If you want to minimize the number of errors, you should try the letter with the most information. In other words, minimize the search space. That is what hythloday suggested.
You must somehow weigh information gain against risk.
It may also be useful to consider how many remaining incorrect guesses there are in the game.
Actually, we should be correct 50% of the time. Which means 22 Bits of information or 4194304 words.
Additionally, we know the length of the word.
The english dictionaries seem to have between 400k and 1000k words [0] of all word sizes. With 22 Bits we get 4000k words. We do not have to worry about getting hanged using the information-reduction algorithm. ;)
More precisely: Given a dictionary of all possible words and a letter to guess, we can split the dictionary into k parts. One sub-dictionary contains all words for which the answer is negative. Additionally, we have k-1 sub-dictionaries for each equivalence class of letter positions.
For each letter, we can compute the partition. Now we need to choose the strategically best partition. I believe it should be the one with "the lowest average sub-dictionary size".