
$$ 
1+P= \frac{1}{(2\pi)^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \frac{2}{2-(\cos \alpha + \cos \beta + i(2p-1)\sin \beta)} d\alpha d\beta. 
$$ 
Note that when $p=\tfrac 12$ this integral diverges, and we recover Polya's theorem that $E=1$.  For any given $p<\tfrac 12$ we may clearly calculate the integral.  Moreover, it is not hard to get asymptotics from the above, and show that  $P$ behaves like $\log (1-2p)^{-1}$ for $p$ near $\tfrac 12$ (forgetting constants here).  
