In the game of Projective Set, it turns out that any seven cards contain a projective set. How can one prove this? And for fewer than 7 cards, how can we determine the probability that one or more sets exist (in terms of the number of cards)?
For those not familiar with Projective Set, here's a description from the wikipedia page linked above:
A Projective Set card has six binary attributes, or bits, generally represented by colored dots. For each color of dot, each card either has that dot or does not. There is one card for each possible combination of dots except the combination of no dots at all, making $2^6 - 1 = 63$ cards total.
Three cards are said to form a "set" if the total number of dots of each color is either 0 or 2. Similarly, four or more cards form a "set" if the number of dots of each color is an even number.
A card and itself could be said to form a two-card set, but as the cards in the deck are all distinct, this does not arise in actual gameplay.
Idea
In a vector space of dimension $6$, any $7$ vectors are linearly dependant.
Details
The $63$ cards can be represented as the elements of $\mathbb F_2^6 \setminus\{\mathbf 0\}$: The six coordinates correspond to the six colors. If a color is present on a card, the corresponding entry is $1$ and otherwise $0$.
Now let $S \subseteq \mathbb F_2^6 \setminus\{\mathbf 0\}$ be a collection of cards. Under the rules of the game, $S$ forms a set if and only if $S$ is not empty and $\sum_{\mathbf x\in S} \mathbf x = \mathbf 0$. Hence $S$ contains a set if and only if $S$ is a linearly dependent subset of the $\mathbb F_2$-vector space $\mathbb F_2^6$.
Now the claim follows by the fact that any $7$-element subset of a $6$-dimensional vector space is linearly dependent.