Here is a game
you can play with a friend. It is a game for two players, with a set
of three dice. These dice are not typical dice however, because
instead of having the values 1 to 6, they display various unusual
values.

Each player picks a die. The two
dice are then rolled together and whoever gets the highest value wins. Seems
fair enough yet,
in a game of say ten rolls, you will always
be able to pick a die with a better chance of winning - no
matter which die your friend chooses. And you can make these dice at home
right now.

Here is the set of three
special dice:

Each die has six faces but only
two values, as follows;

It can be shown (see below)
that in the long run the Red die beats the Blue die, and that the Blue die beats the
Olive die.
So it appears the Red die is the strong die and the Olive die is the weak die. So you
might expect the Red die to beat the Olive die;

If this
is the case we call the dice `transitive' because the winning
property transfers via the die in the middle, the Blue die. [1]

However this is not the case. In fact,
the winning property goes round in a circle - like a game
of `Rock, Paper, Scissors' - with the Red die beating the Blue die, the Blue die
beating the Olive die, and the Olive die beating the Red die. There is no strong
or weak die, so the dice are `non-transitive'. How can this
be?

Let's do an example calculation and show that,
in the long run, the Red die has a better chance of beating the Blue
die.

Notice, when you roll the Red die there are two
possible outcomes; you either roll a 3 or a 6. The probability of rolling a
3 is 5/6, while the probability of rolling a 6 is
1/6.

On the other hand, the Blue die can either roll a 2
or a 5, each with a probability of 1/2. So in total, if we
roll the Red die and the Blue die together, we have four possible
outcomes.

I can
draw a tree diagram of all the possible outcomes. You find the
probability of each outcome by multiplying the probabilities
along the diagram. For example, the probability of
rolling a 5 with the Red die and a 2 with the Blue die is 5/6 x 1/2 =
5/12.

If I want to find the probability that the Red die
wins I add up the probabilities of all the outcomes
where the Red die beats the Blue die. So in this case, the probability the Red die
beats the Blue die is 7/12 - importantly, this is more than
1/2.

In the
same way it can be shown that the Blue die
beats the Olive die, and then remarkably the Olive die beats the Red die. Here, you can
remember the order since the colours are in increasing word-length, until it wraps back to Red.

So, as long
as your opponent picks first, you will always be able to pick a die
with a better chance of winning, with the average winning probability being around 62%
[2] .

This set of three non-transitive dice are particular special because they are
the optimum set of three dice. In other words, we have
maximised the lowest winning
probability. But that is not the only
surprising thing about these particular dice.

After a few defeats your friend may have
become suspicious, but all is not lost. Once you've explained how the
dice beat each other in a circle, challenge your friend to one
more game.

This time you will choose first, in which case
your opponent should be able to pick a die with a better
chance of winning. But let's increase the stakes, and increase
the number of dice. This time each player rolls two of his chosen die,
so that the player with the highest total wins. Maybe two
dice means your opponent has just doubled their chances of
winning. But not so because, amazingly, with two dice the
order of the chain flips!

In
other words, the chain reverses so the circle of victory now
becomes a circle of defeat. Now, the Red die beats the Olive die, the Olive
die beats the Blue die, and the Blue die beats the Red die. Allowing you to win the game
again! [3]

The average winning probability with two dice is around 57% [4] . A word of warning though; although the probability
that the Red die beats the Olive die is greater than 1/2, it is a slim
victory. In the short term, say less than 20 rolls, the effect
is closer to 50-50, so you will still need some luck on your
side!

The idea for non-transitive dice has been around since
the early 1970s [5]
. However, the remarkable reversing property
is not true for all sets of non-transitive dice, for you do
need to pick your values carefully. For example, here is another
famous set of non-transitive dice; it is a set of four
non-transitive dice known as `Efron Dice' and invented by the
American statistician Brad Efron:

This time
the dice use values 0 to 6. Each die has values:

As
before, the dice form a circle where the Blue die beats the Magenta die, the
Magenta die beats the Olive die, the Olive die
beats the Red die, and the Red die beats the Blue die, and they each do so with
a probability of 2/3. This time I have ordered the colours
alphabetically.

Efron Dice are optimal, in the sense that the lowest winning probability achieves
the theoretical upper bound.

We also have
two pairs of dice opposing each other on the circle. In
fact the Magenta die beats the Red die, but the Blue die and the Olive die
are equal, with each having a 50% chance of winning, and neither die dominating [6]
.

Unfortunately, the player picking the Blue die will not have
a very exciting game, with all the values being the number
3. Also, this chain does not display the remarkable property of
flipping when you double the number of dice [7].

