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For anyone not familiar with the game of Set, I'll refer you to the description on this question.

My question is this: The game ends when there are no more cards remaining in the deck and there are no sets visible on the table. I will define a "Perfect Game" as one in which there are no cards at all on the table at the end, i.e. all 81 cards have been removed as part of a set. Can any ordering of cards result in a Perfect Game?

Now, in practice, this is a fairly rare occurrence. Games typically end with 6, 9, or 12 cards remaining. They can even end with as many as 15 (this seems to be even less common than a 0-card ending, but I have seen it happen). 18+ I believe to be impossible. I think that 16 is the maximum number of cards that can be showing without making a set, and there must always be a multiple of 3, so that rules out anything > 15. (Edit: Apparently up to 18 is possible.) A 3-card ending is also impossible, per the question I linked above. (The last 3 must form a set, so this would result in a Perfect Game.)

Now, in a normal game, there are often times where more than one set is visible, and which one you pick can affect the outcome of the game at the end, and you have no way of knowing what those outcomes would be. But let's say that we are given full knowledge ahead of time about the exact order of cards in the deck. Let's say we also have the ability to perfectly predict the outcomes of each choice, or alternatively to "hit Undo" and backtrack all the way to the beginning as often as needed. Could we with this knowledge ensure that any ordering of cards can result in a Perfect Game?

I suspect that this would be easier to disprove than to prove - Simply find a contrived deck ordering which at no time gives you multiple choices of visible sets or which we can trace the game through to all possible endings if there are choices, and prove that they all must leave some cards on the table. Proving that any deck order can end perfectly would be quite the challenge...

Edit: I just noticed this question, but it is not a duplicate since it does not consider the possibility of knowing the order of cards ahead of time.

Darrel Hoffman
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    According to the linked question, 20 (not 16) seems to be the maximal set-less hand ...? – Hagen von Eitzen Jun 11 '18 at 21:02
  • https://en.wikipedia.org/wiki/Set_(game)#Basic_combinatorics_of_Set: "The largest group of cards that can be put together without creating a set is 20." – Martin R Jun 11 '18 at 21:04
  • @HagenvonEitzen - Hmm - I did say "I think" on that one. My reasoning was that if you remove all cards of one color, then all cards of one shape, then shading, then number, what you're left with is 16, e.g. 2^4 instead of 3^4. Add in any one of those removed cards and it's guaranteed to make a set. It's possible some other method might allow more cards? – Darrel Hoffman Jun 11 '18 at 21:05
  • I wonder if a counting argument might work: First, figure out how many ways there are to partition the deck into sets. Then, there are $27!$ ways to order such a partition. Count how many orders of the deck allow the sets to be taken in that order (or get a good upper bound, because this is hard due to the rules for if not sets are on the board). If this is less than $81!$, then there is some deck not allowing a perfect game. (My guess would be that this method does actually work, but I'm not sure how to do all the required computations) – Milo Brandt Jun 11 '18 at 21:11
  • @MiloBrandt Yeah, clearly the ordering of any given sequence of 3 cards in the deck (starting at position 3n+1) is irrelevant since cards are dealt 3 at a time, so that means a lot of deck orderings are going to be equivalent in terms of game-play. – Darrel Hoffman Jun 11 '18 at 21:15
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    2 years and probably hundreds of games later (I have an app on my phone in addition to a physical deck), I'm still convinced that an 18-card ending is impossible. While you can per the above linked article cherry-pick 20 cards with no valid sets, I don't believe you can reach an 18-card ending by following normal gameplay - i.e. if you have 18 cards and no sets, I'd bet that the 63 remaining cards cannot be grouped into 21 valid sets. This might be another good question to ask. – Darrel Hoffman Apr 15 '21 at 19:43
  • @DarrelHoffman About 5 minutes of fiddling around, close to a 'greedy search', yields an example of a collection of 21 disjoint sets with no sets in the remaining 18 cards. – Servaes Aug 01 '23 at 21:14
  • In fact another few minutes of fiddling around shows that you can take 9 blue sets (all blue cards), and then 6 green sets and 6 red sets in a pretty symmetric way, and be left with 18 cards that do not contain a set. – Servaes Aug 01 '23 at 21:19

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