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In a card game called Dobble, there are 55 cards, each containing 8 symbols. For each group of two cards, there is only one symbol in common. (The goal of the game being to spot it faster than the other players, which is not the point of my question).

If I translate that to mathematical language, I would say that:

  • $S = [S_1, S_2, ..., S_{55}]$.
  • $S_n = [n_1, n_2, ..., n_8]$.
  • For $S_n, S_m \in S$ there is one and only one $n_a = m_b$

My double (dobble) question is:

  • Are there a finite or infinite number of sets and elements that allows such a property? I know there is one more with 30 sets containing 6 elements each (because of Dobble Kids, a lighter version of the game).
  • How can I calculate the number of sets, the number of elements in the sets, how many different elements there are in all the sets and which elements go in which sets? Is there a formula or is it simply a step-by-step try and fail method?


I realise that having sets like {1, 2, 3, 4}, {1, 5, 6, 7}, {1, 8, 9, 10}, ... answers the question (with 1 being the only element in common in each set). There is one more restriction:

  • Each element used in the sets must appear the same number of times (for example, in 7 given sets).

In the game, there are 50 symbols altogether. (55 cards, 8 symbols per card, 50 symbols altogether).

I have figured out a simple example with 4 sets of 3 elements (6 elements overall): $$S_1 = [1, 2, 3], S_2 = [1, 4, 5], S_3 = [2, 5, 6], S_4 = [3, 4, 6]$$

Each element is present twice.

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With no further constraints, the problem is trivial: if you want $n$ cards, $S_1,\dots,S_n$, let $\{s_0,s_1,\dots,s_n\}$ be a set of $n+1$ symbols, and let $S_n=\{s_0,s_n\}$. Then each pair of cards will have exactly the symbol $s_0$ in common. This satisfies all of the stated conditions, but I suspect that you want something more elaborate than this. – Brian M. Scott Jul 19 '12 at 8:27
I am not sure that this is exactly the same thing, but lines in a (finite) projective plane enjoy the very same property: any two of them intersect at exactly one point. – Andrea Mori Jul 19 '12 at 8:31
After rereading, I found that one piece of data is missing: how many symbols are there alltogether? (there's 8 on each card, but the total must be considerably bigger.Maybe 57? :) ) – Andrea Mori Jul 19 '12 at 8:47
Your edit improves the similarity with the projective plane. Each card is a line, the symbols appearing on the cards are the line's points, the fact that each symbol belongs to the same number of cards is simply the fact that for each point there's always the same number of lines going through it. – Andrea Mori Jul 19 '12 at 8:52
See also David Madore's recent writeup "Le jeu de cartes Dobble et la géométrie projective expliquée aux enfants"… – Zsbán Ambrus Jul 16 '15 at 11:00
up vote 9 down vote accepted

This is not a complete explanation, but a summing up of my observations in the comments above.

Consider the projective plane ${\Bbb P}^2({\Bbb F}_q)$ over the field with $q$ elements ($q$ must be of the form $q=p^f$ for some prime $p$). Then the following facts follow more or less trivially from the definitions:

  • ${\Bbb P}^2({\Bbb F}_q)$ consists of $1+q+q^2$ points;

  • ${\Bbb P}^2({\Bbb F}_q)$ contains $1+q+q^2$ lines, each of them containing $q+1$ points;

  • every two lines meet at a single point;

  • every point is contained in exactly $1+q$ lines.

Thus, if we call "symbols" the points, and "cards" the lines we have a situation which is exactly thatdescribed in the question.

The problem is that the numeric data do not correspond: if we take $q=7$ so to match the $8$ symbols in each card, the number of cards and of symbols should be $1+7+7^2=57$.

Then, either you lost 2 cards [ :-) ], or I'm left clueless.

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Hehe, I didn't lose any cards, it's the number of cards included in the game, according to the box. That same box also tells me there are 50 symbols altogether, if it helps. – SteeveDroz Jul 19 '12 at 13:47
Also: there are some cards that have the same symbol in common. For example: there are 3+ cards with a heart, as shown here:, which would imply to your calculation that there are points that are on 3+ lines. – SteeveDroz Jul 19 '12 at 13:52
Instead of you being clueless, maybe the manufacturer did it by hand and didn't get it quite right. Or maybe the manufacturer only wanted 55 cards and just skipped a couple. – Ross Millikan Jul 19 '12 at 14:03
@Oltarus: Perhaps you can count them, to see if they are actually 50 - and not count on the manufacturer's word. – ypercubeᵀᴹ Jul 20 '12 at 0:02

You can always define, for a set $I$ (which will be the index set, finite or not), something like this : Define $S_i = \{ \{i,j\} \, | \, j \in I \} $. (Since I chose the two-element subset $\{i,j\}$ instead of the ordered couple, order doesn't matter here.) Clearly $S_i \cap S_j = \{\{i,j\}\}$ for $i \neq j$. But the sets $S_i$ are "just as big" as $I$ itself.

For instance, if $I = \{1,2,3,4\}$, then $S_1 = \{ \{1,1\} , \{1,2\}, \{1,3\}, \{1,4\} \}$, $S_2 = \{ \{2,1\} , \{ 2, 2\} , \{2,3\}, \{2,4\} \}$ and so $S_1 \cap S_2 = \{ \{1,2\} \}$.

Would that be a good example? (The way I came up with this is : imagine for some point in $\mathbb R^2$ on the $x=y$ line that some line goes vertical through this point and another one horizontally. Then for two points $(x_0,x_0)$ and $(x_1, x_1)$ on the $x=y$ line, the intersection of the vertical/horizontal lines of the first point and the vertical/horizontal lines of the second point are precisely $(x_0,x_1)$ and $(x_1,x_0)$. By choosing "non-ordered couples" I get unicity. You can produce an example with some large number of elements in common by simply choosing more "components" and considering ordered tuples instead.)

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Duh, of course. Sorry, my bad. I edited the question. Your answer is perfectly correct but my question was missing something. – SteeveDroz Jul 19 '12 at 8:48

N = number of symbols per card; S = total number of symbols; C = total number of cards;

In stead of the maximum number of cards being S=N*N-N+1 and C=S, where (N-1) is prime, there is also a minimum number of cards, being S=(N+1)*N/2; C=N+1; and this is for all N>1.

SteeveDroz proved it already for N=3 with 6 Symbols and 4 Cards. where 2=(N-1) is prime too, that you do not need 7 Cards and 7 Symbols.

For N=4 there are 10 Symbols needed for the total of only 5 Cards;

For N=8 (the original Spot-It amount of Symbols per card) you need 9 cards and 36 Symbols, and not 57 Cards and 57 Symbols. 9 Cards with 8 Symbols each gives a total of 72 symbols on all cards. So each symbol occurs only 2 times in the pack of cards.

Example of deck of cards for N=4 and N=8

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