[joshbuckler]

The "worst-case number of guesses" objective considered here is pretty insensitive to differences in quality of the decision tree. It seems to me that a better objective would be: minimize the number of secret words that take more than 6 guesses (hopefully to zero), then minimize how many that take 6 guesses, and so on.

[kthejoker2]

This has been done

https://jonathanolson.net/experiments/optimal-wordle-solutio...

Even with adversarial Wordle, the upper limit on guesses is 5 (ie after 3 guesses the maximum remaining word pool is 2)

The hardest words in Wordle are BOOBY / BOOZY no optimizer would include those letters in the first two guesses ...

[Thorrez]

What you described is not what joshbuckler described.

joshbuckler wants an algorithm that has the fewest words that take at least 6 guesses, and also among all algorithms that tie for that number of words that take at least 6 guesses, has the fewest number of words that take at least 5 guesses, and also among all the algorithms that tie for that number of words that take at least 6 and at least 5 guesses, has the fewest number of words that take at least 4 guesses, and also...

The page you linked to doesn't seem to say it does that.

[kthejoker2]

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.