What is the math behind the game Spot It?

Question

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)?

Answer 1

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 $7^2+7+1 = 57$.

Answer 2

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:

Answer 3

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 $57$.

Answer 4

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.

Answer 5

$n^2 -n + 1$ 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.

Answer 6

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 $\Pr[\text{Good}]\ge 0.004$.