Going for greens is a good heuristic, but the general solution is to seek guesses that shrink the size of the set of compatible words the most. https://langproc.substack.com/p/information-theoretic-analys...
This statement (which motivates the rest of the analysis) isn't correct, because it doesn't take into account the highly irregular search space of what you're actually allowed to guess. Sets of possible words can vary a lot in terms of how easily they can be further cut down: for example, if you're on hard mode and your set of possible words is [BILLY, DILLY, FILLY, HILLY, SILLY, WILLY], then you might be in trouble.
I wrote a very simple O(N^3) algorithm to determine that: for each possible actual word and the guessed word, figure out whether or not a word is eliminated. Then calculate a score based on how many words can be eliminated from a particular guessed word, averaged over all possible actual words.
(I originally thought O(N^3) would be way too slow to be useful. But it runs fast enough with a few thousand words.)
Like a sibling comment, I also found that the best initial guess is RAISE, although when I tried a different word list, I got different words like LARES (I don't think it's in the Wordle list).
One thing that surprises me is that by the time you get to the mid-game stage (after 2 guesses or so), occasionally it can be more optimal to guess a word with repeated letters; I did not originally expect that because I thought repeated letters are wasted opportunity to test more letters, but it turns out Wordle's handling of repeated letters can give you helpful information about the number of occurrences of a letter. (For example if you picked a word with 2 a's, the first one can be green the second one can be black, essentially telling you there is exactly one letter "a" in the word.)
I'm making this comment because I started with the same O(N^3) approach and then realized this. :)
That's not true. Suppose the actual word is "abcdx" but the guessed word is "xefgh". The result of testing is Yellow-Black-Black-Black-Black.
Now suppose you have other words like "xijkl" that will test against the solution in exactly the same way (Yellow-Black-Black-Black-Black), but you know won't be the correct solution because in the first round the misplaced "x" already tells you "x" should be present at a different position.
So now we have a word that can be eliminated even if it tests against the solution in the way the guessed word does.
But yeah, given the frequency of letters, in position, in the target list, the answer is SLATE.