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Ruby

A joint submission from user PragTob and myself.

MAX_TURNS = 6

frequencies = {?t=>[3,48,145,214,252,266,249,223,191,142,63,44,16,1,0,1],?h=>[2,14,81,125,85,91,60,42,30,14,11,6,1,1],?e=>[5,49,260,316,456,408,328,279,202,125,50,32,12,0,0,1],?a=>[4,60,211,259,249,266,253,192,152,111,51,42,15,1,0,1],?n=>[4,30,120,136,214,252,238,214,189,128,59,45,16,0,0,1],?d=>[1,25,100,104,131,123,131,81,63,36,14,15,7,1],?f=>[2,13,51,58,64,67,41,40,28,18,11,9,3,1],?o=>[9,44,150,165,195,220,214,168,155,104,46,37,14,1,0,1],?r=>[1,25,140,246,312,310,263,206,150,95,45,32,11,1],?y=>[3,29,41,58,86,94,83,63,52,31,21,12,2],?u=>[2,23,67,117,126,154,107,97,85,48,27,16,2,0,0,1],?b=>[2,22,53,60,72,59,41,30,36,16,7,6,1],?i=>[3,38,143,179,223,299,270,241,205,134,64,44,16,1,0,1],?s=>[3,23,129,176,195,208,177,136,117,71,44,23,13,1],?c=>[0,12,68,122,146,194,180,163,130,85,49,25,7,0,0,1],?l=>[0,18,153,172,190,196,164,131,125,67,35,20,5,0,0,1],?g=>[1,19,42,75,82,104,78,60,39,30,12,10,0,1],?w=>[1,21,56,56,40,41,18,16,6,2,1,0,0,1],?m=>[2,10,77,68,119,94,104,76,68,45,15,17,8,0,0,1],?p=>[1,24,82,84,94,129,105,88,99,56,24,11,7],?k=>[1,6,65,37,28,24,6,10,3,4],?j=>[0,5,5,6,7,6,5,6,2,0,1,1],?x=>[0,6,4,7,15,22,13,16,9,9,2,2],?v=>[0,3,21,39,47,58,63,42,40,23,10,9,2],?z=>[0,0,3,3,3,5,2,3,0,0,1],?q=>[0,0,1,9,5,13,8,3,5,4,3,2]}

while !(input=gets.chomp)['END']
  current_turns = MAX_TURNS
  won = false
  chars = frequencies.keys.sort_by {|c|
    -(frequencies[c][input.length-2] || 0)
  }
  i=0
  while (current_turns > 0) && !won
    c=chars[i]
    i += 1
    puts c
    $stdout.flush
    old_input     = input
    input         = gets.chomp

    if input == old_input
      current_turns -= 1
    # else
    #   frequencies[c][input.length-2] -= 1
    end
    won = !input[?_]
  end
end

Result:

score is 625, totalerr is 23196

This has 1672 characters and isn't golfed yet so we have ample room for algorithmic improvement.

First we store a hash of character frequencies (computed from the word list) grouped by word length.

Then in each round we simply sort all characters by the frequencies for the current length and try them from most common to least common one. Using this approach we obviously fail every single word that has even just a medium common character.