Making two fair dice

Question

You are given two unlabelled standard 6-sided dice. Can you write every number from 1 to 12 on the dice, such that in any given throw there is a 50% chance of the first die being smaller than the second die?

Answer 1

This seems to work:

1,2,3,10,11,12 on one die, 4-9 for the other.

Whichever we throw first,

There's an equal chance of getting 1-3 (smaller than 4-9) or 10-12 (higher than 4-9). In fact we don't even need to roll the 4-9 die at all as its result is irrelevant.

Answer 2

If

The die containing any number i also contains the number 13-i, the result is fair. This sufficiency is easy to see: For a given pair of dice with that condition you can form the complemented pair with all numbers replaced by their difference from 13. A win with one set is a loss for the other set. But with this condition, the complemented dice are the same as the original dice, so only a win probability of 50% is consistent.

I believe

it’s also a necessary condition, but it’s too early in the day for me to prove it.

Answer 3

Surprisingly, this works if and only if you divide the pips evenly between the dices!

Let us take one of the dices and denote its numbers by $\sigma_1,\ldots,\sigma_6$ (in increasing order). Then, the side with $\sigma_i$ wins against all lower sides from the other die. These are $\sigma_i - i$ sides, since of the $\sigma_i$ numbers $1,2,\ldots,\sigma_i$ the $i$ numbers $\sigma_1, \sigma_2, \ldots, \sigma_i$ are on our dice. From all $36$ combinations, our dice wins in $$ \sum_{i = 1}^6 (\sigma_i - i) = \sum_{i=1}^6 \sigma_i - 21$$ combinations. Since this must be $18$, we have to have $39$ pips on each of the dice!

Answer 4

Sneftel's sufficient condition is in fact not necessary. A different construction to achieve 50% win rate is the following.

Divide the numbers $1,\dots,12$ into six consecutive pairs $(1,2),\dots,(11,12)$. Number the dice in any manner satisfying the following two properties: each die has one number from each pair, and each die has three even numbers.

Now roll the dice, and observe only which pair(s) the two numbers rolled come from. If they come from different pairs, then whichever die has the number from the higher pair wins. But since each die bears a number from each pair, this is equally likely to be the first die or the second die. On the other hand, if the two numbers come from the same pair, then whichever die has the even number of that pair wins. But this is the first die for three of the six pairs, and the second die for the other three. So in total, each die has the same probability of winning.

This construction does include some cases that Sneftel's doesn't (and vice versa). For example,

$(1,3,6,8,10,11)$ and $(2,4,5,7,9,12)$ meets my condition but not Sneftel's; $(1,2,3,10,11,12)$ and $(4,5,6,7,8,9)$ meets Sneftel's but not mine.

There are also schemes that work but fit neither pattern, e.g. the following:

Divide $1,\dots,8$ into pairs adding to $9$, and put half the pairs on each die. Then divide $9,\dots,12$ into pairs adding up to $21$ and put half the pairs on each die. An example that fits neither of the previous patterns is $(1,2,7,8,9,12)$ and $(3,4,5,6,10,11)$. If two "low" numbers are rolled we essentially have Sneftel's construction for $4$-sided dice, and each is equally likely to win; similarly for two "high" numbers. But since each die has four low and two high numbers, if we roll one of each they are equally likely to be either way round.

FWIW a computer search reveals there are 29 ways to do it (up to swapping the two dice). 13 fit neither pattern, 10 fit mine and 10 fit Sneftel's (including 4 that fit both). One can easily check that for $2k$-sided dice each pattern produces $\frac{1}{2}\binom{2k}{k}$ ways, so it is not a coincidence that they give the same number here.

Answer 5

I think I built a proof for this :) It's my first one so please give feedback on how I can improve.

There are two things to prove:

One, can the set be divided into 2 equal sets? Two, can sets $A$ and $B$ be arranged such that $A/2 < B$.

Proof 1:

$12/2 = 6$ Therefore the set can be divided equally among the die.

Proof 2:

If there are $6$ elements in each set, and we desire that $1/2$ the elements of one set are less than the other we understand that we need to divide the total range (12) into four 3 (6/2) element subsets. $$\begin{align}W = \{1,2,3\} \\X = \{4,5,6\} \\Y =\{7,8,9\} \\Z =\{10,11,12\} \end{align}$$ Furthermore we can see that $$W < X < Y < Z$$ Given an element from $W, X, Y, Z$ what is the probability of picking an element greater from any other set? $$\begin{align} P(W) = 3/3 (X,Y,Z) \\ P(X) = 2/3 (Y,Z) \\ P(Y) = 1/3 (Z) \\ P(Z) = 0 \end{align}$$ Now we can see that there is only 1 way of arranging these 4 subsets into two sets so that there is an equal probability that after picking one element from 1 set, when we pick an element from the other the new element is greater. $$\begin{align} P(W \cup Z) = P(X \cup Y) = 3/6 \\ = 1/2 \end{align}$$

What was the key intuition?

That a 50% probability translates to $A/2$ (half the set) maintaining a property. If the full set is 12, and we want to divide it into two sets of 6 then each set of 6 needs to be a combination of 2 sets of 3.

Answer 6

I'll try to provide all possible solutions.

First analysis :

However we choose the labels, The first die has 6 possibilities and the second dice has 6 possibilities, making 36 outcomes in total. We want 18 of these outcomes to be winning for the first die.

Construction of the solutions:

Let's start with 1-2-3-4-5-6 on the first die, there is 0 out of 36 outcomes winning for die1. When we raise one of the labels by 1 (if allowed, i.e. no repetition and no going above 12) we get exactly one more outcome winning for die1. 1-2-3-4-5-7 has 1 outcome winning for die1, 1-2-3-4-5-8 has 2, 1-2-3-4-6-8 has 3, etc... We continue this until we reach 18. We can reach any labelling of die1 this way (with any number of steps), so in particular we can reach all correct labelling of die1 (with 18 steps, for example 1-2-3-10-11-12).

Counting solutions:

Let's look at how much we raised all our labels, for example to reach 1-2-3-10-11-12 we have raised 0-0-0-6-6-6. This sequence is always non-decreasing and has a sum of 18, and such a sequence gives us a working solution. We want to count these sequences now. It corresponds to the partitions of 18 with at most 6 parts, I'll just provide a link to the relevant wikipedia page as it is not so simple : https://en.wikipedia.org/wiki/Partition_(number_theory)#Restricted_part_size_or_number_of_parts

If someone calculates it for our case I would be grateful !

A nice necessary and sufficient condition:

It can be a bit surprising, but if we pay attention to the construction of the solutions, a sufficient and necessary condition is that the sum of the numbers on die1 is 39 (for example 1+2+3+10+11+12 = 39 so 1-2-3-10-11-12 is a solution)