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The game called Dobble consists of a deck of cards; each card contains 8 symbols from a set of 50.

The deck is made so that any two cards have exactly one symbol in common. (The idea of the game is that on each turn, the first player to spot the symbol that's on both cards gets the card. The winner is the one who ends up with the most cards.)

My question is,

With the above numbers, what is the size of the largest possible deck?

(According to the rules there are actually 55 cards in a deck.)

My guess is that an explanation would involve an application of the multinomial theorem, but I can't see how to work it out.


marked as duplicate by Gerry Myerson, user147263, Jeel Shah, RghtHndSd, Claude Leibovici Jul 15 '14 at 5:41

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It cannot be done with the numbers originally given. (In other words, with 10 symbols per card chosen from a set of 40, a deck of 50 cards is not possible.)

We will begin to deliberately construct cards so that they obey your constraints. We construct the first card using an arbitrary selection of 10 symbols from the 40 available. Then, for the second card, we select one symbol from the first card and the remaining 9 from the set of unused symbols (there are currently 30 of these).

For the third card, we may either use the symbol already on both of the existing cards (in which case we must use 9 more unused symbols, of which there are now 21, to fill the remaining spots) or we may choose one symbol from the first and a different symbol from the second (in which case we need only to use 8 of our 21 unused symbols). The second seems like the better option, since we use up our symbols less quickly.

We continue constructing these cards using the "better" method and find that we have 13 unused symbols left for the fourth card, 6 left for the fifth card, and... By the time we are trying to construct our sixth card, we have already used up all of our symbols. So that means our sixth card must use a combination of symbols on the existing cards. But we only have five cards; if we are to make a sixth card with 10 different symbols, we must pull more than one symbol from an existing card, which violates the constraint that no two cards share more than one symbol.

  • $\begingroup$ Thanks for the reply, but as I pointed out in my original question (which Andrew Kelley edited out), the numbers were made up, as I didn't have a set of Dobble cards in front of me. Whilst your reply is interesting, I was more interested in the answer to my actual question, which was how many cards can you make using the Dobble rules. $\endgroup$ – Avrohom Yisroel Jul 15 '14 at 14:36

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