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.

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.

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. 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.

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