The game of Dobble (will edit in a link later) involves a set of bespoke playing cards covered in symbols or small pictures - a dog, an arrow, a pencil, a tree etc.

Each card contains eight such symbols, and any two cards will always have exactly one symbol in common.

There are multiple variations of how the game is played but the basic object is to be the first player to spot the common symbol between two cards (typically between one in your hand and one on the table) and shout out the appropriate name to claim the point.

There are 8 symbols on each card.

  • How many different symbols must there be in total?

  • How many cards must there be in the full set?

  • Is there a general formula for the above as the number of symbols on each card changes?

Assume (to avoid degenerate answers) that:

  • each symbol appears on multiple cards;

  • every possible pair of symbols occurs on exactly one card.

  • 1
    $\begingroup$ This question does have the distinct feel of a math problem to it, but since I haven't tried to solve it, I don't know whether the solution might be puzzleriffic enough to make the question on-topic here. In any case, there is no way to give a meaning solution without more restrictions on the use of symbols: is every symbol used on the same number of cards, for example? (If not, then you can have, say, sixty billion cards, each with a teddy bear and 7 other symbols that are not shared with any other cards.) $\endgroup$
    – Bass
    Commented Jun 12, 2018 at 6:45
  • 1
    $\begingroup$ Some links to other places where this is discussed: old XKCD thread, stackoverflow question, Math SE question, blog post $\endgroup$ Commented Jun 12, 2018 at 7:47
  • $\begingroup$ @Bass I've added nondegeneracy conditions, thanks for the prompt. If it's voted off topic I'd be happy to migrate to Maths.SE $\endgroup$
    – IanF1
    Commented Jun 12, 2018 at 9:02
  • $\begingroup$ @IanF1 As Jaap Scherphuis has already mentioned, this was already asked on Math.SE - and your question more or less exactly asked on Math.SE here - so migrating would not be appropriate. $\endgroup$
    – Rubio
    Commented Jun 13, 2018 at 6:01

1 Answer 1


As stated, this problem is indeterminable. For example:

Every card could all share the same one symbol, and 7 other symbols unique to each card. This would satisfy the restrictions provided.

However, suppose we add the additional constraint that any two different symbols appear on exactly one card, and that the configuration isn't degenerate. Ok, I just described a finite projective plane. This turns out to be a surprisingly deep mathematical question!

As it turns out:

If each "line" (here, card) has $N+1$ "points" (here, symbols), it follows that if a finite projective plane exists it will have a total of $N^2+N+1$ points and lines. As $N+1=8$, $N=7$ and $N^2+N+1 = 57$


There are 57 symbols and 57 cards in total.

Now, we can provide a construction.

Let us arrange 49 of our 57 cards in a 7x7 grid. Then, for every "slope", we may draw 7 parallel lines (I let n=5 in this picture because otherwise it would look too cluttered. Ok, it still looks cluttered.)
enter image description here
We may compute that there are 7+1 slopes: we can either take one point from each column, and there are 7 possible row differences, or we can take the vertical columns all at once.
We can do this easily because 7 is a prime number. Now, for each slope, we adjoin a point "at infinity", so each red line connects the 5 points it has, plus the sixth red dot. enter image description here
Finally, to finish up, we connect all points at infinity with a line at infinity. enter image description here

Notably, as stated in the Wikipedia article, a generalization is hard for general n, although as demonstrated above, we can hit all symbol counts that are one more than a prime power. Notably, for example, projective planes of order 6 and 10 do not exist.

  • $\begingroup$ Thank you - I will add the nondegeneracy requirements to my question. $\endgroup$
    – IanF1
    Commented Jun 12, 2018 at 9:00
  • $\begingroup$ In the meantime you have my +1 and probably in time the green tick too. $\endgroup$
    – IanF1
    Commented Jun 12, 2018 at 9:04

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