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

The answer clearly shows why the greedy algorithm is not the best, but doesn't answer how much better we can get if we stray from the greedy path.

If we assume all words are equally likely in case you are playing against a computer opponent. The case of 4 letters, 6 lives case if you have option to look simply the second most popular letter your probability of winning increases from 55.2% to 58.2%, if you are willing to check one more letter then it increases to 59.1%.

Code: https://gist.github.com/anitasv/c9b7cedba324ec852e470c3011187dfc

A full analysis using ENABLE1 (scrabble dictionary which has 172820 words) with 2 to 6 letters, and with 0 to 10 lives, and 1-greedy to 4-greedy gives the following results. Of course at 25 lives every strategy is equivalent with 100% win rate, so not going beyond 10 lives. Going more than 4-greedy was improving probability only slightly.

letters, lives, 1-greedy, 2-greedy, 3-greedy, 4-greedy


2 letters 0 lives   3.1%    3.1%    3.1%    3.1%
2 letters 1 lives   7.2%    7.2%    7.2%    8.3%
2 letters 2 lives   13.5%   13.5%   13.5%   14.5%
2 letters 3 lives   21.8%   21.8%   22.9%   22.9%
2 letters 4 lives   32.2%   33.3%   33.3%   33.3%
2 letters 5 lives   45.8%   45.8%   45.8%   45.8%
2 letters 6 lives   57.2%   57.2%   57.2%   57.2%
2 letters 7 lives   67.7%   67.7%   67.7%   67.7%
2 letters 8 lives   76%     76%     76%     76%
2 letters 9 lives   84.3%   84.3%   84.3%   84.3%
2 letters 10 lives  90.6%   91.6%   91.6%   91.6%


3 letters 0 lives   0.9%    1.1%    1.1%    1.1%
3 letters 1 lives   3.4%    3.8%    3.9%    3.9%
3 letters 2 lives   7.6%    8.4%    8.6%    8.8%
3 letters 3 lives   13.7%   15%     15.1%   15.2%
3 letters 4 lives   21.6%   22.8%   23.3%   23.5%
3 letters 5 lives   30.3%   32.3%   32.8%   32.8%
3 letters 6 lives   40.5%   42%     42.3%   42.5%
3 letters 7 lives   50.2%   51.4%   51.8%   51.9%
3 letters 8 lives   59.6%   60.9%   61.1%   61.3%
3 letters 9 lives   68.7%   69.8%   70.4%   70.5%
3 letters 10 lives  77%     78.3%   78.9%   79.2%


4 letters 0 lives   0.8%    1%      1.1%    1.1%
4 letters 1 lives   3.7%    4.3%    4.4%    4.5%
4 letters 2 lives   9.1%    10.2%   10.6%   10.7%
4 letters 3 lives   18%     19.4%   20.1%   20.3%
4 letters 4 lives   29.6%   31.3%   32.1%   32.3%
4 letters 5 lives   42.2%   44.8%   45.6%   45.7%
4 letters 6 lives   55.2%   58.2%   59.1%   59.2%
4 letters 7 lives   68%     70.4%   71.1%   71.2%
4 letters 8 lives   78%     80.2%   81%     81.1%
4 letters 9 lives   85.9%   87.8%   88.4%   88.7%
4 letters 10 lives  92.1%   93.3%   93.8%   93.9%


5 letters 0 lives   1.5%    1.8%    1.9%    1.9%
5 letters 1 lives   6.1%    7.5%    7.9%    8%
5 letters 2 lives   15.9%   18.3%   18.9%   19.2%
5 letters 3 lives   30.1%   34.1%   34.8%   34.9%
5 letters 4 lives   47.7%   51.5%   52.3%   52.5%
5 letters 5 lives   64.3%   67.4%   68.3%   68.5%
5 letters 6 lives   77.6%   80.2%   80.6%   80.8%
5 letters 7 lives   86.9%   88.6%   89.2%   89.4%
5 letters 8 lives   92.8%   94.1%   94.4%   94.5%
5 letters 9 lives   96.4%   97.1%   97.3%   97.3%
5 letters 10 lives  98.2%   98.6%   98.8%   98.8%


6 letters 0 lives   3.2%    3.7%    3.9%    3.9%
6 letters 1 lives   12.6%   14.3%   14.9%   15%
6 letters 2 lives   29.2%   32.2%   32.8%   33%
6 letters 3 lives   50.1%   53.4%   54.2%   54.4%
6 letters 4 lives   69.2%   72.4%   73.1%   73.2%
6 letters 5 lives   83.1%   85.5%   85.9%   86.1%
6 letters 6 lives   91.5%   92.9%   93.2%   93.2%
6 letters 7 lives   95.8%   96.5%   96.7%   96.8%
6 letters 8 lives   97.9%   98.3%   98.4%   98.5%
...

