Wikipedia:Reference desk/Archives/Mathematics/2016 January 11

As 55 divides neither by 2 or by 3, if the manufacturers thought that 57 was impractical, why not use 54? In my opinion, a little extra effort should have been put in to make the full mathematical complement of 57 cards.

So,what are the cards missing from the 55 card set that would make the set mathematically complete, so that every symbol appears 8 times and every symbol appears with every other symbol exactly once?

The symbol that only appears 6 times is the snowman. This is therefore the symbol common to the 2 missing cards. An analysis of the 440 symbols on the cards (8 x 55) allows us to determine the cards that should be added. These are: -

1) Cactus, dinosaur, flower, gingerbread man, ice cube, four leaf clover, question mark, snowman. 2) Dog, eye, exclamation mark, ladybird, lightning, skull and crossbones, snowman.

= January 11 =

Dobble maths
Dobble is a card game with 55 (it should be 57) special cards, each with 8 symbols on it, such that any and every pair of cards has exactly one symbol in common between them. In the following text, S is the number of symbols per card, N is the total number of symbols in the pack and C is the number of cards in the pack.

It is obvious that a simple set can be made with S=2, N=3, C=3. Namely AB BC CA.

A little thought shows that you could have S=3, N=6, C=4. Namely ABC ADE BDF CEF. But while this is "balanced" (every symbol is on exactly 2 cards), it is not "full" because there are some pairs missing across all cards (namely AF BE CD). This suggests that a better set would be S=3, N=7, C=7 : ABC ADE BDF CEF AFG BEG CDG.

A search on the web has pulled up the statement that a set of order X has S = X+1, C = X^2+X+1, but I have found no formula for N in terms of X.

Using this we have X: S C (N); 1: 2 3 (3); 2: 3 7 (7); 3: 4 13 (?); 4: 5 21 (?); 5: 6 31 (?); 6: 7 43 (?); 7: 8 57 (?).

Can anyone provide a formula for N, and is there an easy way of generating the C groups of S? -- SGBailey (talk) 21:44, 11 January 2016 (UTC)
 * It sounds like what you're describing are finite projective planes. Replace the word "card" with "line" and the word "symbol" with "point" and you get the geometric sounding axioms: 1) for any two points there is one line containing both of them, 2) any two lines have one point in common. (The fact there are no "parallel" lines means we are dealing with projective space.) Finite projective planes have been the objects of much study, in particular the question of which orders are possible is very difficult, so there is no easy way in general of generating the groups. But certain cases can be done easily using modular arithmetic, or finite fields more generally. Set is another card game with a geometric interpretation, but in that case it's as a 4-dimensional affine space. --RDBury (talk) 04:01, 12 January 2016 (UTC)


 * Thanks. The projective plane article seems to be saying that N = X^2+X+1 as well. -- SGBailey (talk) 06:24, 12 January 2016 (UTC)


 * Yes, there is a duality between cards and symbols, just as there is a duality in finite projective planes between points and lines. So in your ABC ADE BDF CEF AFG BEG CDG example you can regard A, B, C etc. as symbols and the groups ABC etc. as cards or you can regard A, B, C etc. as cards and the groups ABC etc. as defining the symbol that is common to those cards. With the alternative interpretation, card A would contain symbols (ABC), (ADE) and (AFG). Gandalf61 (talk) 15:19, 12 January 2016 (UTC)


 * I won't give a direct answer, but let me point you to some posts on the web about the maths of the Dobble card game (also known as Spot it!): Dobble card game - mathematical background (on Mathematics StackExchange), What are the mathematical/computational principles behind this game? (on StackOverflow), and Le jeu de cartes Dobble et la géométrie projective expliquée aux enfants (blog post of David Madore). &#x2013; b_jonas 17:12, 13 January 2016 (UTC)