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IMO, never losing the game is the bare minimum for any reasonable cost function: never take > 6 guesses. Beyond that, you can have different optimization functions based on whether you're trying to minimize the worst-case depth of the decision tree (number of guesses) for each position, or the average depth, or some other reasonable function of the distribution over number of guesses. Although you could have arbitrarily many cost functions, a fairly general class is the following: pick some nonnegative constants (w1, w2, w3, w4, w5, w6, w7), and for a certain strategy, define the cost as: w1*n1 + w2*n2 + w3*n3 + w4*n4 + w5*n5 + w6*n6 + w7*n7
where nk (for 1≤k≤6) is the number of hidden (solution) words for which the strategy takes n guesses, and n7 is the number of words for which the strategy loses the game. Then,• Setting w7 infinite / very high is a way of encoding the condition that you not lose the game. (You can set it finite/small if you don't mind occasionally losing the game for some reason!) • Setting w1 = 1, w2 = 2, …, w6 = 6 will simply minimize the average number of guesses, • Having them grow exponentially, e.g. setting wk = c^(k-1), where c is some constant larger than 12972 (the number of acceptable guess words) will make sure that the minimal-cost strategy first minimizes the maximum number of guesses, then break ties by having the fewest words take that many guesses, and so on. |
I wouldn’t think so. If I get 99.9% of rounds correct with 3 guesses (and fail to find a solution to 0.1% of the rounds), and you get 100% of rounds with 6 guesses, I’d say I’ve soundly defeated you 99.9% of the time.