Answer 2

There are some critical assumptions you have to make as to what a game of "Hangman" is.

• Do you just have one word you need to guess, or do you need to guess a phrase?
• Are some words more likely than others?

One thing to remember is that if you pick a correct letter, you do not lose a guess.

I will provide a solution for the one-word-&-equally-likely case. The two-word case can be generalized by creating a new dictionary equal to the cartesian product of your current dictionary, etc. The more-likely-than-others case can be generalized with a bit of probability.

Before we define our algorithm, we define the concept of a reduction. If we were to guess letters L1,L2,L3,...LN all at ONCE, then we would reduce the dictionary to a smaller dictionary: some words would be eliminated, and additionally some letters may also be eliminated. For example if we had the dictionary {dog, cat, rat} and we guessed a, we would eliminate {d,g} if the guess was true, or also eliminate {c,r,t} if it was false.

The optimal algorithm is as follows:

• Consider the game tree
• Look at all nodes where [#guesses left == 1]
• If there is no node, the game is impossible; if there is a node, that is your answer

Of course that is how you solve any game, and for the most part it is intractable due to the exponential size requirements. You cannot get optimal unless you perfectly replicate this, and I seriously doubt that a strategy which doesn't "look" two or more moves ahead can hope to replicate this. You can however attempt to approximate the optimal strategy as follows.

Repeat the following at each step:

• Consider each letter: choose the letter which will maximize the expected-dictionary-reduction per expected-penalty: that is, pick the letter L which will maximize the (frac words with L#words without L + frac words without L#words with L)/(# words without L/# words total)... note that this may be infinite if all the words have a certain letter, in which case go ahead and guess it since there is no penalty.
• Make your guess, get an updated board state
• Eliminate all words invalidated by the new board

Of course if your dictionary has more than 2^[max number of guesses] entries, the "Hangman" game is mostly impossible in the equal-probability world (unless the dictionary is highly constrained), so you have to work in the unequal-probability world. In this world, rather than maximizing the amount of elimination you do, you maximize the "expected surprisal" (also called entropy). Each word you associate a prior probability (e.g. let's say there is a 0.00001 chance of the word being 'putrescent' and a 0.002 chance of the word being 'hangman'). The surprisal is equal to the chance, measured in bits (the negative log of the chance). An answer to a guess will either yield no letters, a single letter, or more than one letter (many possibilities). Thus:

• For each possible guess, consider the effect that guess would have
• For each possible outcome of the guess, consider the probability of that outcome. For example if you guessed 'A' for a 3-letter word, you'd have to consider each possible outcome in the set {A__, _A_, __A, AA_, A_A, _AA, AAA}. For each outcome, calculate the probability using Bayes's rule, and also the new possible dictionaries (e.g. in one case, you'd have a dictionary of _A_:{cAt, bAt, rAt, ...} and A__:{Art, Ark, Arm, ...} etc.). Each of these new dictionaries also has a likelihood ratio, of the form size(postOutcomeDictionary dictionary)/size(preOutcomeDictionary); the negative log of this ratio is the amount of information (in bits) which the choice conveys to you.
• Thus you want to maximize the ratio of the expected amount of information (in bits) you gain, per the expected cost (the cost penalty is 1 if you fail and 0 if you don't). For each guess, and for each outcome of the guess, the bits-of-information-gained is bitsGainedFromOutcome[i] = -log(size(postOutcomeDictionary)/size(preOutcomeDictionary)). We take the weighted sum of these: sum{i}( prob(outcome[i])*bitsGainedFromOutcome[i] ), then divide by the probability we are wrong: prob(outcome=='___').
• We pick the letter with the minimum sum{i}( prob(outcome[i])*bitsGainedFromOutcome[i] )/prob(outcome=='___'); in case this is infinity, there is nothing to lose, and we automatically pick it.

