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I just purchased the game Spot It. As per this site, the structure of the game is as follows:

Game has 55 round playing cards. Each card has eight randomly placed symbols. There are a total of 50 different symbols through the deck. The most fascinating feature of this game is any two cards selected will always have ONE (and only one) matching symbol to be found on both cards.

Is there a formula you can use to create a derivative of this game with different numbers of symbols displayed on each card.

Assuming the following variables:

  • S = total number of symbols
  • C = total number of cards
  • N = number of symbols per card

Can you mathematically demonstrate the minimum number of cards (C) and symbols (S) you need based on the number of symbols per card (N)?

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6  
A similar question has been posted on Stackoverflow today, if you're interested. –  Thies Heidecke Jun 5 '11 at 7:15
    
The instructions on SpotIt actually say that there are "over 50 different symbols" –  anushr Jan 13 '13 at 3:58
    
Related: Limit to Unique Combinations –  MJD Jul 19 '13 at 2:12

6 Answers 6

up vote 14 down vote accepted

The celebrated Ray-Chaudhuri–Wilson theorem states that $C \leq S$ (contradicting your numbers).

An almost matching construction is as follows. Pick some prime number $n$. Our universe, of size $n^2+n+1$, consists of pairs of numbers in $\{0,\ldots,n-1\}$ plus $n+1$ singletons $\{0,1,\ldots,n-1,\infty\}$ ("points at infinity"). For each $0 \leq a \leq n-1$ and $0 \leq b \leq n-1$ we will have a card of size $n+1$ containing the pairs $\{(x,ax+b \mod{n})\}$ and the singleton $a$. There are also $n$ special cards, for each $0 \leq c \leq n-1$, containing the pairs $\{(c,x)\}$ and the singleton $\infty$. One super special card contains all $n+1$ singletons.

Clearly two cards with the same $a$ intersect only at the singleton. Two cards with different $a$s intersect at the unique solution to $a_1x+b_1 = a_2x+b_2 \pmod{n}$. Two special cards intersect only at the singleton, and a normal and a special card intersect at $(c,ac+b)$. Finally, the super special card intersects the rest at a singleton.

In total, we have $n^2+n+1$ cards and symbols, each card containing $n+1$ symbols, and two cards intersecting at exactly one symbol. In your case $n=7$ and so the number of cards and symbols should be $7^2+7+1 = 57$.

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But does this tell you the fewest number of cards and symbols necessary? For example, with 3 symbols per card (n=2) the math suggests 7 symbols and 7 cards. But, the fewest numbers seem to be 6 symbols and 4 cards: symbols [a,b,c,d,e,f]; cards [abc, cde, eaf, bdf]. –  Javid Jamae May 4 '11 at 14:43
1  
You're right that the Ray-Chaudhury-Wilson theorem provides only one constraint on possible games. There may be other known results in the area. The corresponding mathematical subject is combinatorial block design, and I don't think they have it all figured out. –  Yuval Filmus May 4 '11 at 15:31
2  
You're right and the actual game shows that well: as the French article I quote explains, the cards in Spot It are the elements of the projective plane over the field with seven elements (which is a very elegant 57-point 57-line design) with two points removed. So it's now a 55-point 57-line system of much less (mathematical) interest. But for the game, these considerations are rather irrelevant: the symmetry of the configuration plays no role (it's just desirable to keep the number of symbols to a reasonable amount). –  PseudoNeo May 5 '11 at 23:53
    
The construction you describe works only when n is prime or also when n is a prime power? –  ypercube Jun 7 '11 at 11:42
    
@ypercube: Should work for every prime power (if you use finite field arithmetic), see en.wikipedia.org/wiki/Projective_plane#Combinatorial_definition. –  Yuval Filmus Jun 7 '11 at 14:13

I came to the conclusion it must be $57$ or more symbols the following simple way: the total number of symbols shown on all cards is $55\times8=440$. If it was $50$ different symbols only, each must be shown $440:50=8.8$ times, i.e. some at least $9$ times.

