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For people not familiar with the card game Set, see its entry on Wikipedia and/or one of the related questions here on Math SE. It might be faster to just play the game a couple of times though, see e.g. this web-based version.

As there are $3^4 = 81$ cards, the maximum number of sets that can be obtained is $3^3 = 27$. I have played the game many times, but never used up all cards. This makes me wonder what the odds are of obtaining $27$ sets. The game usually ends with either $23$ or $24$ sets, but $25$ is not uncommon. It is not possible to end with $26$ sets though, as the remaining $3$ cards will also form a set.

Of course, certain types of sets are easier to spot than others. In addition, this seems to vary from person to person. In practice this might lead to a reduced number of obtained sets — the easier sets tend to have multiple features (number, shape, colour, shading) in common, thus upsetting the balance/distribution of features in the remaining cards. Therefore, it seems best to give priority to sets that have no features in common. This is just my intuition though.

[Edit] Based on one of the comments below I'd like to update the question. Are there any tactics that improve the probability of obtaining $27$ sets?

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    $\begingroup$ Quick Post: Anyone who has played the game knows that 12 cards are sometimes not enough to find a SET.Indeed,the rules specify that, if no SET is found, three additional cards must be dealt for the game to continue. This is repeated until a SET makes an appearance. One way to find out how many cards are needed is to do an exhaustive computer search of all the possible combinations. Such a search would reveal a collection of 20 cards that has no SET. Every collection of 21 cards contains a SET. (20 cards forms a Maximal-Cap See : homepages.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf $\endgroup$
    – Alan
    Commented May 12, 2014 at 19:10
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    $\begingroup$ How do you choose which set to remove if there are more than one? The Wikipedia page implies that the cards are dealt in threes and that with twelve cards on the board there are an average of about 2.78 sets. Whatever rule you specify it would not be hard to write a simulation. $\endgroup$
    – Ross Millikan
    Commented May 15, 2014 at 0:09
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    $\begingroup$ This is a nice question, but may need some clarification/refinement before anyone will answer it. One refinement might be to ask, "What is the probability of getting all $27$ sets with random play?" Another might be to ask, "What simple strategy maximizes the probability of getting all $27$ sets, and how much improvement over random play can be achieved?" -- where simple strategy would be suitably defined in the question. (For instance, memoryless and symmetric under permutations of feature types and feature values.) A simulation-based result is probably the most you can hope for. $\endgroup$
    – mjqxxxx
    Commented May 21, 2014 at 23:31

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In the case of play without any strategy, monte carlo runs get about 1% perfect games. https://mathoverflow.net/questions/66400/probability-of-having-a-perfect-game-of-set

If the players are allowed to use suboptimal strategies, they can guarantee taking 27 sets by only taking sets that differ along a single predetermined dimension, or more dimensions if they're careful to keep them oriented in "parallel".

If the players are playing optimally, meaning valid sets on the table must be taken, then the only allowable strategies are based on which set to take when more than one possible set is showing. This is a very limited tool. Players might keep track of the entire card space, and preferentially avoid sets that are more likely to result in unusable holes later on. Nevertheless, shuffles could happen where forced plays with only one option lead to an imperfect game.

So the probability with ideal strategy will be more than 1% (possibly quite a bit more) but less than 100%.

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