Talk:Dobble

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Board and table games Low‑importance

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People playing Dobble.jpg

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Order

So the connection to finite projective planes seems to suggest that similar games can be constructed with only certain numbers of pictures per card - viz. 2-6, 8-10, 12, 14, 17, 18, 20, 24, 26, 28, 30, 32, 33 and 50, and others, but certainly not 7 or 11, among others (all numbers one more than the "order" of the mathematical structure [1]). Obviously, that is not correct - and truly, the article doesn't say so! E.g., taking the "Junior" version of the came with 31 cards (30 cards commercially) and 6 pictures per card, one may add 31 unique pictures, one to each card, and thus create a perfectly playable game with 31 cards with 7 pictures per card, using 62 different pictures of which 31 occur at 6 cards each, and 31 occur only once. The corresponding mathematical structure does not satisfy the criteria for being a finite projective plane, and the creation is much less "optimal" than those that do, using way too many different pictures. But can it be clearly stated, proved, and, for the purposes of wikipedia, sourced, that such games not based on finite projective planes are all much less "economical"? (In fact, the commercial Dobble games, "missing" two cards in the standard version and one in the junior version, are also slightly less economical.)--Nø (talk) 11:15, 27 September 2020 (UTC) revised --Nø (talk) 16:14, 6 October 2020 (UTC) PS. I've created and printed games with 12, 17 and 20 pictures per card; they are perfectly playable, but of course slower than the versions with 6 or 8.--Nø (talk) 16:14, 6 October 2020 (UTC) revised --Nø (talk) 19:43, 24 September 2021 (UTC)[reply]

I'd still be happy to see my question answered in the article. I suspect the answer is that if the number of symbols is n, and m is the largest order of a projective plane strictly less than n, then n-m-1 symbols on each card must be unique to that card.--Nø (talk) 19:43, 24 September 2021 (UTC)[reply]