A Better Strategy for Hangman
datagenetics.com
datagenetics.com
That is, S, when it appears, is heavily skewed towards the last letter in the word, finding it won't help you figure out the word nearly as well as finding the more evenly distributed E or A.
Which, for the ispell words list, appears to be 'e' :( edit: which was the solution found for the suboptimal method in the article, and which is not present for words in the corpus 25% of the time
code at http://pastie.org/3693750
To calculate this, take the sum of -p * log(p) for all patterns of the guessed letter, where p is the proportion of words in the candidate set that have this pattern.
That depends on your opponent. In game theory you often assume that your opponent plays against you and as good as they can, so you'd go with the worst case.
Since you asked about the Nash equilibrium for this case, here it is:
In the event that you did not have enough questions to uniquely identify the possible words (and assuming that words with more letters than guesses are not allowed ie every word can be guessed with the right questions), the game becomes a matching game [http://en.wikipedia.org/wiki/Matching_pennies].
First both players will work out the minimum number of question paths that can identify every allowed word. Any word in more than one set will be ignored as a dominated strategy by the hosting player. Then the guessing player will pick randomly between question paths, and the hosting player will pick randomly among the disjoint sets of words at the end of those question paths, with the specific word not mattering.
You can see that once both players are picking between several options, with the entire game riding on their choice, the actual letter guessing becomes unimportant. The game itself resembles an unfair rock paper scissors.
If the players were betting $1 per round, the expected value to the guesser is ($2 divided by the number of disjoint sets) - $1.
Also, if there are some words that are never picked by your strategy, doesn't that open the door for an alternative set of question paths that only covers the now-reduced dictionary? (cue reference to the Unexpected Hanging Paradox)
In many ways, the entropy is a heuristic for finding a good question path given the known probabilities for each word in the dictionary. The formula that I quoted above can be trivially extended for nonuniform distributions of the chosen words. It sounds like the Nash equilibrium would be designed such that (after eliminating dominated strategies), the expected information gain at each step would be the same for all question paths. I wouldn't want to have to prove it, though.
The unexpected hanging paradox is a great link, thanks :) awesomely enough though, game theory handles this problem! This is the "why" behind minimax. If John Nash were sentenced by that judge, Nash would decide what day to expect by flipping a coin, and the judge would segfault :p
To answer your question though, if there are multiple paths to the same words (eg Tree(1) leads to Union(a,b), Tree(2) leads to Union(b,c), and Tree(3) leads to Union(a,c)), that surprisingly doesn't make the solution any harder... though my links above aren't particularly useful for it.
To find the optimal strategy for the player choosing the word: for every possible set of words (so all 2^words-1 options), calculate what the optimal strategy would be for your opponent if they were playing against someone picking randomly from that group, and then choose the group causing the lowest expected value for your opponent.
Make sense? [edit: NO! I didn't show that optimal would necessarily come from a uniform random selection from the word bag... and a proof doesn't jump to mind, though my intuition tells me so]
[edit 2: YES! Just assume that the strategy doesn't need to be uniform [though I still maintain it will be], and use gradient descent to find the optimal weighting]
You seem like the sort of person who would LOVE reading about the computational roshambo competition run out of UBC a while back (specifically the iocaine powder overview at http://webdocs.cs.ualberta.ca/~darse/rsb-results1.html). You might also like http://wiki.mafiascum.net/index.php?title=WIFOM
Also, if you haven't been corrupted by LessWrong yet I highly recommend it :p
The usual way to find a Nash equilibrium is with linear programming. But for that you have to construct a matrix of size #(deterministic strategies for player 1) * #(deterministic strategies for player 2). Player 1 does not have so many stategies (just the number of words in the dictionary) but player 2 has an enormous number of strategies.
So something else is needed. We can try to write an algorithm for finding expected value of the the optimal counter strategy to when player 1 is playing a mixed strategy, let's call it ev(s) where `s` is a game state. Then we would have ev(s) = max_{s' in successors of s} ev(s'). Perhaps this can be done using dynamic programming, but it could lead to far too many states being evaluated. If you could give an approximated upper bound then perhaps some of the paths could be discarded early. Anyway then the Nash equilibrium is to minimize this ev over all mixed stategies of player 1, i.e. all probability distributions over the words. How one would do that is another tricky question.
