I realize there isn't anything new in my answer but I wanted to convert it to VBA so I could try out the code in an environment I have on hand, Excel. Requirement 3: no symbol appears more than once on a given card. If you solve for $k$, you get $k = \dfrac{2s + 1 \pm 1}{2}$. I was lying in bed this morning trying to think this through in my head (after playing Dobble with my daughter last night), but it was only when I put pen to paper I realised the solution wasn’t as mathematically straightforward as I thought it was going to be, particularly ensuring that all symbols were equally as likely to be the paired one. :) By the way, I translated your code in python and am using it. Find my Dobble. Did COVID-19 take the lives of 3,100 Americans in a single day, making it the third deadliest day in American history? Another interesting parameter to look at is the mean number of times each symbol appears in a deck, $r$. Games For families > Games For kids > Discover the games > Talk with the community. The real Dobble deck has 55 cards, which would require having 54 symbols on each card and a total of 1485 different symbols. Trying to understand what your code is, but don't find the relation with Karinka's code. I've been trying to crack how to generate the symbol arrangements on the "Dobble" cards for months, and have succeeded in generating the sequence as far as N=6, C=31 but I am stuck at N=7 . I guess it's all right with you, I can give you access to the code. k &= (s - 1)^2 @kallikak I see what you are saying. In standard Dobble, there are 55 cards, each with 8 symbols. $$ 6,9,19,23,27,37,41,$$ So far, when creating cards we have chosen to match symbols that have not yet been matched. Start studying DOBBLE symbols (to play the game DOBBLE). Dobble Asmodee Games English Edition 2-8 Players 15 Minute Game Time Ages 6+ Dobble is the award-winning visual perception card game for 2-8 players aged 6 and above that can be played by anyone, regardless of age and interests. Every line goes through three points and every point lies on three lines. We can therefore create a new card using these $s$ unmatched symbols ($CEF$ in the diagram). Dobble Asmodee Games English Edition 2-8 Players 15 Minute Game Time Ages 6+ Dobble is the award-winning visual perception card game for 2-8 players aged 6 and above that can be played by anyone, regardless of age and interests. I have been working on the Dobble problem for a few years. $$ 7,10,15,20,31,36,41,$$ Free Shipping in United Arab Emirates⭐. In Dobble, players compete with each other to find the one matching symbol between one card and another. In doing so, we also end up repeating the remain symbols, so each one occurs exactly three times. Super cool. Sadly, I think it worked in $O(n! This is How I've converted the algorithm in javascript: var res = ''; For primes you can just use normal addition, multiplication and modulus, but that won't work for powers of primes. Genius. I may have gotten that from another Stack post. The players are looking for a symbol on their cards that matches the central card. Every card is unique and has only one symbol in common with any other in the deck. Can we calculate mean of absolute value of a random variable analytically? Here is VBA code inspired from @karinka's and @Urmil Parikh answers but using an arrangement of symbols to match answers from @Urmil Parikh, @Uwe, and @Will Jagy. The plane consists of seven lines and seven points. Requirement 6 (amended): there should not be one symbol common to all cards if $n > 2$. $$ 3,11,18,25,26,33,40,$$ Technically, given the requirements above, you could have infinite cards, each with just an $A$ on it, so we'll add a requirement. Thanks a lot for all the effort in understanding it and put it into such great article. The first four powers of two, $1$, $2$, $4$ and $8$, all have one card, so $r = 1$. Rule 2 corresponds to the fact that we want cards to have at least two symbols. In Dobble, players compete with each other to find the one matching symbol between one card and another. With 14 symbols we finally have enough symbols to scrape four cards together. Here's Dobble . I am still working on the Dobble set for 7 symbols . \qquad\begin{align} Dobble Asmodee Games English Edition 2-8 Players 15 Minute Game Time Ages 6+ Dobble is the award-winning visual perception card game for 2-8 players aged 6 and above that can be played by anyone, regardless of age and interests. The first few Dobble numbers are 1, 3, 7, 13, and 21. Were you able to find a set of cards that would have 11 symbols on each of 111 cards? In the Dobble card game there is a deck of 55 cards. T(s) &= sk - T(k - 1) \\ Where $\lfloor n \rfloor$ means "round $n$ down to the nearest whole number. In Dobble, players compete with each other to find the one matching symbol between one card and another. $$ 5,9,18,21,30,33,42,$$ Thanks a lot Peter for detailed analysis. For $q$ not being prime, but only prime power, these permutation matrices $C_{ij}$ would have to be generated another way (i.e. Thanks for this! Dobble Card Game for - Compare prices of 264189 products in Toys & Games from 419 Online Stores in Australia. For Example you have listed 2,8,14,20,26,32,38 as one card and later 5,8,17,20,29,32,41 as another card and there are three matching numbers ( namely 8,20 and 32). In addition, the game comes with a practical stylish bag in which you can carry the cards. Thanks for contributing an answer to Mathematics Stack Exchange! 