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CmdDot

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CmdDot
·7 maanden geleden·discuss
When it comes to quickly reducing the search space of possible words, it is - that’s how you solve it optimally, even if (or in fact, especially) if the word they chose intentionally does not use the most frequent letters.

The faster you can discard all words containing «e» because of a negative match, the better.

If you want to be really optimal, you’ll use their list of possible words to calculate the actual positional frequencies and pick the highest closest match based on this - that’s what «mostly» was meant to imply, but the general principle of how to reduce the search space quickly is the same
CmdDot
·7 maanden geleden·discuss
«As such, you want to pick a pair where the odds are as close to 50/50 as possible.»

This is incorrect, the correct strategy is mostly to check the most probable match (the exception being if the people in that match has less possible pairings remaining than the next most probable match).

The value of confirming a match, and thus eliminate all other pairings involving those two from the search space, is much higher than a 50/50 chance of getting a no match and only excluding that single pairing.
CmdDot
·7 maanden geleden·discuss
In addition to Mastermind, Wordle also falls into the same category.

Optimal play to reduce the search space in both follow the same general pattern - the next check should satisfy all previous feedback, and included entries should be the most probable ones, both of those previously tested, and those not. If entries are equally probable, include the one which eliminates the largest number of remaining possibilities if it is correct.

For wordle, «most probable» is mostly determined by letter frequency - while in Mastermind, it’s pure probability based on previous guesses. For instance, if you play a Mastermind variant with 8 pegs, and get a 2/8 in the first test - each of your 8 pegs had a 1/4 chance of being correct. So you select 2 at random to include in the next guess.

If you then get a 2/8 from the second - you would include 4 previous entries in the next guess, 2 entries from the first that was not used in the second, as well as 2 entries from the 2nd - because the chance you chose the correct entries twice, is less than the chance the two hits are from the 6 you changed.