5 ms·

I wonder if its feasible to identify the optimal (or at least a near-optimal) strategy if you treat it as a game of pure logic, where the players agree beforeha

I wonder if its feasible to identify the optimal (or at least a near-optimal) strategy if you treat it as a game of pure logic, where the players agree beforehand which attributes from a limited vocabulary apply to which words.

mathgeek 5y ago
The game is sufficiently solvable as a 5-by-5 map without any concern for the words in each cell. Generally, it’s more interesting to study the actual gameplay around word association than to study the optimal winning strategy. There is some room to study whether the association between words could further optimize the naive approach. Another enjoyable side strategy involves optimizing for the opponent’s strategy.
vikingerik 5y ago
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.)
- deleted 5y ago
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- cochne 5y ago
  In order to mitigate the second team winning because of fewer possibilities after guessing, the first team could plan to not guess any on the first round, and then use an 'infinity' hint on the second round.
- anaerobicover 5y ago
  Fascinating, however with English having estimated order of 1e9 words, probably only that two-turn play could be enacted. Making second team to win in certainty in that kind of arms race.

mathgeek 5y ago

The game is sufficiently solvable as a 5-by-5 map without any concern for the words in each cell. Generally, it’s more interesting to study the actual gameplay around word association than to study the optimal winning strategy. There is some room to study whether the association between words could further optimize the naive approach. Another enjoyable side strategy involves optimizing for the opponent’s strategy.

vikingerik 5y ago

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.)

deleted 5y ago
[deleted]
cochne 5y ago
In order to mitigate the second team winning because of fewer possibilities after guessing, the first team could plan to not guess any on the first round, and then use an 'infinity' hint on the second round.
anaerobicover 5y ago
Fascinating, however with English having estimated order of 1e9 words, probably only that two-turn play could be enacted. Making second team to win in certainty in that kind of arms race.