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;

A: 3 3 3 3 3 6 B: 2 2 2 5 5
5 C: 1 4
4 4 4 4

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

If this
is the case we call the dice `transitive' because the winning
property transfers via the die in the middle, die B. [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 die A beating die B, die
B beating die C, and die C beating die A. 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, die A has a better chance of beating die
B.

Notice, when you roll die A 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, die B can either roll a 2
or a 5, each with a probability of 1/2. So in total, if we
roll die A and die B 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 die A and a 2 with die B is 5/6 x 1/2 =
5/12.

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

In the
same way it can be shown that die
B beats die C, and then remarkably die C beats die A. 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] .
Although known before, this set of three dice are currently
sold by toy collector and consultant Tim Rowett [3] . Rowett chose this particular set of dice
over all other choices of non-transitive dice because this is
the optimum set. 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, die A beats die C, die C
beats die B, and die B beats die
A. Allowing you to win the game
again!

The average winning probability with two dice is around 57% [4] . A word of warning though; although the probability
that die A beats die C 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 die A beats die B, die B beats die C, die
C beats die D, and die D beats die A, and they each do so with
a probability of 2/3. 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 die B beats die D, but die A and die C each have
a 50-50 chance of winning, with neither die dominating [6]
.

Unfortunately, the player picking die A 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.

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
A>B>C>D>E>A.

However, within that we now have a second
chain, where
A>C>E>B>D>A.

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]
.

With
two dice the first chain stays the same. But the second chain
now flips so
A>D>B>E>C>A.

The average winning probability for two dice
is 59% [12] .

I will add one caveat: although it is
desirable that E>A with two dice, in the short term (say
less than 20 rolls) the effect is closer to 50-50, with
neither die dominating. In all other cases your choice of
die will give you a healthy advantage over your two
opponents.

Overall, the average winning probability is 61%.
Here is
how you play the game
against
two opponents:
invite each player to pick one of the dice, but do not
volunteer the rules of the game at this point. When the two other players have
made their choice, you may now make your choice - including whether you and
your opponents will be playing the one die or two dice version
of the game. As before we play a
game of say 10 rolls and, no matter which dice your opponents
choose, there will always be a die with a better chance of beating each
player.

For example, if one
opponent chooses die B and the other opponent chooses die C, then you
should pick die A and play the one die version of the game. Then,
according to the first diagram above, you will have a better chance of
beating each opponent.

On the other hand, if
one opponent chooses die C and the other opponent chooses die E, then you should
pick die B and play the two dice version of the
game. Then, according to the second diagram above, you can expect to beat
each opponent again.

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.

You may also like to consider subsets of Grime Dice. There are 5 subsets of three dice and 5 subsets of four dice that make non-transitive chains.

For example, the dice A, C and E form a non-transitive subset of three dice. In fact, this is just a copy of Rowlett Dice with different numbers. This subset also has the reversing property of Rowlett Dice, and the slim victory in the two set version of Rowlett dice explains the slim defeat displayed in the two set version of Grime Dice. This subset is optimal in the sense that the lowest winning probability reaches its theoretical upper bound, with an average winning probability of 62%.

Any subset of four Grime Dice make a non-transitive chain, with the subsets A, B, C and E and A, C, D, E both having an average winning probability of 2/3. However, these subsets are not copies of Efron Dice and are not optimal, although they do have the same average winning probability as Efron Dice.

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 three Rowett Dice and the set of four Efron
Dice from Grand
Illusions
.

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(A >
B) = 7/12, P(B > C) = 7/12 and P(C > A) = 25/36.

[3] Tim Rowett introduced this set of
non-transitive dice at the Gathering for Gardner V (2002). They can
be bought exclusively from Grand Illusions http://www.grand-illusions.com

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

[6] Efron Dice: P(A > B) = 2/3, P(B >
C) = 2/3, P(C > D) = 2/3, P(D > A) = 2/3, P(A > C) = 1/2
and P(B > D) = 5/9

[7] The
probabilities for two sets of Efron Dice are: P(A > D) = 5/9, P(B
> A) = 5/9, P(B > C) =
5/9, P(C > D) =
5/9, P(B>D) = 11/27, P(D > B) =
0, P(B = D) = 16/27 (a draw), P(A >
C) =
1/4, P(C
> A) = 1/4, P(A = C) = 1/2. There is no chain
like before.

[12] Grime
Dice, two dice version: P(A > B) = 7/12, P(B > C) = 5/9, P(C
> D) = 5/9, P(D > E) =
7/12, P(E > A) = 625/1296(*); P(A > D) =
56/81, P(B > E) =
56/81, P(C > A) =
85/144,
P(D > B) =
16/27, P(E > C) =
85/144. (*)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?