Sorry about that. I got too caught up in showing that there was in fact an optimal strategy, and there were no hangman paradoxes. In response to (2), that was d

Sorry about that. I got too caught up in showing that there was in fact an optimal strategy, and there were no hangman paradoxes. In response to (2), that was done in the GGP -- you are selecting the question that for all words gives you the least distance from the answer (and if that doesn't cover every answer, then the minimal number of questions to do so). I posted ruby code in the GGP for a heuristic version, the non heuristic version would look similar. 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)