A game called “Spot It!” has received a lot of attention from recreational mathematicians in recent years. There’s a good video by Matt Parker, and a blog post about it by one of my five readers. (Hi MJD!) The game contains a number of cards each with eight symbols, and the object is to be the first to spot the matching symbol between the pair of cards. The designers of the game, in finding an elegant set of cards where every pair has exactly one match, used a structure that can be understood in terms of finite projective geometry. Parker calls these *monomatch* sets of symbol sets.

This led me to consider monomatching on symbols on the faces of a pair of dice. (I guess I’m on a roll with the dice content here. Sorry, there’s really only one decent dice related pun; if you try to stretch beyond that, things get dicey.) The obvious thing to do with a pair of d6’s is to have a set of 36 symbols. You could consider the symbols as being arranged in a 6×6 grid. The faces on one die would each contain all of the symbols on one row, and the faces on the other die would each contain the symbols on one column. For an added trick, the numbers 1 through 6 could be six of the symbols, and if they are on a diagonal of that grid, the dice could be used as regular dice by ignoring the non-numerical symbols.

This seems like an idea worth exploring, once my laser engraver arrives. There is a version of Spot It! with 30 cards and 6 symbols per card that is marketed to be played by young children, so it seems like it would be somewhere near the realm of playability. (Adding dice beyond the first two might help.) And although Spot It! already comes in a compact tin, there’s not much that’s more portable than a pair of dice. But there’s not really any kind of interesting puzzle to be found in it, so I was hoping to find something else to do with symbol matching on dice.

And I did come up with another idea, and it is good, and it is dumb. Imagine, if you will, that you could use a pair of dice as… 2d6!

But not, of course, 2d6 as we know it. Instead of adding numbers on the two dice, we’d have the numbers 2 through 12 as symbols for matching. The frequencies in which the numbers occur on the two dice would have to be such that the probability of getting a number as a match would be the same as the probability of getting that number as a sum using a regular pair of d6’s.

2d6 roll | Frequency | Faces per die |
---|---|---|

2 | 1 | 1, 1 |

3 | 2 | 1, 2 |

4 | 3 | 1, 3 |

5 | 4 | 2, 2 or 1, 4 |

6 | 5 | 1, 5 |

7 | 6 | 2, 3 or 1, 6 |

8 | 5 | 1, 5 |

9 | 4 | 2, 2 or 1, 4 |

10 | 3 | 1, 3 |

11 | 2 | 1, 2 |

12 | 1 | 1, 1 |

Now, finding a set of number sets for the symbols on the faces of each die becomes an interesting puzzle, especially if we add constraints to make our dice more nice. One type of constraint we might care about is on the quantity of numbers on each die. Minimizing the total quantity on both of the dice would be good, as would be balancing the quantities on the dice.

Unfortunately, we cannot do both. The minimum total quantity is 43, which is odd. So in order to balance the quantities, we need to use the inefficient alternative for either 5 or 9. (Using the inefficient alternative for 7 doesn’t change the parity of the total, so I’ve dismissed that option.)

We could try to go further in our pursuit of balance. The solution I found above has balanced number quantities on the two dice, but at the level of faces, there are issues. The first set has face quantities of {2, 3, 3, 4, 5, 5}, while the second has {2, 3, 4, 4, 4, 5}. Having the same face quantities between the two dice would be desirable, especially if it could be done while minimizing the number of faces with five numbers, since those faces look more cluttered.

You could also drill down to the numbers themselves. The sum of all of the numbers on the upper die above is 158, while the sum on the lower die is 148. Ideally, we’d make those sums equal, or at least closer.

Problem **#50**: Find a “nicer” numbering for a pair of monomatch 2d6 dice than the one I found. One potential flaw that I haven’t mentioned already is having a pair of faces on the same die with identical number sets. When I was manually looking for numberings, they seem to want to have pairs like this, so they are harder to avoid than you might think.

Well, let’s see. The number of printed values must be in the range [43, 47] to yield valid dice, and only 44 and 46 allow the two dice to have the same number of printed values.

Consider 6, which must appear on one face of the “first” die and five faces of the “second” die. If we roll 6 on the first die and not the second, how many values are possible? Just one, since the above specifies unique faces for both dice. So 6 appears on a face with two values; similarly, so must 8. 6 and 8 might share such a face, but there must ultimately be at least two two-value faces.

To avoid five-value faces, the other ten faces must have a maximum four values each, for a maximum 44 values. This sounds good, but there’s one final, unfortunate consequence to consider.

Values other than 5, 7, and 9 must appear 30 times on any pair of valid dice. 7 must appear an odd number of times, so to reach 44 values, 5 and 9 must have a combined odd frequency: one with an odd frequency (1 and 4), one an even frequency (2 and 2). With, say, 9 appearing once on the “first” die and four times on the “second,” a similar argument to that regarding 6 and 8 shows that the 9-face on the first die can have at most three values, not the four required to avoid some face having five values.

Tl;dr: for a pair of dice so constructed, the properties…

1. Gives the right roll frequencies

2. No one face has five values printed

3. The count of values (with multiplicity) over the six faces of each die is the same

…cannot all hold.

I can get the dice differently balanced, though. Here’s one possible table of outcomes (as opposed to the labeling of the twelve sides; it was easier to work with, and won’t be too hard to check):

…sorry, that table was wrong. I’ll fix it later.

Odd. I might have missed something, but it looks like you can’t get both “number of values” and “sum of values” to match, either.

All right, I found the impossibility.

For each result 2-12, I looked at the difference of the numbers of times it appears on the two dice. I also looked at the difference of that value’s contributions to the

sumof the dice’s values. It may be possible to prove this result without looking at differences, but this gives me fewer numbers to handle.The value 7 can be printed either once on one die and six times on the other, or twice and thrice. This corresponds to profiles of (5, 35) and (1, 7), respectively. 5 and 9, needing to be printed nine times as described above, have candidate combined profiles of (3, 15) and (3, 27). This comprises all decisions at this point in construction. One important detail is that

every profile is either two odd numbers or two even numbers(where the even numbers are sometimes both zero).One set of choices, having the smallest total profile of the four, uses (1, 7) and (3, 15). This total profile is (18, 120). To get the same number of values and sum of values, the profiles must be distributed into sets with shared total (9, 60). This is the problem; because of the important detail above,

no sum of profiles can have one odd number and one even number.And since other profiles only change the totals by multiples of 4, this is inescapable.Anyway, here’s a results table for a pair of dice with the same sum, along with their face sums because I was paranoid this time: