## Puzzle #6: Probabilities of dice comparisons

Puzzle: Suppose there are 4 dice: Blue, Green, Red, and White. These dice have different numbers than usual printed on their faces. After observing a long sequence of experiments rolling pairs of these dice, you conclude that

• When both are rolled simultaneously, the blue die gives a higher number than the green die 2/3 of the time.
• When both are rolled simultaneously, the green die gives a higher number than the red die 2/3 of the time.
• When both are rolled simultaneously, the red die gives a higher number than the white die 2/3 of the time.
You are now asked to consider rolling the blue and white dice simultaneously. What can you conclude about the probability P that the blue die will produce a higher value than the white one? (Is P>50%?)

Answer: We cannot conclude P>50%. In fact, P can be as small as 1/3 (i.e. 33.33%) and as large as 100%! A set of dice that yield a Blue>Green>Red>White>Blue "probability cycle" (where "A>B" means "rolling A will produce a higher roll than B with probability 2/3") is given in the top row of this table.

How to construct dice.
blue green red white
004444 333333 222266 111555
446666 555555 222266 111333

However, if the numbers painted on each die's 6 faces instead are those given in the bottom row of the table, then Blue>Green>Red>White and the Blue die will beat the White die 100% of the time.

The fact that 1/3 and 100% both are best possible bounds is obvious for 100%, and not-so-obvious for 1/3 – but proven by S.Trybula: On the paradox of N real variables, Zastos. Matem. (Applied) 8 (1965) 143-154 and also independently by Z.Usiskin: Max-min probabilities in voting paradoxes, Annals of Math'l Statistics 35,2 (1964) 857-862. Usiskin indeed showed that with 2K "dice" (i.e. independent probability distributions over the reals), the maximin intransitivity probability P(2K) was bounded by (3K-1)/(4K)≤P(2K)≤3/4, and P(3)=(√5-1)/2=0.61803... (recognize the "golden ratio"?) and P(4)=2/3.

The exact maximum possible intransitivity probabilities P(n) for n=2, n=3, and n=4 dice, namely P(2)=1/2, P(3)=(√5-1)/2, and P(4)=2/3, all meet the following upper bound claimed by [Ilya I. Bogdanov: Intransitive Roulette, Mathematical education, ser. 3, no. 14 (2010) 240-255; this journal and this paper both are in Russian].

P(n) ≤ 1 - (1/4) sec(π/[n+2])2 = 1 - 1/[2+2cos(2π/[n+2])] < 3/4.

Since my Russian is poor (actually, nonexistent) I am not sure whether Bogdanov is claiming the "≤" is tight for all n.

Other references:
Colin R. Blyth: Some Probability Paradoxes in Choice from Among Random Alternatives, J.American Statistical Association 67, 338 (June 1972) 366-373
R.P. Savage Jr: The Paradox of Nontransitive Dice, American Mathematical Monthly 101,5 (May 1994) 429-436.
H.Steinhaus & S.Trybula: On a paradox in applied probability, Bull. Acad. Polo. Sei. 7 (1959) 67-69.