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by hervature 1610 days ago
Funnily enough, I think neither of those notions are the correct thing to optimize. For me, you want to minimize the maximum number of guesses. This leads to an equilibrium. Otherwise, your opponent can just abuse your strategy.
5 comments

My point is exactly that though - the choice of cost function is subjective. There is no best cost function, just a mapping from costs functions to optimal strategies.
The (exact) optimal EV-minimizing strategies (assuming each target word is equally likely) always solve in 5 of fewer guesses in normal mode, and always solve in 6 guesses in hard mode. If you wish to guarantee 5 guesses in hard mode, there's a different (optimal) strategy with slightly higher EV. See: http://sonorouschocolate.com/notes/index.php?title=The_best_...

I don't think a maximum-number-of-guesses optimizer leads to an equilibrium (if you mean Nash equilibrium), assuming you're playing the game where the score of the game for the setter is the number of guesses. In particular, if the solver's strategy is deterministic, the setter will always pick one of the words that needs 5 guesses. There's no reason to think the nash equilibrium can be easily found.

Adversarial strategies are another type of problem entirely (as in, if you're playing Word Mastermind). Wordle is a 1-player game, so I find hgibbs's comment totally reasonable. I would think that 'optimal' is more accurately his second definition - minimizing avg guesses but always winning (as in, losing is +inf guesses).
Isn't that just #2, but changing the definition of losing to be the least number of rounds where you can always win (which for Wordle is 5)? These small changes to the definition don't seem to change ev (or the code needed to generate it) that much.

If you don't care about optimizing the EV at all, the greedy tree already has the property you're interested in.

What do you mean by opponent?

The entity choosing the word in a game like Wordle or Hang Man is more like the dungeon master in a game of D&D than an opponent in the traditional sense? They are there to help you have fun.