Designer Dice
In this activity, students will design dice with the aim of beating other dice in a rolling face-off.
Setup
For this activity, you will want dice that can be edited. Ideally, they would come in different colors to make them easily distinguishable from one another. Here are some options that require dry erase markers:
Blank dice (100 for $15)
Dry erase blocks (12 for $15)
While the latter is a little spendier, it seems like it would be easier to handle without smudging.
I haven't seen a great out-of-the-box manipulative that doesn't require writing, but something like Lego dice could be good with a set of numbered 2x2 flat pieces. There are some pipped pieces for use with these dice, but stickers are likely needed for numbers greater than 6. I don't know a great way to get Lego dice without buying an expensive, possibly discontinued kit, but here are some off-brand options that can be combined into a more budget-conscious version:
Strictly Briks Brik Dice (2 for $10)
Strictly Briks Mathbriks (100 tiles for $11)
Strictly Briks 2x2 Pixels (288 tiles for $12)
Number stickers (50 copies of 1-40 for $6)
Number stickers (50 copies of 1-50 for $6)
The rules of the game are simple. An opponent die will be given. It is your job to create a die whose faces have the same sum so that your die will beat the opponent die more often than not, which we'll test via a dice face-off. In each round, both dice are rolled and the winner of the round is the die that rolls the higher number. In the case of a tie, the dice are rerolled until there is a winner for the round. The face-off is won by the player who wins the majority of nine rounds.
Challenges
Each challenge will take the form of a specified opponent die, designated by the numbers on its faces. The first few will have faces summing to \(21\), as on a standard \(6\)-sided die:
\((1, 1, 1, 1, 1, 16)\)
\((2, 2, 2, 2, 2, 11)\)
\((3, 3, 3, 3, 3, 6)\)
\((1, 4, 4, 4, 4, 4)\)
\((1, 1, 1, 6, 6, 6)\)
\((2, 2, 2, 5, 5, 5)\)
\((3, 3, 3, 4, 4, 4)\)
\((1, 1, 4, 5, 5, 5)\)
\((2, 2, 2, 2, 6, 7)\)
\((1, 2, 3, 4, 5, 6)\)
The next several have faces summing to \(30\):
\((1, 1, 1, 1, 1, 25)\)
\((1, 1, 1, 1, 13, 13)\)
\((1, 1, 1, 9, 9, 9)\)
\((1, 1, 7, 7, 7, 7)\)
\((1, 1, 6, 6, 8, 8)\)
\((1, 1, 5, 5, 9, 9)\)
\((2, 2, 6, 6, 7, 7)\)
\((2, 2, 5, 5, 8, 8)\)
\((2, 2, 4, 4, 9, 9)\)
\((3, 3, 6, 6, 6, 6)\)
\((3, 3, 5, 5, 7, 7)\)
\((3, 3, 4, 4, 8, 8)\)
\((1, 5, 6, 6, 6, 6)\)
\((5, 5, 5, 5, 5, 5)\)
\((4, 5, 5, 5, 5, 6)\)
\((3, 4, 5, 5, 6, 7)\)
\((2, 3, 4, 6, 7, 8)\)
\((2, 3, 5, 5, 7, 8)\)
\((2, 4, 5, 5, 6, 8)\)
\((1, 2, 5, 5, 8, 9)\)
Notes
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The main tool for investigating these dice in a systematic manner is laying out the sample space as a table. Here is an example of a table for the dice \(A = (2, 2, 2, 5, 5, 5)\) and \(B = (2, 2, 2, 2, 6, 7)\):
2 2 2 5 5 5 2 - - - A A A 2 - - - A A A 2 - - - A A A 2 - - - A A A 6 B B B B B B 7 B B B B B B By comparing every possible roll of \(A\) against every roll of \(B\), we can see that \(A\) wins \(12/36\) of the time, \(B\) wins \(12/36\) of the time, and the other \(12/36\) of the time, they tie. We say that these two dice tie, on average. Whenever the probability that \(A\) beats \(B\) is higher than the probability that \(B\) beats \(A\), we can say that "\(A\) beats \(B\) on average" and write \(A > B\). We might also say things like "\(A\) is better than \(B\)."
