Following Douglas' comments, here is the IID case. Let $a$ be the prize ball probability and $b$ be the empty ball probability (we shall choose them later) and $c=1-a-b$ be the probability of the penalty ball.
Case 1. $p>1$. Here, if the prize ball is there at all, we'll get it with the column strategy. The only chance to get $-1$ is when everything is $-1$. The row strategy cannot give more in the all $-1$ case, gives $0$ if the prize ball is not there and there are empty balls, and can yield less than the prize ball in some ohter cases too. So, it is worse regardless of the probability assignments.
Case 2: $p<1$. If we go rows first, we get at least $[1-(1-a)^n]^2p$ chance to get the prize ball (the probability that both columns have at least one) and the probability $\le 2c^n$ for the penalty ball (at least one column is pure penalty). So, if $a\gg 1/n$, $b\gg 1/n$, and $p$ is close to $1$, we get almost $1$ on average.
Now, the column strategy cannot yield more than the prize ball but fails to yield it if all prize balls are coupled with the penalty ones (the chance that it is not the case is that we have one prize-prize or one prize-empty pair, which amounts to roughly speaking $na(a+b)$) and there is an empty-empty pair (that fails with probability $(1-b^2)^n$). So, we cannot hope to get more than $na(a+b)+(1-b^2)^n$, which can be easily made small under our restrictions.