If you took the $9$ cards with one common symbol, all the other symbols would need to be different, i.e. you'd need $(8-1)\times9+1=64$ different symbols.

If we reduce the max. usage per symbol to $8$, you only need $(8-1)\times8+1=57$ Symbols. As $57\times8=456$, this also exceeds the #of symbols shown thus being a valid solution.

To be able to use fewer symbols (e.g. $56$), usage per symbol would require reduction to 7 per (each individual) symbol, which would limit the no. of cards to $56\times7:8=49$.

Thus, with $55$ cards, the minimum number of different symbols is $57$.

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Here's an article (in French) that aims to explain the mathematics behind the game to a wide audience.

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I have the game myself. I took the time to count out the appearance frequency of each object for each card. There are 55 cards, 57 objects, 8 per card. The interesting thing to me is that each object does not appear in equal frequency with others ... the minimum is 6, max 10, and mean 7.719. I am left curious why the makers of Spot It decided to take this approach. Apparently they favor the clover leaf over the flower, maple leaf, or snow man.

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$n^2 -n + 1$ where $n$ is the number of images.

This is the simplest formula to arrive at the number of both individual symbols and total number of cards required to display them (these are the same).

I derived this formula logically but not necessarily mathematically as follows:

I picked a random card and focused on a single image. Assuming eight images per card as are found in this game, this image can only be found $8$ times, once on the card you're holding and $7$ more times.

The same holds true for the next image. It can only appear $8$ times if it to remain unique - once on the card you are holding and once over each of $7$ more cards.

I noticed the trend. Each image appears once on the card you're holding and requires $7$ more cards. So, you need the 1 card you are holding and 7 more per image. Mathematically, I guess that's: $1 \text{card} + (7\text{cards}\times 8\text{images})$. That's $1+(7\times8)$ or $1+56 = 57.$

Logical, so far.

Then, I ran the same logic and considered a card with only $4$ images. Each card would require one base card and $3$ additional cards per image. Mathematically, that would be $1+ (3x4)$. That's $1+12$ or $13$ cards.

Then, I tried to tie these observations together. I asked myself "Is there a formula that would arrive at the right answer no matter the number of images?" The answer is yes.

I remembered that in the examples above I started with 1 card then added (one less than the number of images) $\times$ (the number of images). That's $1+ (n-1)(n)$ if $n$ is the number of images. Then I just kinda rearranged a little:
$$\begin{eqnarray*}1+ (n-1)(n) \\ 1+ (n)(n) - n \\ 1+ n^2 - n \\ n^2 - n + 1 \end{eqnarray*}$$

I tested it and it works out every time. I was very happy before I got yelled at by my wife for taking so long on the computer.

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N squared minus N + 1 is the correct formula for calculating both the number of images and the number of cards. Here is the specification for generating the cards themselves: Below are the specifications for generating each card of Spot-it. Although the commercial Spot-it game contains 8 symbols per card, these specifications work for any desired number of symbols/card. The rules are that every card contains one and only one symbol matching every other card. Also, all symbols have an equal probability of being the matching symbol.

Let N = number of symbols per card Let C = the total number of symbols as well as the total number of cards to be generated. Then C = N2 – N + 1

Generate a matrix with N columns and C rows Column 1: Symbol 1 for N rows Symbol 2 for N-1 rows . . . Symbol N for N-1 rows Column 2: Row 1 = Symbol (Column #) Row 2 = Symbol N+1 Row 3 = Symbol 2N Row 4 = Symbol 3N-1 Row 5 = Symbol 4N-2 Etc. through row 2N-1 Repeat above rows N-2 times Column 3: Row 1: Symbol (Column #) Row 2: Previous column Row 2 plus 1 Row 3: Previous column Row 3 plus 1 Etc. through row 2N-1 Repeat above rows N-2 times Column 4: Same as Column 3 . . . Column N

The Spot-it Cards are the Rows

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