A Lazy Way to Make Two Dice from Many

Question

Recently, I stumbled upon the puzzle: Make Two Dice out of Three, which asks for a solution to a case of the following question:

Suppose I roll $n$ indistinguishable dice. From purely the $n$ numbers showing, and no information about the ordering of the dice, what algorithm can I use, spitting out an unordered pair of numbers $a$ and $b$ such that the distribution of the unordered pair $(a,b)$ is the same as the distribution of a roll of two unordered dice?

After reading through the solutions posed on the question, none of which seemed particularly easy to remember, I sent the question along to one of my mathematician friends, Cassandra. She proposed the following general form for a more elegant strategy:

Well, obviously if you wanted to make one die from many, you could do so by taking the sum of the dice mod $6$. That is, if the dice come up as $x_1,x_2,\ldots,x_n$ you combine them into one value as $$x_1+x_2+\ldots + x_n \pmod6.$$ Rather than adopt some complicated scheme for getting two rolls, it would make more sense to me to try and salvage this strategy. In particular, choose two functions $f$ and $g$ and then say that if the dice come up as $x_1,x_2,\ldots,x_n$, you read the two rolls as the following: $$f(x_1)+f(x_2)+\ldots + f(x_n)\pmod6$$ $$g(x_1)+g(x_2)+\ldots + g(x_n)\pmod6.$$ I'm sure this is possible for large enough $n$ and some $f$ and $g$.

I like this idea, since one would only have to memorize two functions on a domain and range of six elements and it avoid any casework. To ease my work a bit, I have decided to only consider whether this is possible for even $n$. After much trying, I have been unable to find such functions. I'm sure I could go back and ask her for clarification, but I'm too embarassed - so I'll ask all of you: Can Cassandra's prediction be true for any even $n$?

---

As a couple answers below interpret the question differently than intended, let me write out the condition formally: Let $$F=f(x_1)+\ldots +f(x_n)\pmod 6$$ $$G=g(x_1)+\ldots+g(x_n)\pmod 6.$$ We wish to have that for any suitable $a$ and $b$ we have that $$P(F=a\text{ and }G=b) + P(F=b\text{ and }G=a)=\frac{1}{18}.$$ where $P$ is the probability of an event happening. This is what is meant by saying that $(F,G)$ has the same distribution as a roll of two unordered dice.

As a hint, let me say that the intended solution is not a counting argument, nor does it have to do with examining probabilities too carefully; it is rather more analytical in nature.

Answer 1

[This answer formerly claimed to prove that the thing can't be done for any $n$, odd or even, but there were at least two big holes in the proof. The following now claims to be no more than what might some day be the beginning of a proof. I'm still thinking about it.]

Write $p=x^{f(1)}y^{g(1)}+\cdots+x^{f(6)}y^{g(6)}$. Then the probability of getting sums $i,j$ after rolling $n$ dice is given by the coefficient of $x^iy^j$ in $(p/6)^n$. To reduce the sums mod 6 we need to work in polynomials mod $(x^6-1,y^6-1)$. And the question is then whether we can arrange that $p^n/6^{n-2}=(1+\cdots+x^5)(1+\cdots+y^5)$ mod $(x^6-1,y^6-1)$ or, equivalently, that $p^n/6^{n-2}=(1+\cdots+x^5)(1+\cdots+y^5)+(1-x^6)a+(1-y^6)b$ where $a,b$ are polynomials.

... Except that we're supposed to be looking at unordered pairs of results, which means what we want is that a corresponding thing holds not for $p(x,y)^n$ but for $p(x,y)^n+p(y,x)^n$. If we put $x=y$ then this is just $2p(x,x)^n$.

Suppose $x$ is a 6th root of unity other than 1. Then clearly the RHS is zero, so the LHS is too. So $p(x,x)^n=0$ for every such choice of $x$; so $p(x,x)=0$ for every such choice of $x$. In particular, when $\omega^6=1$, $(x-\omega)$ is a factor of $p(x,x)$ and therefore so is the product of all these factors, namely $(x^6-1)/(x-1)=1+\cdots+x^5$.

Now, what exactly is $p(x,x)$? It's $\sum x^{f(i)+g(i)}$; and remember that we only care about $f,g$ mod 6. So what this tells us is that each residue mod 6 appears exactly once among the $f(i)+g(i)$. So we can write $p=\sum x^i(y/x)^{h(i)}$ with the usual caveats about exponents being mod 6. (If you're queasy about non-polynomials, say $x^5y$ instead of $y/x$.)

Answer 2

If the statement holds for any even $n$, it holds for $n=2$.

