Does anybody know what the lowest amount of tiles that can be put on a Scrabble board such that all words that are on the board are legal, but it is now impossible to play another word with the remaining tiles is?
I don’t know, but I’ll warrant it involves the word “qi”
What sort of assumptions are you making about the tiles each player has? Can they have a rack with all the worst tiles, or assuming they have access to all the tiles?
I’m going to guess 5 pieces in a plus shape can result in no legal plays.
I don’t think so. Cross the words EGO and AGE and I see several plays right off the bat, for example.
I would be assuming that you can take any 7 tiles you like from those not on the board.
Well that’s Using two of the more advantageous 3-letter words, don’t you think?
Is the same word allowed to appear more than once on the board?
A 3X3 cross of five tiles might work, but watch out for ways to extend words. For instance, with LEG and BEG, I don’t think you could put a single tile in any of the corners, but you could still stick an S onto the end of either of them, or build, say, GIFT off of one of the Gs (the cross-word, “gi”, is the loose robe worn in martial arts).
Yes, but the question was about number of tiles, not the actual play.
Hehe yeah. I started out looking at scrable databases, looking for odd words, and thought I at found one with ZUZ x ZUZ.(Z is cool because it is part of no 2 letter words). Unfortunately somebody decided to allow MeZUZoth 
(although I have no idea if there are 4 Zs in a scrabble box on second thought, probably not.)
Yes.
However, it’s not on the official list for tournament play.
And I see from that list ‘Za’ is now allowable, so I was just really really offbase.
There’s one Z and two blanks (which can be any letter but have zero point value).
Man I wish I was still TAing. The OP would be a very interesting problem for computer students to think about on algorithms for, or running a deep stacked brute force search. I just might do that myself.
Although IF you don’t find an answer by 15 or so tiles you might be looking a complexity close to solving chess.
Aha! That reminds me of a technicality that makes a trivial answer possible. While of course tournaments need to have an official dictionary, the rules for the game itself do not specify any particular dictionary, only that it must be one that is mutually agreeable to all players. A group of players could, therefore, choose a sufficiently-restricted dictionary that there exists a two-letter word which cannot be further extended.
For the five-letter cross, VAC crossed by VAC would not be buildable within it, since V and C are the two letters with no two-letter words; you could, however, do any of dozens of words around one or the other VAC, or do VACS and build up or down from the S.
Jeez…took me over an hour,but I think this is a dead board using six tiles.
V
VOX
…XU
I’m having trouble understanding here. Are you saying there’s a two letter word in a dictionary somewhere, in which no other words exist that contain either of the two letters of the original word? I seriously doubt it.
Clever.
Very good. Bravo.
+1
As the saying goes – Mad taught me everything I know about playing Scrabble competitively – alas, he hasn’t taught me everything he knows … 