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So to answer your question:

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

Clearly not: if the dictionary was

cATs
bATs
rATs
vATs
sATe
mole
vole
role

your algorithm would guess a or t, which have a 5/8 chance of reducing your dictionary to 5/8 size for free, and a 3/8 chance of reducing your dictionary to 3/8 size for a cost of 1. You want to choose letters which reveal the most information. In this case, you should guess S, because it has a 4/8 chance of reducing your dictionary to 4/8 size for free, a 1/8 chance of reducing your dictionary to 1/8 size for free, and a 3/8 chance of reducing your dictionary to 3/8 size for a cost of 1. This is strictly better.

edit: I wanted to use an English dictionary example (above) to demonstrate how this is not artificial, and assumed that people could extrapolate from the example without being hung up on the non-strict equality. However, here is an unambiguously clear counterexample: You have 2000 words. 1000 words contain the letter A in the first place. The other 1000 words contain a unique combination of Bs embedded elsewhere. For example, ?B???, ??B??, ???B?, ????B, ?BB??, ?B?B?, etc. The ?s represent randomly-chosen characters. There are no As in the first ?, except for one word (whose ? is an 'A'), so that the frequency of As is strictly greater than the frequency of Bs. The proposed algorithm would guess A which would result in {50%: choices_halved, 50%: choices_halved & lose_one_life}, whereas this algorithm would dictate the choice B which results in {50%: YOU_WIN, 50%: choices_halved & lose_one_life}. Percentages have been rounded very slightly. (And no, a word with double letters does not contribute twice to the 'frequency', but even if it did under an insane definition, you could trivially modify this example by making the words begin with AAA...A.)

(regarding comments: It is unreasonable to complain about strict equality in this example, e.g. "999/2000", since you can make the probabilities arbitrarily close to each other.)

(Which points out an amusing side-note: If the dictionary is large enough to make hangman impossible sometimes, a strategy ought to throw away guesses that it does not expect to be able to guess. For example if it only has 2 moves left, it ought to make the highest-probability assumption it can which eliminates subtrees with more than 2-moves worth of bits of surprise.)

Answer 3

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.

Answer 4

I have written a script that solves hangman optimally [github].

My basic strategy is this:

• For a pattern such as ..e.. with tried letters: e,s,t
• Check against words of only n digits (in this case 5)
• Create a list of possible words
  • Create a regex from the supplied info
  • In this case it would be [^est][^est]e[^est][^est]
  • Parse your word list for words that match this regex
• Cycle through each word, counting up the number of times every letter appears
• Your optimal next guess is the most likely letter

Answer 5

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.

Answer 6

About your idea to try and guess the word instead of trying to guess letters, there sure may be some isolated cases where you guess the word from the first try or something like that, but this doesn't make that algorithm better on the average case. It's about expected probability.

Some improvement that could be done to that algorithm (in the version posted by Lajos) is some more informed pick of letter to be tried.
This could be achieved by adding one more weight: consider the usage of each word the vocabulary of the language.

For example, technical, medical, juridical etc. words would have much lower chances.

Take this dictionary (with some example usage weights):

frost    6
guilt    5
genes    1
fever    2
meter    1

Here e is the most frequent letter, so it would get chosen. This would mean leaving only the medical terms, which are very unlikely. But if the decision is taken by:

weight(letter) = w * frequency +
              (1 - w) * sum( usage_weight(word) where letter in word )

then, most probably t would be chosen.

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Because (let's say w = 0.2):

weight(e) = 0.2 * 3   +   0.8 * (1 + 2 + 1)
          = 3
weight(r) = 0.2 * 3   +   0.8 * (1 + 2 + 6)
          = 7.8
weight(t) = 0.2 * 3   +   0.8 * (1 + 5 + 6)
          = 10.2

Note: We should, of course use normalized weights, like frequency / num_words to get accurate weight measuring.

Note: This algorithm can and should be adapted to the opponent:

• when playing against human, more usual words get higher weight
• when playing against AI, it depends on the difficulty level:
  • on easy level aim for usual words
  • on hard level aim for unusual words