One positive news is that we can consider each word length separately, as the final Nash equilibrium would always choose a word of the same length for player 1, i.e. the optimal strategy for hangman would always choose a word of size n (but we don't know n yet, and n will depend on the dictionary).
What follows is my solution to this game after having a night to sleep on this. It's O(26^26 times words) [because it requires you construct the optimal move for the guesser], so perhaps only of interest to myself.
Problem restatement: Imagine instead that we are playing a game where, for doors 1 through m, we must pick a door and hide some prize behind that door. Our opponent must then choose from a list of moves that open ranges of doors.
To solve that problem, it is relatively easy. First form the set of transitive closures on moves, such that any move that shares a door with another move is in the same closure. Next, for each closure, mark every door by the number of moves that can reach it, and progressively remove the highest marked number until doing so would then leave no doors. If at any point this process led to those moves no longer sharing a closure, reform the closures, add them to the list, and process them again in this manner.
Observe that for any closure remaining, for every pair of doors (a,b) reachable by the same moves, we can discard the second without changing the optimal strategy (because any move the guesser made would be the same to us for both a and b).
Now observe that for any closure remaining, if the dictionary only included words reachable from that closure, picking with a uniform randomness among the doors not removed by counting or pairing would be optimal (as the same number of moves can reach any remaining door). To prove that moves with higher marks would only decrease our EV, simply observe that if we did include any door with more marks, then those moves could be favored by the guesser to increase the guessers chance of winning.
To complete the solution, we work out the the expected value of playing each closure independently, and select between them by the ratio of their inverse EV with the whole. (So if we had two closures, one with EV -1 [if we could only choose from words behind this door we would always loose], and the other with -4/5 [all doors can can be reached by 4 out of the 5 moves in this closure], we would pick between the two with a ratio of 4:5).
For a proof that the above mixing is optimal, see the analysis for the game "Bluff".
Anyone who understood all that, lives in San Fransisco, Santa Barbara, or San Diego, and would like to meet up to discuss economics, game theory, algorithms, machine learning, computer science, or world domination over coffee is cordially invited to email me at dls@lithp.org (yes, that's "lisp with a lisp", perhaps puns should have been on that topic list as well :p)
For example, which letter do you guess if the puzzle is _og? You still have to go through bcdfhjln even after you've burned however many guesses getting that o & g.
It had one problem in that it always tried to make sure you guessed wrong, but by carefully choosing your letters you could cause it to exclude tons of words because they contained the letter you guessed, leaving only a small number of possible words.
It needed a refinement to occasionally allow you to guess correctly if that maximized the number of possible words remaining that still fit the prior guesses.
Remember, if you pick a correct letter you will not loose any guesses, in other words you get another guess if your guess is right. It's not about finding the word in a fixed number of guesses.
But otherwise you could also extend your method to increase the weight of letters appearing more than once in a word since that gives even more information gain.
The thing is, at least for the dictionaries I had available, trying to weight for information gain only considering the guess correctness wasn't all that great a strategy. Even if you know 100% certain that the letter will be correct, its very likely to still cut the search space down by virtue of the pattern of letters, and in aggregate my strategy made no statistical difference. (+/- 1% depending on random seed).
If I were going to try and improve my scores, I guess the next thing would be look at the distribution of letter-patterns for each remaining letter, but my intuition is that it just wouldn't be a big deal.
Their homepage says this "DataGenetics is a technology consultancy specializing in unlocking the value stored in large databases. Using a variety of techniques we can mine the trends in your data to help you maximize your marketing and advertising campaigns."
Damn.
Yet another group of smart folks focused on more-effective advertising.
Are you able to consider the idea that someone's responsibility to themselves and their families might lead them to make decisions other than maximizing financial gain?
Also - I'm curious as to why you think family members are a special case?
Of course I'm able to consider the idea, rbarooah. What I reject is the notion that we can project our own values on a person and then blame him for not fulfilling them (thus my original comment).
What were you trying to convey with the phrase "It's a shame?" Was that not your judgement of their choice?
Also, how do you propose to prevent other people from blaming one another for things?
You didn't answer my question about why family members are exempt from your general view - which I am genuinely curious about.
This is great if you're playing hangman against a computer. Probably, you're not.
You're playing against a person, who probably has some sense of your strategies.
Even when I was a kid, we didn't play hangman by choosing random words. What's the fun in that? You notice how the other person plays, and in the next rounds you pick words that break their strategies.
