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I was playing the game Dobble and I became extremely curious to figuring out a formula to make a similar game.

There is a deck of cards.

Rules.

  • Each card must contain a multiple number of unique symbols
  • Each pair of cards has exactly 1 matching symbol.
  • 8 unique symbols per card.

Unknowns:

  • Not sure how many cards
  • not sure the total of different symbols for the entire game.

$x$: symbols per card. $y$: total symbols. $n$: number of cards.

Is this enough information? How can I solve for $n$ and $x$?

There are 3 variables and I would like to change them down the road, so can we just assume that:

8 unique sybols per card. 15 total unique sybols. $n$: number of cards.

I am not sure if 15 works. We can change that if needed as well.

Michael Bruce
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    $x = 8$ and $y = 15$ is a trivial case: the first cards contains 8 units, the second one contains one of this 8 and 7 more, the third card would violate rules. So $n \le 2$. – Smylic Feb 09 '17 at 21:38
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    I started writing up a solution before thinking that there was something that did not make sense in the formulation. This requirement is troublesome: "Each card must have ONLY 1 matching symbol with each other card". It seems to say that any two cards have exactly one matching symbol, which implies that the number of cards altogether is $x+1$: for any one card chosen, there must be exactly $x$ other cards, each with one symbol matching the $x$ symbols on the one card chosen. – Lee Mosher Feb 09 '17 at 21:46
  • There are two more reasonable conditions: $x \ge 2$ and each symbol presents on at least two cards. – Smylic Feb 09 '17 at 21:48
  • Certainly it is not x+1. Perhaps I am explaining it wrong. Only 1 matching per card and you can have any number of symbols and cards. 8 symbols can be used at once per card. Only 1 can match for each card. – Michael Bruce Feb 09 '17 at 21:49
  • Maybe it would be better to say that each pair of cards has exactly one common symbol. – Smylic Feb 09 '17 at 21:51
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    Are you aware that this game implements projective geometry in a finite field (http://www.madore.org/cgi-bin/comment.pl/showcomments?href=http%3A%2F%2Fwww.madore.org%2F~david%2Fweblog%2F2015-07.html%23d.2015-07-15.2307) ? Some exchanges are in French, but most in English. – Jean Marie Feb 09 '17 at 21:57

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There are already few partially duplicate questions about Dobble (SpotIt cards) :

What is the algorithm to generate the cards in the game "Dobble" ( known as "Spot it" in the USA )?h

Dobble card game - mathematical background

How many combinations can be made with these rules? (game of Dobble)

Here is the summary that answers this question :

Assuming that we have cards with $x = q+1$ symbols, the set of Dobble cards is based on Projective planes (see Wikipedia for Projective planes) of $q$th order. I.e. if there is prime number $p$ and positive integer $i$ such that $q = p^i$, then maximal number of cards is $q^2+q+1$ and the same is necessary number of symbols. If $i = 1$, then Incidence matrix for generating Projective plane in Paige-Wexler normal form may easily be generated using Galois Filed $GF(p)$. If $i$ is not 1, then it is a little bit problematic to generate corresponding Galois Field $GF(p^i)$ consisting of polynomials, so it is much harder to generate corresponding projective field and game of Dobble. Available Dobble sets (with $x = 8$ and $x = 6$) have two and one missing card respectively (i.e. 55 instead of 57 and 30 instead of 31).

There are additional interesting properties that are fulfilled by the projective field and would be fulfilled for set of Dobble cards if it had all 57 (or 31) cards:

  1. Every pair of symbols appears on exactly one card
  2. Every symbol appears on exactly $x$ cards
  3. Role of symbols and role of cards may be interchanged so that the incidence between symbols and cards is still maintained (principle of self-duality).

For $q$ not equal to prime power, the projective plane does not exist, so more general Steiner system specifically $S(2,x,y)$ have to be used, which is more difficult to construct generally.

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ludvik02
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