| As I noted, adversarial wordle shows that no words take 6 guesses and only 2 words take 5 guesses (BOOZY and BOOBY.) If you removed just 1 of those 2 words (leaving 2312 possible Wordles) all Wordles can be solved in max 4 guesses. Ignoring the 2 scenarios where your first two "optimal" guesses are the Wordle X% of the time you have 1 word left (guesses= 3) Y% of the time you have 2 words left (guesses =3.5) (100-Y-X)% of the time there is a pool of words left, requiring one more "optimal" guess And either your optimal guess #3 is the Wordle Z% of the time (guess=3) or there is now one word remaining (guess ≈ 4) Where Z in this case is just a function of the number of words remaining (ie Y is just a special case of Z) So you want to maximize X and the weighted average of Z across pools. By sheer brute force you can do so 1) for each 2 guess combo
2) discard any "suboptimal" combo where for any remaining response state word pool there is no optimal guess #3 (ie not possible to definitively guess in 4 guesses)
3) calculate avg remaining guesses
4) identify optimal word 1 (minimum of sum of step 2 per first guess)
5) within combos with word 1, identify optimal guess 2 for each response state And the weighted average of step 3 for these combos is the global minimum for Wordle. I believe this is the algorithm you're after? In this case, we're making a first guess that maximizes the chance we will get 3 guesses instead of 4. Per this algorithm, optimal guess 1 is RAVED. |