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Interesting. Thanks for your comment here and the one above. Although I firmly believe that hangman is a game, and hence the assumptions of game theory are in f

by dlss 15y ago

Interesting. Thanks for your comment here and the one above. Although I firmly believe that hangman is a game, and hence the assumptions of game theory are in full effect, I had not considered the information theory take on things :) 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 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.

kd5bjo 15y ago
How does this work if there are multiple covering sets with the same (minimal) number of question paths? Does it matter which minimal covering set you evaluate, or does your strategy need to be some kind of superposition derived from all the minimal sets? 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.
- dlss 15y ago
  I buy that information gain is the same as the optimal strategy, so long as you are taking into account your opponent's optimal strategy with a weighted probability distribution. 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 http://webdocs.cs.ualberta.ca/~darse/rsb-results1.html). You might also like http://wiki.mafiascum.net/index.php?title=WIFOM http://wiki.mafiascum.net/index.php?title=WIFOM Also, if you haven't been corrupted by LessWrong yet I highly recommend it :p
  - jules 15y ago
    That's not even close to giving a practical algorithm because (1) the set of words is huge, the powerset of words is huger (2) you didn't describe how to find the optimal counter strategy. And actually I don't think it's necessary to consider any subset of words. 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).

by dlss 15y ago

kd5bjo 15y ago

How does this work if there are multiple covering sets with the same (minimal) number of question paths? Does it matter which minimal covering set you evaluate, or does your strategy need to be some kind of superposition derived from all the minimal sets? 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.

dlss 15y ago
I buy that information gain is the same as the optimal strategy, so long as you are taking into account your opponent's optimal strategy with a weighted probability distribution. 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 http://webdocs.cs.ualberta.ca/~darse/rsb-results1.html). You might also like http://wiki.mafiascum.net/index.php?title=WIFOM http://wiki.mafiascum.net/index.php?title=WIFOM Also, if you haven't been corrupted by LessWrong yet I highly recommend it :p
- jules 15y ago
  That's not even close to giving a practical algorithm because (1) the set of words is huge, the powerset of words is huger (2) you didn't describe how to find the optimal counter strategy. And actually I don't think it's necessary to consider any subset of words. 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).