Dicey Dice
This problem follows naturally from X-Dice, although it may be attempted independently.
They are to be made with the following properties:
1. The faces are to be numbered using only whole numbers 1 to 6.
2. Some of the numbers 1 to 6 can be left out or repeated as desired on each dice.
3. Apple Green is expected to beat Bright Pink on a single roll
4. Bright Pink is expected to beat Cool Grey on a single roll.
5. Cool Grey is expected to beat Apple Green on a single roll.
Invent a set of such dice.
What is the probability that each of your dice wins or loses over each of the other dice?
Is it possible to create a totally fair set of such dice with $P(A> B) = P(B> C) = P(C> A)$?
NOTES AND BACKGROUND
These dice exhibit a property called non-transitivity. Other examples of non-transitivity are found in voting systems. You can read about this in our article Transitivity.
You can use common sense for simple comparisons between dice, but where the relative probabilities are not obvious you can use the formula for conditional expectation:
$$
P(A> B) = P(A> B|A=1)P(A=1)+\dots + P(A> B|A=6)P(A=6)
$$
Steve writes this about his problem:
We will need to use a mixture of common sense and conditional probability to solve this problem, as there is no obvious systematic point to get started with this anaysis. The key equation we need is:
$$
P(A> B) = P(A> B|A=1)P(A=1)+\dots +P(A> B|A=6)P(A=6)
$$
An obvious starting point would be to consider the case where dice B has only one number on it, 3, say.
For A to have a greater than 50% chance to beat B, A must have 4 higher numbers, Say 4444xx.
For B to have greater than 50% chance to beat C, C must have 4 lower numbers, say xx2222. Can C beat A? If C = 662222 and A = 444411 then P(C beats A) = 1/3 x 1 + 2/3 x 1/3 = 5/9
To find dice with equal probabilities after a bit of fiddling around we settle on the answer 5/9. The trick is to adjust down any dice which are too strong and up any dice which are too weak. There is some flexibility to adjust numbers without altering the probabilities for any other wins/losses.
An answer is
| A | 5 | 5 | 4 | 4 | 1 | 1 |
| B | 4 | 4 | 3 | 3 | 3 | 2 |
| C | 6 | 6 | 2 | 2 | 2 | 2 |
Explicitly, we have
$$
\begin{eqnarray}
P(A> B) &=& P(A> B|A=5)\times \frac{1}{3}+P(A> B|A=4)\times \frac{1}{3}+P(A> B|A=1)\times \frac{1}{3}\\
&=&1\times \frac{1}{3}+ \frac{2}{3}\times \frac{1}{3}+0\times\frac{1}{3}\\
&=& \frac{4}{9}
\end{eqnarray}
$$
Why do this problem?
Along with X-Dice, this is a great problem for thinking about conditional probability.
Possible approach
Key questions
Possible extension
Possible support