For the dice with faces summing to \(21\), every die that can be bested has a better die among the others. Of the \(110\) possible dice with faces summing to \(21\), the only die where no other die is better-on-average is surpisingly the standard \(6\)-sided die \((1, 2, 3, 4, 5, 6)\). There are \(78\) dice that lose to it and \(32\) that tie it, on average.
An anomaly that students might stumble upon is that one die beating another on average is a nontransitive relation. For example,
\((1,4,4,4,4,4)\) beats \((3,3,3,3,3,6)\) with probability \(25/36\)
\((3,3,3,3,3,6)\) beats \((2,2,2,2,5,5)\) with probability \(7/12\)
\((2,2,2,2,5,5)\) beats \((1,4,4,4,4,4)\) with probability \(7/12\)
We'll shorthand this
\[ (1,4,4,4,4,4) > (3,3,3,3,3,6) > (2,2,2,2,5,5) > (1,4,4,4,4,4) \]The above nontransitive triple is extremal in the sense that only one other nontransitive triple has the minimum of its three probabilities \(7/12\) but all three probabilities for it are \(7/12\):
\[ (1,1,4,4,4,7) > (3,3,3,3,3,6) > (2,2,2,2,5,5) > (1,1,4,4,4,7) \]We can increase the involved probabilities by allowing for more dice in a nontransitive cycle. For example, in the cycle
\[ \begin{eqnarray*} (3,3,3,3,3,6) & > & (2,2,2,2,6,7) \\ & > & (1,1,1,6,6,6) \\ & > & (1,1,4,5,5,5) \\ & > & (1,4,4,4,4,4) \\ & > & (3,3,3,4,4,4) \\ & > & (3,3,3,3,3,6) \end{eqnarray*} \]each die is the die that beats the next with maximal probability and all probabilities are \(\geq 3/5\).
For the dice with sides summing to \(30\), most of the dice listed again have dice that beat them somewhere else in the list. One exception is \((1,1,1,9,9,9)\). All dice that beat it must have a face labeled \(\geq 9\), but it's not too hard to come up with examples. \((2, 2, 2, 2, 11, 11)\) works nicely.
There are again nontransitive cycles. For people hip to nontransitive dice, two nice triples are:
\[ (1,1,5,5,9,9) > (3,3,4,4,8,8) > (2,2,6,6,7,7) > (1,1,5,5,9,9) \]and
\[ (1,1,6,6,8,8) > (3,3,5,5,7,7) > (2,2,4,4,9,9) > (1,1,6,6,8,8) \]
Unlike with our dice with faces summing to \(21\), there is no unbeatable-on-average die when the faces sum to \(30\). The closest to it is \((2,3,5,5,7,8\), which is beatable, but the best probability any other die has of beating it is \(9/17\), which is a slim margin. There are four dice that can beat it \(9/17\) of the time: \((3,3,6,6,6,6)\), \((1,1,6,6,8,8)\), \((1,5,6,6,6,6)\), and \((1,3,6,6,6,8)\).
Extra Challenges
Here's a version of the challenge with two opponents: Take two dice with faces summing to \(12\) from the list. Can you design a die that, when all three dice are rolled together each round, wins the majority of nine rounds? Again, if the highest roll is a tie, the dice must all be rerolled. (Do things change if only the tied dice are rerolled in this case?) Try this also with two dice whose faces sum to \(30\).
If a triple of nontransitive dice \(A,B,C\) with \(A > B > C > A\) dice is a nontransitive \(3\)-cycle, can you find a nontransitive \(4\)-cycle? A nontransitive \(5\)-cycle?
Is it possible to find a set of \(5\) dice so that every one of them beats two and loses to the other two, on average?
For some other implementation options, check here.
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