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Non-transitive dice
You will learn: Compute strict wins and ties, and distinguish win probability from expected score.
Start with: Events and probability · Expectation and variance
Four dice can form a cycle: A usually beats B, B usually beats C, C usually beats D, and D usually beats A. “Usually beats” means a strict win probability greater than one half in a single independent contest. It does not order dice the way ordinary numbers are ordered.
Use four fair six-sided dice with these labels: A has 0,0,4,4,4,4; B has six 3s; C has 2,2,2,2,6,6; D has 1,1,1,5,5,5. Repeated labels remain distinct physical faces. All six faces are equally likely, and the rolls of different players are independent.
Choose a matchup
Player A uses die A; player B uses die B. Independent fair faces; each player sums 1 roll(s). A tie remains a tie.
Filled left: exact. Open right: sample. Exact counts: 24 first-player wins, 0 ties, 12 second-player wins among 36 equally likely ordered roll outcomes.
| Outcome | Exact probability | Sample frequency |
|---|---|---|
| A wins | 0.6666667 | 0.5700000 |
| Tie | 0.0000000 | 0.0000000 |
| B wins | 0.3333333 | 0.4300000 |
If ties were rerolled until resolved, the first player's win probability would be 0.6666667. That is a different denominator from the strict-win matrix below.
Every matchup: P(row score > column score)
| Die | A | B | C | D | Mean total |
|---|---|---|---|---|---|
| A | 0.222222 | 0.666667 | 0.444444 | 0.333333 | 2.666667 |
| B | 0.333333 | 0.000000 | 0.666667 | 0.500000 | 3.000000 |
| C | 0.555556 | 0.333333 | 0.222222 | 0.666667 | 3.333333 |
| D | 0.666667 | 0.500000 | 0.333333 | 0.250000 | 3.000000 |
Start with player A using die A and player B using die B. Die A wins precisely when it shows 4, which occurs on four of six faces. Its win probability is 2/3. There are 24 winning ordered face pairs out of 36, with no ties.
Follow the cycle in the matrix. B beats C when C shows 2, also with probability 2/3. C beats D whenever it shows 6, or when it shows 2 and D shows 1. That gives 1/3 + (2/3)(1/2) = 2/3. D beats A whenever D shows 5, or when D shows 1 and A shows 0: 1/2 + (1/2)(1/3) = 2/3.
Thus each die has another die that beats it two times out of three. If someone publicly picks a die before their opponent chooses, the opponent can take its predecessor in this cycle. This is an advantage in win probability, not a guarantee on any individual roll.
Why the means do not settle the contest
The die means are 8/3, 3, 10/3, and 3 for A, B, C, and D. B has a larger mean than A, yet A wins their single-roll comparison more often. B’s wins are larger: when A rolls zero, B wins by three; when A rolls four, B loses by one.
A game’s payoff matters. If the payoff equals the numerical score difference, expected scores determine expected payoff. If the payoff is one for a strict win and zero otherwise, the whole distribution of comparisons matters. A higher mean is not a substitute for a win probability.
Sum two rolls
Set rolls per player to two. Each player now rolls their chosen die twice and compares totals. This enumerates 6⁴ = 1,296 equally likely ordered face outcomes. The original cycle is not a rule that survives every scoring change.
In particular, B always totals six. A totals eight only when both its rolls show four, with probability (2/3)² = 4/9. All other A totals are below six, so B now beats A with probability 5/9. The single-roll advantage has reversed. Inspect the other matrix entries rather than assuming every edge reverses.
Make a prediction
If both players choose die B, is each player's strict win probability one half?
Explore the answer
No. Both scores are identical, so every contest ties and both strict win probabilities are zero. Splitting a tie’s payoff equally would give expected payoff one half; rerolling these dice would never resolve the tie. These are different game rules.
Exact probabilities and sampled contests
The filled bars enumerate every outcome; the open bars show a seeded sample. Increasing the sample count retains earlier contests. Frequencies vary and need not equal the exact probabilities, especially in small samples. Neither the simulation nor the matrix silently discards ties. A separate figure gives the win probability conditional on a resolved contest, where defined.
Reference and next step
Hulko and Whitmeyer, A Game of Nontransitive Dice studies dice comparison games and their strategy assumptions. The four dice here are the classical Efron set; the face counts above establish their cycle directly. Continue to Penney’s game for a related advantage created by overlapping coin patterns.