54 is of course exactly divisible by 2 and 3 (plus the much less useful 6, 9, 18 and 27) which are likely to be the most frequent number of players, whereas 56 is divisible by 2 and 4 but not 3 (plus the much less useful 7, 8, 14 and 28) so it does allow for 4 people, but this may be less frequently required than 3 [Benford's law may help suggest how more likely 2 players would be than 3?]. Thanks for providing a Dobble set for 5 symbols per card. A tiny free promotional demonstration version of real-time pattern recognition game Spot it!. Dobble card game - mathematical background, Create 55 sets with exactly one element in common. The real game of Dobble has 55 cards with eight symbols on each card. When $n$ one less than a Dobble number, the number of repeats is one less than for that Dobble number, i.e if $n = D(s) - 1$, then $r = s - 1$. I will need to write a computer program to compare the different cards. What about 7 cards on 43 cards? $$ 1,14,15,16,17,18,19,$$ Making statements based on opinion; back them up with references or personal experience. The numbers $2$, $4$ and $8$ are also powers of two. $$ 7,9,14,25,30,35,40,$$ I'm hoping this can help someone else. This is an example of the pigeonhole principle, which is an obvious-sounding idea that is surprisingly useful in many contexts. $$ 1,38, 39,40,41,42,43,$$, $$ 2,8,14,20,26,32,38,$$ Can we be more efficient by having symbols appear on more than two cards? I didn't really use any of them to write this article; I've mainly put them here so I can remember what I should read when I get the chance. The total number of symbols in a deck is equal to the number of symbols multiplied by the average number of repeats. See prices & features . Once the deck size gets into the teens, it becomes hard to be sure that you've found the best solution using pen and paper. Triplete Se juega una ronda. $$ 1,8,9,10,11,12,13,$$ The page gives a long list of properties for this sequence. In addition, each triangle above or below the diagonal, contains each symbols once. With three symbols, $\{A, B, C\}$, we have something more interesting: three cards, each with two symbols: $AB$, $AC$ and $BC$. So if this pattern does hold, the total number of symbols in these decks, $N$, is: $\qquad \begin{align} The match can be difficult to spot as the size and positioning of the symbols can vary on each card. 10 symbols per card is also easy (p = 3^2) but there is no finite field of order 6 or 10, so 7 and 11 symbols per card cannot be generated (unless you allow more symbols than cards). We already know when $n$ is a triangular number, $r = 2$, and when $n$ is the Dobble number, $D(s)$, $r = s$ ($21$ is both a triangular number and a Dobble number, but the Dobble number wins out since we want the largest deck). I imagine that the reason they decided to have 55 rather than 57 cards is that once the cards are dealt and the face up card is removed this leaves 54 cards to be dealt rather than 56. With four symbols, you could have three cards: $AB$, $AC$ and $AD$. The second rule is there to rule out situations where all the points lie on the same line. I had been trying to make one using Excel and my own brain power (thinking like. If we use the triangular number method to get seven cards, we need 21 symbols, each appearing on two cards. Actually the last card needs to be "for I = 0 to N" instead of "for I = 0 to N-1". Now the problem is one of incidence geometry: the study of which points lie on which lines. The first thing to notice is that with $s = 3$, when now need $n$ to be at least seven symbols: one repeated symbol and three lots of two symbols. This has been explored extensively in the linked question "What is the Math behind the game Spot it". Wonderful, thank you, I understand how you have arrived at the sequences. Since this is a triangular number each symbol appears on exactly two cards. How does it work? We might expect that if $n$ is the triangular number $T(s)$, then we could have $s$ cards, e.g. I found an algorithm, as I was doing this it seemed right, but maybe... Below see the $43$ cards, symbols are the numbers from $1$ to $43.$, $$ 1,2,3,4,5,6,7, $$ In other words $k = s$ and $k = s + 1$. You can even arrange them a bit like dominos, joined by their common symbols. It will work for N power of prime if the computation of "(I*K + J) modulus N" below is made in the correct "field". In Dobble, players compete with each other to find the matching symbol between one card and another. r=r+1 Does Texas have standing to litigate against other States' election results? $$ 3,10,17,24,31,32,39,$$ Given $n$ different symbols, how many cards can you make, and how many symbols should be on each card? 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