The game of Set consists of 81 cards. Each card has one, two, or three identical symbols of one of three shapes (oval, diamond, or squiggle), in one of three colors (red, green, or purple) and one of three textures (solid, hollow, striped). A “set” consists of three cards where each of the four attributes (number, shape, color, texture) is represented either once or three times. For instance, a green hollow oval, two purple hollow diamonds, and three red hollow squiggles are a set, as are two red striped ovals, two red solid diamonds, and two red squiggles. The deck of 81 cards represent each unique configuration of attributes (that is, \(3^4 = 81\)).

For the standard version, a dealer lays out twelve cards, and players look for a set. If no sets can be found, three more cards may be laid out. This led me to the question: How many cards are needed to guarantee that there is a set?

Every card is involved in exactly 80 sets, but every pair of cards is involved in exactly one set. So if we have two cards, there is a 1/79 chance that the third card will form a set. If we have three cards that do not form a set, there are three ways to create a pair, so there are three other cards we could deal next to form a set (a 3/78 chance that the fourth card will form a set). If we have four cards without a set, there are six ways to create a pair, so there are six other cards we could deal next (a 6/77 chance that the fifth card will form a set).

Let’s consider the pattern now. Given \(n\) cards that do not contain a set, how many ways are there to pair the cards? Each card can be paired with each other card, and we divide this by two because pairings are symmetrical. Generally, this is called the Handshake Problem, and the formula is \(p = n(n-1)/2\), where \(p\) is the number of pairs. Let \(d\) equal the number of cards not in the tableau (that is, \(81 – n\)).

Let’s create a table:

n | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |

p | 3 | 6 | 10 | 15 | 21 | 28 | 36 | 45 | 55 | 66 | 78 |

n+p | 6 | 10 | 15 | 21 | 28 | 36 | 45 | 55 | 66 | 78 | 91 |

d | 78 | 77 | 76 | 75 | 74 | 73 | 72 | 71 | 70 | 69 | 68 |

If there are twelve cards that do not contain a set, there are 66 cards that would create a set, of 69 available. That means there are three cards we could deal out andÂ *not* create a set. If there are thirteen cards that do not contain a set, there are 91 cards that would create a set, of 68 available. That means the next card, the fourteenth, must create a set.

Hence, any fourteen Set cards must contain at least one set.

An additional question to explore: What is the probability that twelve cards will contain a set? It is higher than 55/70, which is the probability that, if a tableau of eleven has no sets, the twelfth card will create a set.