Right, that point of view makes sense if you imagine two players separately playing Wordle and comparing their results, and indeed that's what the OP may have been thinking. But as far as I know, Wordle is primarily a game for one player to play against the computer, where eventually getting to the word (or not) is the most prominent outcome (the only notion of "win" or "defeat" in the game itself), and solving it in fewer guesses is just a bonus. But sure, if you don't care about occasionally losing, you can just set an appropriate weight for w7 within the same family of cost functions.

Indeed, it’s essentially a single player game, but we’re talking about which bot a player would want to use which is inherently a notion of bots competing against each other.

If you look at the failed attempts in the article, you'll see it's not so easy to get 100% correct with 6 guesses. For example:

SLATE, CRONY, HOUND, BOUND, POUND, MOUND (correct: WOUND)

SLATE, SHARE, SHAME, SHAPE, SHAKE, SHADE (correct: SHAVE)

...so sometimes even if it gets 4 of 5 letters right on the second try, it still has too many options left. Of course in cases like this it could go "one step back" and try a word with (most of) the letters that are different between these words (HBPMW or MPKDV). So try different words until the number of options is narrowed down sufficiently.

> I’d say I’ve soundly defeated you 99.9% of the time.

Except it wouldn't be 99.9% of the time. Just the games were you guessed fewer than GP.

GP's solver would expect to guess the correct word 100% of the time in fewer than 6 guess, but not that it would always take 6 guesses.

I was saying if you hypothetically got every correct guess in exactly 6 tries. You would pass the suggested “bare minimum” while my 99.9% strategy doesn’t, yet I claim my strategy is better.

For some value of better, sure.

Sure, where "some value of better" is "wins in the overwhelming majority of rounds."