You notice they are doing the "common letters" or even "common vowels" strategy, and you give them "lynx".
Or you trick them with a word that has a few easy-to-get letters, but is going to be hard to guess the last few because there are so many possible matching words -- I had a great one that I forget now... maybe "budder"?
Did everyone really play with randomly-chosen words? What a waste.
Of course, this analysis has a counterpoint. If I try to pick words strategically, I suspect that my words would begin to follow patterns...
We did have one minor rule change -- definitely more than 11 line segments to hang the man. If you had a really good word, and wanted to draw out the torture, you'd start drawing in facial features, fingers...
It also adds a bit of artistic fun to the game; you don't even have to be hanging a "man" if you want to get creative; it could be a giant insect, horse, whatever.
It would still be pretty clear who was "winning" by how long the string of crossed-out letters were by each round.
In general, shorter words are harder to guess in hangman.
* assuming as he does at the bottom the existence of a computer.
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".
But right and wrong guesses are not equally valuable in hangman. A wrong guess will eliminate all words of a given length containing the given letter. But a correct guess gives you more: you get the location (or locations) of the letter in the word as a bonus. This will reduce your remaining search space drastically.
Thus, you should optimize for the success of your next guess.
- you are correct (happens 99% of the time). No penalty, and you find out at which positions your guessed letter occurs, a decent information gain.
- you are incorrect. You suffer a penalty but eliminate 99% of the remaining words, a massive information gain.
How would 50% be preferable in either case?
If we get a hit, it's not just the presence, or not, of the letter, we learn both the number of hits, and the positions of these hits. These are incredibly valuable pieces of information (more valuable that dividing the remaining words into two equal piles).
If we miss, we've carved away and removed the biggest non-possible set of words with one guess.
I look forward to seeing your analysis of how to deal with that :)
Does 'qi' count? http://www.merriam-webster.com/dictionary/qi
I know it counts when playing Words with Friends...
I also think 'za' is valid when playing Scrabble type games, but 'za' technically isn't a word.
Hmmm... Yak butter tea.
I could make good arguments for both "most likely to be present" and "most information gain", so I implemented a mixed strategy. I calculated the information gain (in bits) and the probability of a correct guess, then used a linear combination of them to determine which letter to guess. Additionally, the weights changed over the course of the game so that correct guesses were valued more as incorrect guesses became more scarce.
To determine appropriate weights, I ran a crude Monte Carlo simulation. The final weights I ended up with were 0.60:0.03 with no incorrect guesses and 0.54:0.69 once there aren't any strikes left. (bits information gain : p(correct guess))
I'm not arguing that they are, or are not, words. I'm just saying that they were not in the dictionaty file I used :)
Vs a week player the ideal word list should only include 'common' words.
Vs an ideal player you need to assume they will pick from the list of words that you least likely to guess using optimum play.
Generically, optimum play ends up with a somewhat random list, aka 90% of the time pick E, 10% of the time pick I etc.
PS: Actually generating this list is a 'hard' problem and vary dictionary dependent, but you can probably get reasonably close using some sort of genetic algorithm and a enough simulation time.
Sounds like a challenge. I don't think, given a dictionary, finding the ideal strategy for both parties will be that challenging. It's a fairly straightforward two person zero-sum game. You can either model it assuming hidden information, or concurrent play. (Which is basically the same here.)
As an interesting variation, you might allow the chooser to cheat: I.e. don't make them write down the word in the first place, just require their play to be consistent.
1) A hit or miss of a letter tells you a lot about what the new subsequent optimal guess is. You could make a flow chart (a really big one) that shows you the optimal letter to call out next based on what has hit and missed.
2) the position that a letter has hit tells you a lot about the target word. if you have a computer, the regex becomes trivial to identify the next optimal letter to guess. an 'optimal' table could be generated based off this pattern, and it would be a huge table.
I kept feeling like every step of the way it was an infomercial, 'but wait... there's more!' and I like that, it got me thinking.
I'd love to have my hangman bot go head to head with yours on random words. that would be a fun little project.
Put another way, what are the letters I should guess when I don't know the length of the word?
While it wouldn't be perfect strategy, it would be easier to execute.
We should have a hangman AI tournament.
For example, if I have guess T and E and see _ _ E as the word I know it's not the word "The".