The combinatorial of 25-choose-9 is only 2,042,975. If you have a vocabulary space of that size, you could guarantee always winning in one turn.
Doing it in two turns of 25-choose-4 and 21-choose-5 is easy, you need a vocabulary space of only 12k and 20k for the two steps. However, the second team could win in one turn in a smaller combinatorial space after the first team makes some guesses. 21-choose-8 is only 203k.
How you encode the combinatorial space to the game grid is an open question. The rules do disallow using the position of the words or the letters within them. It works if you have any legal method of comparatively sequencing the 25 words on the board.
(Of course, I'd never advocate actually playing the game like this, this is just mathematical speculation.)
Doing it in two turns of 25-choose-4 and 21-choose-5 is easy, you need a vocabulary space of only 12k and 20k for the two steps. However, the second team could win in one turn in a smaller combinatorial space after the first team makes some guesses. 21-choose-8 is only 203k.
How you encode the combinatorial space to the game grid is an open question. The rules do disallow using the position of the words or the letters within them. It works if you have any legal method of comparatively sequencing the 25 words on the board.
(Of course, I'd never advocate actually playing the game like this, this is just mathematical speculation.)