Spot It! Deck Design

My colleague Asmita Sodhi has been presenting some of the mathematics of the game Spot It! in the problem incubator this week and last, which reminded me of an activity I've run before, charging students with designing their own Spot It!-like game. Today we'll look at an iteration of that, using Asmita's "Double Spot It!" for an added challenge.

Setup

I think the last time I ran this I used stickers but will advocate for stamps this time around. You can get bundles of stamps for a reasonable price and they are more reusable than stickers. Here are some possibilities:

• Zoo Animal Stamps (26 for $9)

• Gejoy Wild Animal Stampers (40 for $14)

• Crtiin Animal Stampers (48 for $16)

I would use something like index cards or a worksheet with circles to represent Spot It! cards for the deck.

A Spot It! deck consists of 55 cards, each featuring 8 images, which we'll call stamps. In each round, two cards are placed on the table and the winner of the round is the first player to identify the stamp that the two cards have in common. When introducing this activity, I've started by playing a few games with a commercial Spot It! deck so students understand the mechanics.

Some important aspects of a Spot It! deck:

• Every pair of cards must have a stamp in common, since otherwise some rounds are unwinnable.

• A pair of cards can't have too many stamps in common, since otherwise a round with those cards will be too easy. For example, if there are \(8\) stamps on each card some pair of cards has \(4\) in common, this wouldn't pose much of a challenge.

• One stamp shouldn't be shared way more frequently than another stamp. An extremely bad deck would be one where every card has a clover and then \(7\) stamps that appear on that card only. In this case, the stamp in common is always a clover and the game is simply to say "clover" the fastest. If half of the cards in the deck have a clover on them, there is still a problem, but not as extreme.

From these conditions, it seems like a good Spot It! deck could require:

• Every pair of cards has exactly one stamp in common.

• Each card has the same number of stamps.

• Each stamp appears on the same number of cards.

The original Spot It! deck could meet this third condition but doesn't, which we'll address later! For the purposes of the challenges below, try to check all three of the above boxes.

Challenges

Spot It!

1. Choose \(3\) stamps and make a Spot It! deck with \(3\) cards and \(2\) stamps per card.

2. Choose \(6\) stamps and make a Spot It! deck with \(4\) cards and \(3\) stamps per card.

3. Choose \(7\) stamps and make a Spot It! deck with \(7\) cards and \(3\) stamps per card.

4. Choose \(12\) stamps and make a Spot It! deck with \(9\) cards and \(4\) stamps per card.

5. Choose \(13\) stamps and make a Spot It! deck with \(13\) cards and \(4\) stamps per card.

6. Choose \(21\) stamps and make a Spot It! deck with \(21\) cards and \(5\) stamps per card.

Double Spot It!

(Same rules, but every pair of cards must share exactly two stamps and no pair of stamps is shared by more than two cards)

7. Choose \(4\) stamps and make a Double Spot It! deck with \(4\) cards and \(3\) stamps per card.

8. Choose \(7\) stamps and make a Double Spot It! deck with \(7\) cards and \(4\) stamps per card.

9. Choose \(11\) stamps and make a Double Spot It! deck with \(11\) cards and \(5\) stamps per card.

10. Choose \(16\) stamps and make a Double Spot It! deck with \(16\) cards and \(6\) stamps per card.

Notes

• Many of these decks are straightforward to assemble through trial and error. Since stamps can be permuted to get another solution, if the requirement is \(k\) stamps per card, then we can choose any \(k\) distinct symbols for the first card. Since two cards share exactly one stamp in spot it, we can choose a stamp \(v_1\) from the first card and \(k-1\) new stamps to be the stamps on the second card. Since we're requiring that it not be the same stamp all cards have in common, a choice for the third card could be \(v_2 \neq v_1\) from the first card, \(v_3 \neq v_1\) from the second card, and \(k-2\) new stamps. Something like this can get the ball rolling toward a solution. Sometimes guessing another card or two with \(v_1\) and \(k-1\) new vertices is useful.

• Since every stamp will be on the same number of cards, there's some algebra that can be done to help find a solution. Using the traditional variables of combinatorial design theory, if

  • \(v\) is the number of stamps,
  • \(r\) the number of cards per stamp,
  • \(b\) is the number of cards, and
  • \(k\) is the number of stamps per card

  then

  \[ vr = bk. \]

  Since all of our puzzles give \(v,b,k\), we can compute \(r\).

  When solving \(\#4\), for example, we could note that \(v = 12\), \(b = 9\), and \(k = 4\), so \(r = 3\). If we use stamps A, B, C, D, E, F, G, H, I, J, K, L, then since A must appear on \(3\) cards, we might as well choose ABCD, AEFG, and AHIJ as our first three cards. We can then build from there, careful to never use A again.

• Many of the arrangements have interesting geometric objects lurking in the background. For example, one way to solve \(\#3\) is to take stamps A,B,C,D,E,F,G and focus on four of them:

  A		B
  C		D

  We can think of this grid as having three pairs lines: horizontal AB and CD; vertical AC and BD; and diagonal AD and BC. No two in a pair intersect, but each two from distinct pairs do. To fix the problem of lack of points in common in each pair, we'll add a point to each pair: horizontal ABE and CDE; vertical ACF and BDF; and diagonal ADG and BCG. Now all pairs from these six have a point in common. However, A,B,C,D all appear on three lines while E,F,G each appear on two. We can fix this by creating a seventh line EFG, which conveniently shares a point with each of the other six. This geometry on seven points is known as the Fano plane.

  Several of the solutions to Spot It! problems also arise as finite projective planes that can be constructed in a similar manner, though finding the parallel lines can be trial and error without some higher algebraic insight.

  If we consider the Fano plane ABE, CDE, ACF, BDF, ADG, BCG, EFG, the complementary objects CDFG, ABFG, BDEG, ACEG, BCEF, ADEF, ABCD form what's called a biplane and happen to be a solution to \(\#8\).

Extra Challenges

• According to the lore I was told, Spot It! originally consisted of \(57\) cards, since the deck designed based on a finite projective plane of order \(7\). However, the instruction booklet had been overlooked, so two cards had to be removed in order to allow the instruction booklet to fit in the tin. Determining what the two cards that were omitted is a nice exercise, if you have a copy of the game.

• Spot It! Jr has cards that feature \(6\) stamps each. Before looking at it, what are some possibilities for what that deck could look like?

• Relaxing some conditions allows for more types of decks. For example, since two cards were left out of Spot It!, every stamp does not appear on the same number of cards! What sorts of decks can you create if instead of each stamps appearing on \(r\) cards, some are allowed to appear \(r\) times and others \(r+1\) times? What if instead of every card featuring \(k\) stamps, some have \(k\) and others \(k+1\)?

This post was sponsored by the Julia Robinson Mathematics Festival