Here's a solution with ~3.27 average tries total (including the last guess). It works by spreading the possibilities across each of the 100 "score" buckets as e

Here's a solution with ~3.27 average tries total (including the last guess). It works by spreading the possibilities across each of the 100 "score" buckets as evenly as possible (aka minimizing the variance). It can identify more than half of all colors in just 2 guesses. If you edit the code to minimize the size of the largest bucket instead, you get an algorithm that has a slightly worse average case, but never needs more than 4 guesses. Paste it in the site's browser console to try. function solve() { let possibleTargets = new Array(0x1000).fill(0).map((_, i) => i); while (true) { // for each possible guess, calculate the score-buckets const guessResults = possibleTargets.map((guess) => { // for better brute-force, iterate through all colors, but this performs better (probably because we don't reward for guessing correctly in the current turn) const buckets = new Array(101).fill(0).map(() => []); for (const possibleTarget of possibleTargets) { buckets[calcScore(possibleTarget, guess)].push(possibleTarget); } return { guess, buckets }; }); // find guess with lowest variance const best = guessResults.sort((a, b) => calcVariance(a.buckets) - calcVariance(b.buckets))[0]; // make the guess & update possible targets const sc = makeGuess(best.guess); if (sc >= 100) return "Success!"; possibleTargets = best.buckets[sc]; } } function calcVariance(buckets) { const maxBucketSize = Math.max(...buckets.map(b => b.length)); return buckets.reduce((acc, b) => acc + (b.length - maxBucketSize) ** 2, 0); } function deconstruct(color) { return [(color & 0xF00) >> 8, (color & 0x0F0) >> 4, (color & 0x00F)]; } function calcScore(target, guess) { return score(...deconstruct(target), ...deconstruct(guess)); } function makeGuess(guess) { [rr, gg, bb] = deconstruct(guess); submitInput(); return score(r, g, b, rr, gg, bb); }; solve();