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Are there any words that are impossible on Scrabble even if the blank tiles are used?

I mean that the words can't be played because there are not enough tiles to play it.

Scratch---Cat
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2 Answers2

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According to this reddit thread, the complete list of words in the allowed dictionary (which doesn't list words with more than 15 letters), the Collins Scrabble Words list from 2015, that you can't make with the available tile set is:

BAZZAZZ
BEZZAZZ
PAZZAZZ
PIZZAZZ
PIZZAZZY
BAZZAZZES
BEZZAZZES
PAZZAZZES
PIZZAZZES
KNICKKNACK
RAZZMATAZZ
KNICKKNACKS
RAZZAMATAZZ
RAZZMATAZZES
RAZZAMATAZZES
STRESSLESSNESS
CLASSLESSNESSES
POSSESSEDNESSES
SENSELESSNESSES
SUCCESSLESSNESS

Forget I was ever here
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L. Scott Johnson
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The Scrabble board is 15 squares square. That means that no word longer than 15 letters can possibly be played. So 'absentmindedness', 'counterbalancing', and 'antidisestablishmentarianism' will never be played in a Scrabble game.

There is only 1 'z' and 2 blanks. That means that any word with 4 'z's cannot possibly be played. This is not a particularly long list, but 'pizzazz', 'razzmatazz', and their conjugates are on it.

There may be other words formed using multiples of other rare letters, or combinations of rare letters, but I think this answers your question in the affirmative.

Arcanist Lupus
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    Of course, if playing the Dutch tile set with 20 "e"s and three "z"s worth only 2 each, the game is very different. – Forget I was ever here Sep 14 '19 at 07:40
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    What's the shortest? – Scratch---Cat Sep 14 '19 at 08:12
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    "What's the shortest?" is likely best made as a new question or an edit/clarification/addendum to the original question rather than a comment on this answer. – L. Scott Johnson Sep 14 '19 at 14:07
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    There is another much harder to enumerate class of unplayable words due to the limitation of only being able to play 7 letters at a time. In order to prove an 8-15 letter word is in this class, you'd have to show that there is no configuration of the other letters that forms a connected grid on which the word can be played with < 8 tiles in a turn without running out of any letters. This would be pretty challenging to prove, but there probably are words for which this is the case. – Darrel Hoffman Sep 15 '19 at 19:19
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    I'm know I'm late to this, but intrigued by @DarrelHoffman's comment I'm looking into it (using Python). We can assume that all 8-letter words are playable as they can cross another word containing any letter. I'm starting with 15-letter words from a list that's been linked here before (5925 of them). Of those, 1070 have valid 8-letter words in them and can be trivially dismissed. The rest will have to wait until tomorrow. From some examples towards the end of the list, ZIGZAGEDNESSES is easily formed from ZIGZAG and ESSES inserting EDN; ZYGOPHYLLACEOUS would be a little harder – Chris H Oct 22 '20 at 21:15
  • On further reflection though, it would be possibly to set things up so that every other letter was placed via a (e.g.) vertical word, and the 15-letter word was played by placing 7 tiles in a horizontal line, none consecutive. Thus all 15-letter words are theoretically playable, except those we can dismiss for having 4 Zs (or equivalent) as in the answer – Chris H Oct 22 '20 at 21:19
  • @ChrisH I think Darrel Hoffman already understood your latest comment. It's not quite a proof until you explicitly exhibit a sequence of plays from the start of the game that result in the given word. (For the purposes of this question, I guess we assume that no unchallenged phonies are played.) – Timothy Chow Oct 01 '23 at 14:24
  • @TimothyChow if I'd had a proof, it would have been an answer - except an answer to a slightly different question! I just offer an example that could be constructed from a fairly trivial setup, and one that couldn't by such an easy method. I got thoroughly bogged down trying to code a versatile demonstration, but maybe I should have just got the board out and worked backwards – Chris H Oct 01 '23 at 15:32