Optimal Algorithm for Winning Hangman

Question

In the game Hangman, is it the case that a greedy letter-frequency algorithm is equivalent to a best-chance-of-winning algorithm?

Is there ever a case where it\'s worth

Answer 1

Choose a letter that divides the remaining valid words in 2 sets of nearly equal size. With positional information there could be more than 2 sets. Repeat until your set has size 1. That is the best way. The proof is left as an exercise.