I like to play a game with my family and friends. It goes like this:

The players take turns to say a letter. The letters taken in order must be the start of some word, and the person who finishes a word without there being a continuation on the word loses. If you are challenged on the letter you say, you lose if you don't have a word, otherwise the challenger loses.

For simplicity, plurals are allowed.

In the two player version of this game, assuming each player plays optimally who has a winning strategy (the player who goes first or the player who goes second)?

Note that by the fundamental theorem of combinatorial games, we are guaranteed that one player has a winning strategy cannot be a draw.

You are welcome to specify your own dictionary that you would use for 'adjudication' in your answer as long as it is reasonably well-known.

  • $\begingroup$ Isn't this a solved game, or am I thinking of something else? $\endgroup$ – greenturtle3141 Jan 21 '17 at 22:59
  • $\begingroup$ @greenturtle3141 It may be, I don't know. I couldn't find it online. $\endgroup$ – boboquack Jan 21 '17 at 23:01
  • $\begingroup$ Nevermind, this is basically the game Ghost with a slight rule change. $\endgroup$ – greenturtle3141 Jan 22 '17 at 1:13
  • $\begingroup$ I'm guessing that given any first letter, there is a letter which can be added to it so that creates the first two letters of a three-letter word that cannot be extended. Therefore, I'd guess Player 2 would have the winning strategy if both players played optimally. But since this is only a conjecture, I'm not posting it as an answer. $\endgroup$ – wildBillMunson Jan 22 '17 at 2:02
  • $\begingroup$ @greenturtle3141 What's ghost? Never heard of it. $\endgroup$ – boboquack Jan 22 '17 at 7:47

If the players use the Official Tournament and Club Word List as their dictionary, then the first player can always force one of the following words:

(I chose j because there's only so many letters it can be continued with.)

I think this strategy also works if they play according to the OSPD5, but I couldn't find a plaintext version for that, so it's hard to verify.


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