I'm referring to the Ebert's version where the solution is closely related to Hamming codes. What happens if there are not 2^^K-1 players ? I wrote an algorithm to try to find how many N long code-words are there where each two has a minimum distance of 3. (It's greedy, so not definitely correct, but seems to work for small numbers.) For example with 6 players, there seem to be 8 code-words, which means that 8 out of the 64 times all of them will guess incorrectly, 8*6=48 times only one will guess and correctly, and 8 times none of them will guess. This is a 75% success rate.
Is there a general formula for how many N long words are there where each two differs in at least 3 positions ?