Non-transitive dice

Question

Consider three six-sided dice with these faces:

Die A: $2, 2, 4, 4, 9, 9$
Die B: $1, 1, 6, 6, 8, 8$
Die C: $3, 3, 5, 5, 7, 7$

Two players each pick a die (no repeats), roll, and the higher number wins.

Show that A beats B, B beats C, and C beats A, each with probability greater than $\tfrac12$ . In a game where your opponent picks first, is that good or bad for you?

Accepted Answer

Each die has three distinct values, each appearing twice, so any matchup has 6 	imes 6 = 36 equally likely outcomes. We count wins. A vs B. A wins when its face beats B's: - A shows 2: beats B's 1 (loses to 6, 8). 2 wins. - A shows 4: beats B's 1 (loses to 6, 8). 1 win among B's three values... but counting all pairings, let's tally over the 9 value-pairs (each pair = 4 of the 36 outcomes since both faces are doubled): A\B 1 6 8 -------------- 2 A B B 4 A B B 9 A A A A wins in 5 of the 9 value-pairs, so P(AB) = 	frac{5}{9} 	frac12. B vs C. B\C 3 5 7 -------------- 1 C C C 6 B B C 8 B B B…

Non-transitive dice

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