At a recent North Coast Puzzle Club meeting, one of our members came up with a dice game loosely derived from the target game of Krypto, a card game in which the players attempt to arithmetically arrange \(5\) cards in ways to reach a sixth target card, all chosen at random from a deck of \(54\) cards, carefully designed so that there are very few "hands" that are impossible to make reach the target, and there are usually many ways to reach each target in any dealt hand.
The dice game, subsequently named Take Sides by me, is a clever variation on the target games in that for Take Sides, there is a no particular target. Instead, the goal of the game is to design an equation using only the numbers on the dice. (Like in Krypto, you have to use all the dice – only once each – to make your equation.) In the original version, concatenation of numbers into larger numbers was not thought of, but we have decided that concatenation should be allowed.
Since dice, like playing cards, are quite portable, we think this game could be an excellent one to play in schools, at social gatherings and elsewhere. Since the possibilities of the dice roll with five dice are somewhat limited (only \(252^\ast \)), the game may get repetitive after long periods, but the nature of social game playing makes this not too much of an issue, since you would only be playing for a half-hour at most anyway. I have not yet determined how many of the possible rolls of the five dice have no solutions, but so far, when you include exponentiation in addition to the typical arithmetic operations (addition, subtraction, multiplication and division), we have not had any rolls that did not have at least one solution.
Let's take an example roll as follows:
⚁ ⚁ ⚂ ⚂ ⚄
Here's one solution to this roll:
\[3^{3-2} + 2 = 5\]
Here's another solution (assuming the standard order of operations conventions or PEMDAS):
\[3 \times 3 - 2 \times 2 = 5\]
Here's one more, using division:
\[2 \times 2 + \frac {3}{3} = 5\]
All of these solutions have a single digit on the RHS of the equation. Here's a solution that has multiple terms on each side, just re-ordering the previous solution:
\[2 \times 2 = 5 - \frac {3}{3}\]
It's starting to appear that there might be multiple solutions for many of the possible rolls of \(5\) dice. That would make the game much more exciting and interesting, sort of like Krypto, where there are often dozens of solutions to a particular "hand". Like Krypto, for Take Sides we could have a winner based on their calling out "Take These Sides!" first. Or, as is our penchant in the North Coast Puzzle Club, we could have all solutions for a given roll be counted, with each unique solution getting a point or points.
To demonstrate the use of concatenation, let's look at the following roll:
Here's a solution using concatenation of some of the numbers:
\[26-24=2\]
One more example of concatenation for this roll:
One solution is:
\[\frac{46}{23}=2\]
And looking at this solution, another one (without concatenation) sort of pops out:
\[\frac{4+6}{2+3}=2\]
So, even with just the four standard arithmetical operations plus exponentiation and concatenation, there seems to be at least one solution to many of the possible dice rolls. I'd like to prove that there is (or is not) at least one solution for each roll, and I'd also like to create an app that finds all solutions for any roll, sort of like the amazing dCode app, which already has a Krypto solution finder (with many options). That would in some ways make the game less interesting, but I find that with an exhaustive list of solutions, you can sometimes find patterns that answer additional questions about the problem at hand, or maybe even different questions entirely. For instance, it might be interesting to see what happens if we add the factorial operation to the list of allowed operators. Or maybe see if the game is better or worse with 6 dice? 4 dice? Another amount of dice? And what about dice with a different number of sides?
Whether or not my nerdy goals can be reached, I think Take Sides has some potential as a recreational game that people of all ages can play anywhere with minimal investment. I'll be updating this page with more examples and the experience of playing the game with kids in the near future. In the meantime, enjoy playing Take Sides! If you can't find five dice to play with, you can go online to the RANDOM.ORG Dice Roller page and get truly random rolls. (If you want to try using less or more dice, RANDOM.ORG Dice Roller allows between \(1\) and \(60\) dice!)
*There are formulas for calculating the number of combinations both with and without considering order for rolling dice. (These are from a Perplexity AI query.)
With Order (Permutations)
The formula for the total number of possible outcomes when rolling \(n\) dice with \(s\) sides each, considering order, is \(s^n\). For 5 six-sided dice this is \(6^5 = 7,776\). This is because for each die, we have \(s\) choices, and we make this choice \(n\) times independently.
Without Order (Combinations with Repetition)
The formula for the number of unique combinations when rolling \(n\) dice with \(s\) sides each, not considering order, is \(\binom{n+s-1}{n} = \frac{(n+s-1)!}{n!(s-1)!}\). For 5 six-sided dice this is \(\binom{5+6-1}{5} = \binom{10}{5} = 252\).
Explanation
1. With order: Each die can show any of the \(s\) faces, and this choice is made \(n\) times independently. This is a fundamental counting principle, resulting in \(s^n\) possibilities.
2. Without order: This is equivalent to choosing \(n\) items from \(s\) types with replacement and without regard to order. It's often referred to as the "stars and bars" or "balls in bins" problem in combinatorics [1][3].
- Imagine \(n\) stars (dice rolls) and \(s-1\) bars (dividers between face values).
- We need to arrange these \(n+s-1\) positions, choosing \(n\) positions for the stars.
- This is equivalent to choosing \(n\) items from \(n+s-1\) items, hence the combination formula.
This formula accounts for all possible ways to distribute \(n\) indistinguishable objects (dice rolls) into \(s\) distinguishable bins (face values), allowing empty bins [5].
Citations:
[1] Posit Community: all possible combinations of two dice
[2] Reddit, subReddit r/MathHelp: Dice rolling, permutations or combinations?
[3] Stack Exchange, Mathematics: Formula for working out the number of dice combinations resulting in a given value
[4] Statistics How To: Dice Roll Probability: 6 Sided Dice
[5] Combinatorics and Graph Theory by David Guichard, Department of Mathematics, Whitman College: 1. 2 Fundamentals, Combinations and Permutations
[6] freeMATHhelp: Rolling the dice...
[7] Reddit, subReddit r/theydidthemath: How many possibilities can you make with 3 six-sided dice?
[8] Worlds of Math and Physics: The dice roll with a given sum problem