$x_1'=f(x_1)+f(x_2) \mod 6\\ x_2'=g(x_1)+g(x_2) \mod 6$

Since $+$ is commutative, the order of the two input dice is not important (e.g. $(3, 5)$ will give the same result as $(5,3)$). This means that the number of unique ordered pairs $(x_1',x_2')$ is at most $21$ (the number of unique inputs). However, the number of unique ordered pairs of two dice roll results is $36>21$, which means that not every result can be produced, and the statement is false.

Answer 3

(Check the edit history of this post to see my old answer, which was missing the point.)

Now that I see what's going on, I'm not confident that such a function does exist of the form described.

That is, we need it to be the case where there are at least six $\bf{x}$ vectors such that $\sum f(x_i) \mod 6 = 0$, which also satisfy the property that $\sum g(x_i) \mod 6$ is uniformly distributed from 0 to 5.

The number of input and output states line up--for basically any number of dice--but I think the functional constraint that the function is applied to individual rolls, and then summed, destroys your ability to sufficiently distinguish between states. It's not clear how you're going to get rid of the correlation between the sum of $f$ and the sum of $g$, but it's also not immediately obvious to me how to prove they're always linked. Perhaps there's a group theory argument here that's neat?

Answer 4

Let $x_i$ be the $i$ th element of the sequence of results sorted into increasing order.

Let $k$ be the roll iteration.

Using Iverson bracket notation:

$F = \Big( {\large{\scriptsize [k \in \{2j:j \in \mathbb N\}]} \sum_{i=1}^n {\scriptsize [i \in \{2j−1:j \in \mathbb N\}]} ~ x_i} \Big )(mod~6) + \Big( {\scriptsize [k \in \{2j-1:j \in \mathbb N\}]} \sum_{i=1}^n {\scriptsize[i \in \{2j:j \in \mathbb N\}]} ~ x_i \Big) (mod~6)$

$G = \Big( {\large{\scriptsize [k \in \{2j-1:j \in \mathbb N\}]} \sum_{i=1}^n {\scriptsize [i \in \{2j−1:j \in \mathbb N\}]} ~ x_i} \Big )(mod~6) + \Big( {\scriptsize [k \in \{2j:j \in \mathbb N\}]} \sum_{i=1}^n {\scriptsize[i \in \{2j:j \in \mathbb N\}]} ~ x_i \Big)(mod~6)$

Here $n$ can be odd or even. $(mod ~6)$ makes it irrelevant.

Answer 5

Let $x_i$ be the $i$th element of the sequence of results sorted into increasing order.

Let $k$ be the roll iteration.

Let $E = \Sigma x_i \ (mod ~6)\ $ for even $i$

Let $O = \Sigma x_i \ (mod ~ 6)\ $ for odd $i$

Let $F = E$ if $k$ is even, otherwise $F=O$

Let $G = O$ if $k$ is odd, otherwise $G=E$

Answer 6

Perhaps I'm being too lazy... perhaps I misunderstand the question.

Let f(x) be the sum of rolls modulo 6 of all the dice i <= n/2: $f(x) = \sum_0^{n/2} x_i \ ,mod ~6$

Let g(x) be the sum of rolls modulo 6 of all the dice i > n/2: $g(x) = \sum_{(n/2)+1}^n x_i \ ,mod ~6$

For any even n, an equal number of dice are combined for F and for G, but as each probability on a single roll is equal, you can simply split the rolled dice into two arbitrary piles and take the sum modulo 6 of each pile separately.

What remains is trivially equivalent to two single rolls.

Answer 7

Doesn't the following yield normal distribution, for any $n$, any number of unordered outputs, and any type of dice?

Let $d$ = number of virtual output dice
Let $s$ = number of honest die sides
Given $n>d$

$$x'_1=f(x_1)+...+f(x_{n-d}) \text{ } [Mod \text{ }s] $$ $$x'_2=f(x_2)+...+f(x_{n-d+1}) \text{ } [Mod \text{ }s] $$ $$x'_3=f(x_3)+...+f(x_{n-d+2}) \text{ } [Mod \text{ }s] $$ $$\cdot$$ $$\cdot$$ $$\cdot$$ $$x'_d=f(x_d)+...+f(x_{n}) \text{ } [Mod \text{ }s] $$

---

Answer 8

$F = f(x_1)+…+f(x_n) = x_1$

$G = g(x_1)+…+g(x_n) = x_2$

Since the die are indistinguishable by design, $x_i$ is a random assignment. Use whatever assignment the OP used in framing the the puzzle. The formulas satisfy the problem. It's not a counting argument, and it doesn't examine probabilities too carefully.