Yes. The actual values on the dice faces don't matter, only their ordering. WLOG we may assume that no two faces of the same or different dice are equal. We can assume "generalised dice", where the faces need not be equally probable. These can be approximated by dice with equi-probable faces by having enough faces and marking some of them the same.

Take the case of three dice, called A, B, and C. Picture the different

values on the faces of the A die. Suppose there are three

A A A

The values on the B die must lie in between those of the A die

B A B A B A B

With three different A values, we need only four different B values.

Similarly, the C values must lie in between these

C B C A C B C A C B C A C B C

Assume we want A to beat B, B to beat C, and C to beat A. Then the above

scheme for the ordering of values can be simplified to

B C A B C A B C A B C

since for example, the first C in the previous arrangement can be moved to the second with the effect that the probability that B beats C is increased, and the probabilities that C beats A or A beats B are unchanged. Similarly for the other omitted faces.

In general we obtain for n dice A...Z the arrangement

B ... Z A B ... Z ...... A B ... Z

where there are k complete cycles of B..ZA followed by B...Z. k must be at least 1.

CONJECTURE: The optimum can be obtained for k=1.

So the arrangement of face values is B ... Z A B ... Z. For three dice it is BCABC. Thus one die has just one face, all the other dice have two (with in general different probabilities).

CONJECTURE: At the optimum, the probabilities that each die beats the next can be equal.

Now put probabilities into the BCABC arrangement
B C A B C x y 1 x' y'

Clearly x+x' = y+y' = 1.

Prob. that

A beats B = x' B beats C = x + x'y' C beats A = y

Therefore x' = y = x + x'y'

Solving for these gives x = y' = 1-y, x' = y = (-1 + sqrt(5))/2 = prob. of each die beating the next = 0.618...

For four dice one obtains the probabilities
B C D A B C D x y z 1 x' y' z'

A beats B: x' B beats C: x + x'y' C beats D: y + y'z' D beats A: z

CONJECTURE: for any number of dice, at the optimum, the sequence of probabilities abc...z1a'b'c...z' is palindromic.

We thus have the equalities
x+x' = 1 y+y' = 1 z+z' = 1 x' = z = x + x'y' = x + x'y' y = y' (hence both = 1/2)

Solving this gives x = 1/3, z = 2/3 = prob. of each die beating the next.

Since all the numbers are rational, the limit is attainable with

finitely many equiprobable faces. E.g. A has one face, marked 0. C has two faces, marked 2 and -2. B has three faces, marked 3, -1, -1. D has three faces, marked 1, 1, -3. Or all four dice can be given six faces, marked with numbers in the range 0 to 6.

Finding the solution for 5, 6, or n dice is left as an exercise.

-- Richard Kennaway __\_ / School of Information Systems Internet: jrk@sys.uea.ac.uk \ X/ University of East Anglia uucp: ...mcsun!ukc!uea-sys!jrk \/ Norwich NR4 7TJ, U.K.

References:

Martin Gardner (of course!) wrote about non-transitive dice, see the Oct '74 issue of Scientific American, or his book "Wheels, Life and Other Mathematical Amusements", ISBN 0-7167-1588-0 or ISBN 0-7167-1589-9 (paperback).

In the book, Gardner cites several results: (1) the above result, to wit, that with three dice the maximum probability is the golden ratio phi = (sqrt(5) - 1) / 2. (2) Bradley Efron of Stanford U. found three sets of four 6-sided dice, where each set has 2/3 as the advantage probability. (3) .75 is the limit approached as the number of dice increases.

"The Paradox of Nontransitive Dice," Richard P. Savage, Jr., AMM, 101, 5, May 1994, p. 429.

Savage proves that for three n-sided dice it is always possible to arrange the numbers on the dice such that each of the probabilities exceeds phi * (1 - O(1/n)). Thus one can arrange that the probabilities are as close to phi as one wants by increasing n.

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