3 ms·
It's the way to win in the fewest words on average. If you want to maximize your number of three-guess-wins even if your 6-guess-failure rate is higher, it is
by esrauch 5y ago
It's the way to win in the fewest words on average.
If you want to maximize your number of three-guess-wins even if your 6-guess-failure rate is higher, it is obviously can't be a correct strategy to use your third guess on a word that you know can't be correct.
- willis936 5y agoIt's a gamble. 20% 3-guess win/80% 4-guess to FAIL or 100% 4-guess win. Maximizing information from each guess until the answer is down to two candidates minimizes the number of guesses needed to win.
- esrauch 5y agoSure, my point is that the metric "minimize mean number of guesses to win" isn't axiomatically the objective function: the 6 threshold already suggests that many people would rather have a "guaranteed 6 and average 5" rather than "average 4 but fail 5% percent of the time". If someone considers a 4 to be a soft-failure then they will prefer (3,6,6) to (4,4,4) even though the latter is a lower mean number of guesses: it's a higher count of 4+ 'failures'. The strategy that considers (3,6,6) a better result than (4,4,4) wouldn't guess "dream" in the above example.