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

Assume the following dictionary: ABC ABD AEF EGH. (I'll capitalize unguessed letters.)
Assume you have only 1 life (makes the proof so much easier...).

The algorithm proposed above:

Starting with A, you lose (1/4) or are left with aBC aBD aEF (3/4).
Now guess B and lose (1/3) or are left with abC abD (2/3).
Now guess C or D and you lose (1/2) or win (1/2).
Probability to win: 3/4 * 2/3 * 1/2 = 1/4.

Now try something else:

Starting with E, you lose (1/2) or are left with AeF eGH (1/2).
Now you know what to guess and win.
Probability to win: 1/2.

Clearly the proposed algorithm is not optimal.