Set Game -- Statistics

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Version of April 30, 2002

Some Probabilities. From the standard Set pack of 81 cards, there are 85320 = 81 x 80 x 79 / 6 possible collections containing exactly three cards. Of these, 1080 are sets. Given a random collection of three cards, the probability that it is a set is only 1/79 = 1080/85320. However, under the standard rules of the Set Game players seek to find one set out of a collection of twelve cards, and the odds are much better.

Enumerating the Possibilities. It is often helpful to describe a collection of cards by mentioning how many properties are broken, how many are diverse, and how many are uniform. Indeed, an important feature of a set is that the number of broken properties is zero.

We can write a sequence of letters, a string, to summarize the configurations of the properties of a collection of at least two cards. In a string, the letter B appears once for each broken property, D once for each diverse property, and U once for each uniform one. For convenience, we place the letters in alphabetical order.

One can determine from either its string or its disposition how many properties are in a configuration. The distinction between the two is the the string will not tell us which properties are in which configuration, and the disposition will.

For example, this first collection:

has one uniform property (color), two broken ones (shape and number) and one diverse property (texture). Hence this card's string is BBDU. This second collection: has a different disposition, but is still represented by the string BBDU.
Three values per property, three cards per set. The first set of tables gives, string by string, the quantities of three-card collections and the quantities of sets that can be produced from the standard Set pack, as well as several other packs that have three values per property, but not four properties.

We obtained these numbers by writing a simple computer program that generated all the combinations and categorized them.

General Formulas. If we limit ourselves to strings that contain only D's and U's (in other words, combinations that qualify as sets), we can easily write a comprehensive formula for the figures in the tables. First we mention a few mathematical functions with which some readers may not be familiar.

The factorial function of a positive integer n, written as n!, is n multiplied by all the lesser positive integers. For instance, 5! = 5 x 4 x 3 x 2 x 1 = 120. Meanwhile, 1! = 1; and 0!, by special definition, also equals one. The exclamation mark is a well-chosen indicator because factorials can be very large; consider that 12! = 479,001,600.

C(m,p) denotes the number of combinations of m things taken p at a time. For instance, from five objects, A, B, C, D and E, there are ten ways to select two of them if the sequence of selection does not matter: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. By formula, C(m,p) = (m!) / ((m-p)! x p!).

Because superscripts do not always work in browsers, we write P(b,e) to denote b raised to the e power. To illustrate, P(5,3) = 5 x 5 x 5 = 125. We will assume that b is a positive integer, and e is a nonnegative integer.

Now we can get to the point. If within some string:

then the number of different sets described by that string will be

We require that d be greater than zero because if D never appears in a string, the string must then be all U's and all the cards be identical. This is impossible, hence the number of sets turns out to be zero, hence this formula does not work.

The general formula for combinations with broken properties turns out to be much more complicated. If some three-card combination has two broken properties A and B, they might be in phase:

Here the minority value of A is on the same card as the minority value of B. Alternatively, the two minority values might be on different cards, rendering the broken properties out of phase: In the first case, properties C and D cannot both be uniform, or the first two cards would be identical. In the second case, C and D can each be uniform without duplicating any cards. Two different combinatorial analyses are required.

With three broken properties, matters get worse. In this first example, any two broken properties are in phase with each other; we call this uniform phasing.

In the second example, any two broken properties are out of phase with each other. Naturally, this is labeled diverse phasing. And in the final example, property A is out of phase with each of B and C, but B and C are in phase with each other. This is (inevitably) called broken phasing. With three broken properties, three combinatorial analyses are required. The situation becomes even worse if all four properties are broken, and in packs with more than four properties, complication grows at an alarming rate. This is why we have not calculated formulas for the quantites of combinations with broken properties.
Four values per property, four cards per set. The second set of tables gives information analogous to that of the first set of tables. In the strings, symbols B, D and U still appear, but to them has been added P representing a paired property.

Paired properties generate phasing complications in much the same way as broken properties, so we have not attempted to produce formulas to yield the numbers in the tables.