It is said
the successful American investor Warren Buffett is a fan of
non-transitive dice. When he challenged his friend Bill Gates to a
game, with a set of Efron dice, Bill became suspicious and insisted
Warren choose first. Maybe if Warren had chosen a set with a
reversing property he could have beaten Gates
- he would just need to announce if they were playing a one die or two
dice version of the game after they had both
chosen.

I wanted to know if it was possible
to extend this to make a three player game, a set of dice
where two of your friends may pick a die each, yet you can pick a die
that has a better chance of beating both opponents - at the same
time!

It turns out there is a way. The Dutch puzzle inventor M.
Oskar van Deventer [8] came up with a
set of seven non-transitive dice, with values from 1 to 21 [9]
. Here two opponents may each choose a die from the
set of seven, and yet there will be a third
die with a better chance of beating each of them. The probabilities are
remarkably symmetric with each arrow on the diagram illustrating a probability
of 5/9.

This means we can play two games simultaneously, but there is still the
question of whether you can beat your two opponents at the
same time.

If the dice were regular fair dice,
with two competing dice having a 50-50 chance of victory (ignoring draws), then
the chance of beating two opponents at the same time would
stand at just over 25%. It isn't exactly 25% because beating one player is not
independent of beating the other player - when you roll a high number you are likely to beat both!

However, these
dice are not fair. Even
though you have the advantage against both opponents,
beating both players is still a challenge, with the probability
of doing so standing around 39%.

This
set of seven dice form a complete directed graph. In the same way, a four
player game would require 19 dice. It is not known if
such a set exists.

However, if
it is possible to construct a set of non-transitive dice with a
reversing
property, then it might be possible to have a three player
game with fewer than seven dice. And potentially such a set might improve our
chances of beating two players at the same time. I have devised such a
set below.

This set of five dice is similar to other
sets of dice we have seen, in that we have a chain where Blue>Magenta>Olive>Red>Yellow>Blue.

However, we also have a second
chain, where Red>Blue>Olive>Yellow>Magenta>Red.

Notice the first chain is ordered alphabetically, while the
second chain is ordered by word-length.

In general, the chain in alphabetical order is stronger than
the chain in order of word-length. But if your friend discovers you using one chain, you can switch
to the other.

Overall, the average winning probability for one
die is 63% [10] .

If
our original set of three non-transitive dice was like a game
of `Rock, Paper, Scissors', this diagram is closer to the
related, but more extreme, non-transitive game `Rock, Paper,
Scissors, Lizard, Spock' [11]
.

But not only that, because if you consider subsets of these five dice you will find
other non-transitive chains! In particular, the first three dice in the chain ordered by word-
length; that's Red, Blue and Olive; is a copy of the optimal set of three non-transitive dice that I
describe above. So with this set of five dice you get the optimal set of three dice for free! In
fact, any three consecutive dice in the chain ordered by word-length makes a set of three non-
transitive dice.

And not only that, because any four consecutive dice in the alphabetical chain makes a
set of four non-transitive dice. The first four in this chain; that's Blue, Magenta, Olive and Red;
is not a copy of the optimal set of four dice, but they do have the same average winning probability
as Efron dice.

With
two dice the chain ordered by word-length
now flips so
Magenta>Yellow>Olive>Blue>Red>Magenta.

On the other hand, the alphabetical chain stays the same - with
one exception, the Red-Olive arrow reverses. This is to be expected since Red, Blue and Olive form a
copy of the optimal set of three.

With two dice, the chain ordered by word-length is stronger than the alphabetical
chain. The average winning probability for two dice
is 59% [12] .

However, the probability that two Red dice beats two Olive dice
is close to 50%. If we treat these dice as equal we can now play two games simultaneously.

Invite two opponents to pick a die each, but do not volunteer whether you are
playing a one die or two dice version of the game.

If your opponents pick two dice that are next to each other alphabetically then
play the one die version of the game. Use the diagram for the one die game to pick a die that can
beat each opponent.

If your opponents pick two dice that are next to each other in the chain ordered
by word-length, then play the two dice version of the game. Again, use the diagram for the two dice
game (that treats Red and Olive as equal) to pick a die that can beat each opponent.

But, can we expect to beat the two other
players at the same time? Well, we have certainly improved the
odds, with the average probability of beating both opponents
now standing around 44% - a 5% improvement over Oskar
Dice, and a 19% improvement over fair dice!
[13] So, if the odds of beating two players isn't over 50%
then how do we win? Consider the following gambling
game:

Challenge two friends to a dice game, where
you will play your two opponents at the same time. If you lose
you will give your opponent £1. If you win your opponent gives
you £1. So, if you beat both players at the same time you win
£2; if you lose to both players you lose £2; and if you beat
one player but not the other then your net loss is zero. You
and your friends decide to play a game of 100 rolls.

