Spot It! is a fast-paced matching card game. Each of the 55 cards in the deck features eight symbols, and there is always exactly one matching symbol between any two cards in the deck. Your goal is to be the quickest to find the match between two cards. There are a total of 57 different symbols throughout the deck. While playing the game with several family members, my son questioned how it was possible that any two cards always have exactly one match. I tried to explain using the following examples:
If there were 2 symbols on a card, then the deck would contain 3 cards and a total of 3 different symbols.
Let the symbols be A, B, and C.
Then there would be 3 cards: AB, AC, and BC.
Note that between any two of the cards, there is one and only one matching symbol.
If there were 3 symbols on a card, then the deck would contain 7 cards and a total of 7 different symbols.
Let the symbols be A, B, C, D, E, F, and G.
Then there would be 7 cards: ABC, ADE, AFG, BDF, CDG, CEF, and BEG.
Note that between any two of the cards, there is one and only one matching symbol.
Also, note that each symbol occurs on 3 cards.
If there were 7 symbols on each card, what is the maximum number of cards that could be in the deck so that any two cards always have exactly one and only one matching symbol?
Bonus: If there were n symbols on each card, what is the maximum number of cards that could be in the deck so that any two cards have exactly one match?
Solution to the Problem:
The answer is 43 cards (and 43 symbols).The answer to the bonus is n2 - n + 1 cards.
If there were 4 symbols on a card, then the deck would contain 13 cards and a total of 13 different symbols.
Let the symbols be A, B, C, D, E, F, G, H, I, J, K, L, and M.
Then there would be 13 cards:
ABCD AEFG AHIJ AKLM
BEHK BFIL BGJM
CEIM CFJK CGHL
DEJL DFHM DGIK
Note that between any two of the cards, there is one and only one matching symbol.
Also, note that each symbol occurs on 4 cards.
I was able to create the 13 cards very easily by using a table similar to the one I use in solving logic puzzles.
I used the method of elimination to solve for the 13 unique cards where there was exactly 1 match between any two of them.
I used X's to eliminate the possibilities, and O's for the symbols on the cards.
Here is the table:
| A | B | C | D | E | F | G | H | I | J | K | L | M |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| O | O | O | O | X | X | X | X | X | X | X | X | X |
| O | X | X | X | O | O | O | X | X | X | X | X | X |
| O | X | X | X | X | X | X | O | O | O | X | X | X |
| O | X | X | X | X | X | X | X | X | X | O | O | O |
| X | O | X | X | O | X | X | O | X | X | O | X | X |
| X | O | X | X | X | O | X | X | O | X | X | O | X |
| X | O | X | X | X | X | O | X | X | O | X | X | O |
| X | X | O | X | O | X | X | X | O | X | X | X | O |
| X | X | O | X | X | O | X | X | X | O | O | X | X |
| X | X | O | X | X | X | O | O | X | X | X | O | X |
| X | X | X | O | O | X | X | X | X | O | X | O | X |
| X | X | X | O | X | O | X | O | X | X | X | X | O |
| X | X | X | O | X | X | O | X | O | X | O | X | X |
Now, let us jump to cards with 8 symbols on each card (which the makers of SPOT IT! used in their game).
If there were 8 symbols on a card, then the deck could contain 57 cards and a total of 57 different symbols.
Let the symbols be the numbers from 1 to 57.
Then the cards which contain the number 1 would be:
1 and the numbers from 2 to 8
1 and the numbers from 9 to 15
1 and the numbers from 16 to 22
1 and the numbers from 23 to 29
1 and the numbers from 30 to 36
1 and the numbers from 37 to 43
1 and the numbers from 44 to 50
1 and the numbers from 51 to 57
Note that between any two of the cards above, there is one and only one matching symbol -- the number 1.
These are the 8 cards that contain the symbol 1, and it will not occur on any of the other cards.
The next thing that I did was to create a table showing the number of symbols on a card and the corresponding maximum number of cards in that deck.
Note that the total number of times each symbol is used is the same as the number of symbols on each card.
Also, the total number of symbols needed throughout the deck is equal to the maximum number of cards for the deck.
| # of symbols on a card | # of cards in that deck | |
|---|---|---|
| 2 | 3 | |
| 3 | 7 | |
| 4 | 13 | |
| 5 | 21 | |
| 6 | 31 | |
| 7 | 43 | |
| 8 | 57 | |
| 9 | 73 | |
| 10 | 91 | |
| n | n2 - n + 1 |
Hopefully, you found a pattern in the table above for the number of cards needed for a deck:
Starting at 2 symbols on a card, there are 3 maximum cards for the deck.
Then for 3 symbols, add 4 to obtain 7 cards in the deck.
Then for 4 symbols, add 6 to the previous number to get 13 cards.
Then for 5 symbols, add 8 to the previous number to get 21 cards.
Then for 6 symbols, add 10 to the previous number to get 31 cards.
Then for 7 symbols, add 12 to the previous number to get 43 cards.
Then for 8 symbols (which is used in SPOT IT!), add 14 to the previous number to get 57 cards.
Mathematically, the formula for the number of cards for n symbols on a card is given by n2 - n + 1.
I don't know why the makers of the game of SPOT IT! did not use 57 cards. And since two cards are missing, not all of the symbols are used equally! So, there is a greater chance of certain symbols being the match than other symbols! You could use this fact to your advantage in playing the game! Check the symbols that are used less frequently last! I do not have a deck of SPOT IT! or I would check to see which symbols are used less than 8 times, which is the maximum that any symbol could be used.
After writing out the cards needed for 4 symbols on a card, I tried to create a 5 symbol deck but failed every time.
James Alarie sent in the following analysis:
I'm still getting nowhere with set-5, so I tried set-6. Using A-Z and 1-5, it was easy; the 31 cards are:
ABCDEF AGHIJK ALMNOP AQRSTU AVWXYZ A12345
BGLQV1 BHMRW2 BINSX3 BJOTY4 BKPUZ5
CGMSY5 CHNTZ1 CIOUV2 CJPQW3 CKLRX4
DGNUW4 DHOQX5 DIPRY1 DJLSZ2 DKMTV3
EGPTX2 EHLUY3 EIMQZ4 EJNRV5 EKOSW1
I'm thinking that set-5 may be impossible (I haven't given up, but....) because 4 is non-prime and any kind of pattern loops back on itself. The set-6 was easy specifically because 5 is prime. If the prime/non-prime idea holds, set-7 may be impossible because 6 has so many factors, but set-8 should be easy and should give the 57 cards we expect.
Correctly solved by:
| 1. James Alarie (solved the bonus) | Flint, Michigan |
| 2. Chad Fore (solved the bonus) | Scott County, Virginia |