What is the math behind the game Spot It? 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)?
 A: 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.
A: $n^2 -n + 1$ 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.
A: 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 $7^2+7+1 = 57$.
A: Here's an article (in French) that aims to explain the mathematics behind the game to a wide audience.
In the interest of link rot prevention, here are two diagrams from the article that may be of interest even to non-French speakers:


A: Here's my explanation for the algorithm for making the cards.
We can describe the cards as a matrix, with a row for each card and a column for each symbol. A cell will have a 1 if the card corresponding to the row has the symbol corresponding to the column, and 0 otherwise.
It turns out that for each prime number $p$ we can create such a solution, with $p^2+p+1$ cards and a total of $p^2+p+1$ symbols, with $p+1$ symbols on each card. The Dobble game has $p=7$, which gives 57 cards and 57 symbols with 8 symbols in each card. Here is the matrix for $p=5$, meaning there are 31 cards with 6 symbols in each:

Here's Python code for producing this solution:
def dobble(p):
    cards = [[] for i in range(p**2 + p + 1)]
    cards[0].append(0)
    for i in range(p+1):
        for j in range(p):
            cards[1 + i*p + j].append(i)
            cards[i].append(1 + i*p + j)
    for i in range(p):
        for j in range(p):
            for k in range(p):
                cards[1 + p + i*p + k].append(1 + p + j*p + (i*j - k) % p)
    return cards

from itertools import combinations
for card0, card1 in combinations(dobble(7), 2):
    assert len(set(card0) & set(card1)) == 1

The matrices $C_{ij}$ are all permutation matrices, that is, each row and each column have exactly one 1. They are all just a cyclically shifted reverse-diagonal, with the diagonal shifted by $ij \mod p$.
It is quite quite easy to see that any two rows have exactly one column with a common 1, except for two rows that come from different rows in the large matrix of matrices. Let's see why they have exactly one common column.
A matrix $C_{ij}$ has a 1 in position $k,l$ if $ij = k + l$. All the arithmetic is done modulu $p$. The first row can be described by $i_0$ and $k_0$ (k is the row in the $C_{i_0}j$ matrices), and the second row by $i_1$ and $k_1$. Since we come from different rows in the large matrix, $i_0 \ne i_1$. A column is described by $j$ and $l$. If both rows have a 1 in the column, it means that:


*

*$i_0 j = k_0 + l$

*$i_1 j = k_1 + l$
Subtracting the equations gives us:
$(i_1 - i_0)j = k_1 - k_0$
Now, since $p$ is prime, and $i_0 \ne i_1$, we can solve this equation, and get a single result for $j$, namely
$j = (k_1 - k_0)(i_1 - i_0)^{-1}$
which we can put back in equation 1 to get $l$. So there is exactly one column which has 1 in both rows.
So, thanks to the integers modulu $p$ being a field, we have exactly one symbol which appears in each pair of cards.
Notes:


*

*This would have also worked with regular diagonals, but this way the matrix is symmetric.

*I based my understanding on Paige L.J., Wexler Ch., A Canonical Form for Incidence Matrices of Projective Planes, which was referenced in this answer to a related question.

*The code for generating the figure is here.

A: 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 $57$.
A: Here's what I think is a more intuitive way to reach same result:
let:
p = icons per card
t = total number of unique icons in the system
n = total possible cards in the system
r = total times/cards every icon appears in.
take the case of p=3:
all possible cards (every line is a possible card):
A|B|C
A|D|E
B|D|F
C|D|G
A|F|G
B|E|G
C|F|E
The first batch of 4 cards is the core because:
I. We took the first 3 icons from first card (A|B|C).
II. To create more possible cards, we introduced the fourth icon D to each one of the icons in the first card.
III. we filled the other positions with new icons (E, F & G)
We now see that it is impossible to introduce any new icon to the system, as it will not be able to have exactly 1 common icon with EVERY other card.
So the first 4 cards determines 't' - the total # of unique icons in the system. The other 3 cards are merely the remaining permutations.
We also learn that r = p,  and  n = t.
To determine t we take the first matrix (excluding the first card) and see that the # of unique icons in the system is p square minus p + 1. 
When p = 3 → t = n = 7
When p= 8 (as in the game) → t=n=57.
in the game they used only 55 cards,  I guess simply because its a cleaner number.
A: Here is an alternative approach that uses the probabilistic method but does not give us the optimal result.
Consider the $8$ slots in each of the $55$ cards and follow the following randomized
process. For each of the $8$ slots pick a random size-8 subset of the $57$ symbols
and fill in the slots. What is the probability that each pair of cards shares a
single symbol? In particular, is this probability non-zero? If so, then we are
sure there is such an assignment.
So, let Good be the event that each pair of cards shares a unique symbol and
consider $\Pr[\text{Good}]$. The complement of this event is that there is at least one
pair that does not share exactly one symbol. For a pair $p$, let $\text{Bad}_p$ be the
event that $p$ does not share exactly one symbol:
$$\Pr[\text{Good}] = 1 - \Pr[\exists p: \text{Bad}_p].$$
Applying union bound, we can find an upper bound on $\Pr[\exists p: \text{Bad}_p]$ as
$Pr[∃ p: \text{Bad}_p] ≤ ∑ \Pr[\text{Bad}_p]$, where the sum is over all pairs of cards $p$. The
number of pairs is given by the binomial coefficient $\binom{55}{2}$. So it is
enough to compute $\Pr[\text{Bad}_p]$ for an arbitrary p. The number of all possible
assignments of symbols to two cards is $\binom{57}{8}^2$. The "bad"
assignments are all these assignments that share 0,2,3,4,5,6,7 or 8 symbols;
i.e., any number but 1. For a number $c$ of common symbols, the number of
assignments that share $c$ symbols are $\binom{57}{c}\binom{57-c}{16-2c}$.
Putting everything together we get
$\Pr[\text{Good}]\ge 1-\binom{55}{2}\frac{\binom{57}{16}+\sum_{c=2}^8 \binom{57}{c}\binom{57-c}{16-2c}}{\binom{57}{8}^2} = -0.0351$.
This is not positive but the above is only a lower
bound so we lost a little bit there. By picking 58 symbols instead, we get $\Pr[\text{Good}]\ge 0.004$.
A: 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