If the dice were fair then each player will
expect to win zero - each player wins half the time and loses
half the time.

However, with Oskar Dice, you should expect
to beat both players 39% of the time, and lose to both
players 28% of the time, which will give you a net profit of
£22.

But even better, with Grime Dice, you should
expect to beat both players 43.8% of the time, but only lose to both players
22.7% of the time, giving you an average net profit closer to £42! (And
possibly the loss of two former friends).

This set is the best set of five
non-transitive dice with these properties that I have
found.

I invite you to try out these games
yourself [14]. They are easy to make by either writing on blank
dice, or
modifying some old dice. They are also inexpensive to buy and you can
get the set of five Grime Dice from Maths Gear. Remember, the set of five Grime dice contains subsets of three and
four non-transitive dice for free, including the optimal set of three. Or, you can buy the optimal
set of three separately.

Try them out on your
friends and enjoy your successes and failures!

Footnotes:

[1] Other examples of transitivity: If a, b
and c are real numbers then if a > b and b > c, then we know a
> c. For example, if 9 > 4.6 and 4.6 > 5/3 then 9 >
5/3. If a, b and c are integers such that a stictly divides b
(with no remainder) and b strictly divides c, then we know a
strictly divides c. For example, if 3|15 and 15|60 then
3|60. The last two relations are transitive. On the
other hand, if a,b and c are integers such that a+b > 7 and b+c > 7, we
cannot deduce that a+c > 7. This relation is not
transitive.

[2] In full the probabilities are: P(Red >
Blue) = 7/12, P(Blue > Olive) = 7/12 and P(Olive > Red) = 25/36.

[3] Tim Rowett introduced this set of
non-transitive dice at the Gathering for Gardner V (2002).

[4] This
time the full probabilities are: P(Red > Olive) =
671/1296 (a slim victory!), P(Olive > Blue) = 85/144, and P(Blue > Red) =
85/144.

[5] Gardner, M. "Mathematical Games: The
Paradox of the Nontransitive Dice and the Elusive Principle of
Indifference." Sci. Amer.
223 ,
110-114, Dec.
1970.

[12] Grime
Dice, two dice version: P(Magenta > Yellow) = 16/27, P(Yellow > Olive) = 56/81, P
(Olive > Blue) = 85/144, P(Blue > Red) = 85/144, P(Red > Magenta) = 56/81. P(Blue
> Magenta) = 5/9, P(Magenta > Olive) = 7/12, P(Olive > Red) = 625/1296(*), P
(Red > Yellow) = 7/12, P(Yellow > Blue) = 5/9. (*)This is a slim defeat! However, when
rounded to the nearest whole number, the expected number of wins is the same as 50% for anything
less than 28 trials. Also, 95% of the time we expect the number of wins to be plus/minus two
standard deviations from the mean. When we take this into account, the effect is little different
from 50%.

[13]
In the ideal three player game, we would like the average
probability of beating two players to be over 50%. This would mean,
on average, each winning die should beat another with a probability
of over 70%. It has been shown that for three non-transitive
dice, the die with the smallest winning probability has an upper bound, namely the
Golden Ratio Conjugate Φ = 0.618. The upper bound may not necessarily be obtainable,
and it will depend on the number of sides the dice have. You will come closest to the upper bound
when the number of sides is a Fibonacci number. Rowett Dice are the optimum set of three six-sided
dice. For four non-transitive dice the upper bound is 2/3 - making Efron dice optimal in this sense,
since the die with the smallest winning probability obtains its upper bound. For five dice the upper
bound is 0.692. This sequence of upper bounds increases with the number of dice and converges to
3/4. Naturally, if the weakest die achieves the upper bound then the average winning probability
will be even higher. References; The Paradox of Nontransitive Dice, Richard P. Savage, Jr. The
American Mathematical Monthly, Vol. 101, No. 5 (May, 1994), pp. 429-436
http://www.jstor.org/stable/2974903 and; S. Trybula, On the paradox of n random variables, Zastos.
Mat. 8 (1965), 143-154.

[14] For
the very keen, here's a set to try that uses mathematical constants:
A: 1 1 1 1 1 π ; B: Φ Φ Φ e
e e; C: 0
φ φ φ φ φ ; where e is Euler's Number, φ is the
Golden Ratio and Φ
is
the Golden Ratio Conjugate. Do these dice form a non-transitive set? What
about with two dice, does the order reverse as before?