Breaking Perfect Hangman in reverse, in reverse

Question

this is the follow-up question to Perfect Hangman in reverse

golfers are now asked to develop a program that uses all rules in Perfect Hangman in reverse, instead, playing the player.

the program shall ask how many letters you want the word to be, a single required letter, then pick what word is LEAST LIKELY to be guessed by the Perfect Hangman in reverse program.

Example:

you state 6 letters, and the letter M, the program knows that its counterpart will guess E first, so it rules out any words with E, it knows the next letter is G, so it also rules out all words with G, the 3rd letter is A, however no words remain without an A, E, and G, and also contains M, so it assumes that its counterpart will get the A correct, and starts analyzing the next step to determine the best way to fool the program.

Example may not be exactly correct, if anybody wants to edit the example with a real-world result, go for it.

bonus example:

program sees letter E is most common, however by taking the word enigma, it realizes that the program is LEAST likely to guess the total word.

tiebreaks:

all tiebreaks in code should assume that the result that expands all lives, with the most remaining undiscovered letters, is the best. if the tie is still unbroken, you should select the alphabetical first.

A winning program will be scored based on the following criteria.

length - 1 point per byte
smarts - if your program can accept a letter, when it is only acceptable because a LATER step becomes harder than if it didn't accept, award yourself -300 points
incorrect - to stop cheating and 0 length, if your program doesn't solve the problem, award yourself 10 million points, and refrain from posting.

Contest ends 8/17/2012, noon CST.