Another week, another dice-y Classic!
When you roll a pair of fair dice, the most likely outcome is 7 (which occurs 1/6 of the time) and the least likely outcomes are 2 and 12 (which each occur 1/36 of the time).
Annoyed by the variance of these probabilities, I set out to create a pair of “uniform dice.” These dice still have sides that are uniquely numbered from 1 to 6, and they are identical to each other. However, they are weighted so that their sum is more uniformly distributed between 2 and 12 than that of fair dice.
Unfortunately, it is impossible to create a pair of such dice so that the probabilities of all 11 sums from 2 to 12 are identical (i.e., they are all 1/11). But I bet we can get pretty close.
The variance of the 11 probabilities is the average value of the squared difference between each probability and the average probability (which is, again, 1/11). One way to make my dice as uniform as possible is to minimize this variance.
So how should I make my dice as uniform as possible? In other words, which specific weighting of the dice minimizes the variance among the 11 probabilities? That is, what should the probabilities be for rolling 1, 2, 3, 4, 5 or 6 with one of the dice?
Because of symmetry, let the probabilities for rolling be:
| Dice Roll | Probability |
|---|---|
| $1$ or $6$ | $x$ |
| $2$ or $5$ | $y$ |
| $3$ or $4$ | $z$ |
It follows that $x+y+z =0.5$
Looking at all sums, we get the following table:
| Sum | Probability |
|---|---|
| 2 | $x^2$ |
| 3 | $2xy$ |
| 4 | $y^2+2xz$ |
| 5 | $2xz+2yz$ |
| 6 | $z^2+2xy+2yz$ |
| 7 | $2(x^2+y^2+z^2)$ |
| 8 | $z^2+2xy+2yz$ |
| 9 | $2xz+2yz$ |
| 10 | $y^2+2xz$ |
| 11 | $2xy$ |
| 12 | $x^2$ |
Expanding the squared differences $\left( \dfrac{1}{11} - P \right)^2$:
| Sum | Squared Difference |
|---|---|
| 2 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{2x^2}{11} + \left(x^2 \right)^2$ |
| 3 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{4xy}{11} + \left(2xy \right)^2$ |
| 4 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{2y^2+4xz}{11} + \left(y^2+2xz \right)^2$ |
| 5 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{4xz+4yz}{11} + \left(2xz+2yz \right)^2$ |
| 6 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{2z^2+4xy+4yz}{11} + \left(z^2+2xy+2yz \right)^2$ |
| 7 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{4x^2+4y^2+4z^2}{11} + \left(2 \left(x^2+y^2+z^2 \right) \right)^2$ |
| 8 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{2z^2+4xy+4yz}{11} + \left(z^2+2xy+2yz \right)^2$ |
| 9 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{4xz+4yz}{11} + \left(2xz+2yz \right)^2$ |
| 10 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{2y^2+4xz}{11} + \left(y^2+2xz \right)^2$ |
| 11 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{4xy}{11} + \left(2xy \right)^2$ |
| 12 | $\left(\dfrac{1}{11} \right)^2 - \dfrac{2x^2}{11} + \left(x^2 \right)^2$ |
The left most term is $11 \cdot \dfrac{1}{11^2} = \dfrac{1}{11}$.
The sum of all $P$s from the 11 expressions is $(2x + 2y + 2z)^2$ to account for all two dice pairs. The middle terms simplify to $-\dfrac{2}{11}$
The sum of the right most terms is complicated. I ended up with
I attempted to determine an algebraic solution but was unsuccesful.
If $x = 0.2438$, $y = 0.1374$, and $z = 0.1188$, then the variance is $\approx 0.0012176$.