Now asked at MO:
The proof by strategy-stealing that CHOMP on a rectangular board is a first-player win involves player 1 taking the top-right square on their first move. Of course given the proof-by-contradiction nature of the argument, this does not mean that the top-right square is in general a good opening move, and there are plenty of rectangular boards on which it is in fact a losing move for the first player.
Note: different versions of CHOMP place the "poison square" at different locations. For simplicity, I'm sticking with the convention from the above-linked notes.
I'd like to know how often it is in fact a good move. (Note: by a strategy stealing argument, if it's winning in a given board it is in fact the only winning move in that board. I missed this at first.)
There are a couple ways to phrase this problem:
Question 1: For fixed $n$, let $c_n$ be the upper asymptotic density of the set of $m$ such that moving in the top-right square is the winning move for the first player on the $m$-by-$n$ board. What is $c_3$?
(It's easy to see that $c_1=0$ and $c_2=1$, so this is the first interesting case. Of course the higher $c_i$s are also interesting, but I worry that they are intractable.)
Question 2: Is it the case that, for every $k$, there are $m,n>k$ such that the top-right square is the winning move for the first player in the $m$-by-$n$ board? (Is this even true for $k=2$?)
Partial answer too big for a comment:
For a $3×n$ board, the "do almost nothing" move is never the right move. Using the recurrence from this page:
The author of that page draws a table:
He then notes that elements on the diagonal are always the largest in their column:
This is easily extended to show that $f(q,q-1)$ is at most third greatest. But there are $q-2$ numbers below $f(q,q-1)$ - and neither $1$ nor $3$ are in the column, so $f(q,q-1)$ is at least $q+1$.
Finite for any $m>2$ on boards $m×n$
A similar table can be created for any $m$ - just much more complicated because it would have more dimensions. And the same argument would apply to the diagonal of the table - $f(k,k,k)$ would be the greatest value in its column, and so on. $f(k,k,k-1)$ would be at least the fifth greatest, and so on.
Thus, for every $m>2$ there exists an $N$ such that "do almost nothing" is a losing move on all boards $m×n$, where $n>N$.
Given that $N$ increases with $m$, I think it's safe to say that there are no solutions for $m=4,5$ (I have with computer search checked boards up to $6×65$).