The Life of a Mathematician

My sister has a card game called Set. Each card has a picture on it, which can be uniquely described in terms of four properties. Each property has three possible values. The deck has exactly one card with each combination of property values, giving a total of eighty-one cards. To play, you deal twelve cards face up, and then everyone looks for "Sets". A Set (capitalized to distinguish it from the math term) consists of a (lowercase) set of three cards which are either all the same or all different with regard to each of the four properties.
When someone identifies a Set, they remove those cards from the table and three more are dealt. If there are no Sets available, then you deal an extra three.

I had played this game once before, and thought I was competent, but no one comes even close to competing with my sister. She usually sees Sets before I have a chance to start thinking. Anyway, that's not what I'm posting about.

The instructions made claims about the probabilities of not having a Set when there are twelve or fifteen cards on the table. Over the past few days, I've tried to verify these figures mentally and have had much more difficulty than I anticipated in determining how many sets of a given size can be constructed which contain (or do not contain) a Set. I think I'm going to be obliged to work this out on paper.

A couple of interesting ideas:

For any given pair of cards, there is exactly one card which can be added to form a Set. Any given card is therefore a member of exactly forty Sets. This gives rise to an equivalence relation on pairs of cards, where two pairs are equivalent if they require the same card to make a Set. There are (81 choose 2) = 81*80/2 = 81*40 pairs of cards, which are thereby divided into 81 equivalence classes with 40 elements each.