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Foundational

Non-Transitive Dice

A set of dice where A beats B, B beats C, and C beats A — a concrete counterexample to the intuition that 'beats' must behave like a normal ordering, and a favorite interview trap for testing whether you assume transitivity.

If die A tends to beat die B, and die B tends to beat die C, surely die A tends to beat die C — that's just how "bigger than" works for numbers. It's not how it works for dice. Non-transitive dice are a set where "beats more than half the time" forms a cycle instead of an ordering: A beats B, B beats C, and C beats A, all with probability greater than 1/21/2. There is no "best" die in the set. This is a favorite interview question precisely because it breaks an assumption people don't realize they're making.

A concrete set of three dice

One standard example uses these face values:

  • Die A: 2,2,4,4,9,92, 2, 4, 4, 9, 9
  • Die B: 1,1,6,6,8,81, 1, 6, 6, 8, 8
  • Die C: 3,3,5,5,7,73, 3, 5, 5, 7, 7

Each die has faces averaging around 5, so none is "bigger" in the usual sense — the trick is entirely in how the values are distributed relative to each other.

Worked example: checking A vs. B

Roll A and B and compare, face by face, all 36 equally likely outcomes. A beats B whenever A's face exceeds B's:

  • A = 2: beats B = 1 (2 of B's 6 faces are 1s: both). Loses to B = 6, 6, 8, 8 (4 faces). So A=2 beats B on 2 of 6 rolls of B.
  • A = 4: beats B = 1, 1 (2 faces), loses to 6, 6, 8, 8 (4 faces). Same: 2 of 6.
  • A = 9: beats every face of B (1, 1, 6, 6, 8, 8 — all less than 9). 6 of 6.

Each face of A occurs with probability 1/31/3 (two of six faces share each value), so

P(A>B)=13(26)+13(26)+13(66)=13(2+2+66)=101855.6%.P(A > B) = \frac{1}{3}\left(\frac{2}{6}\right) + \frac{1}{3}\left(\frac{2}{6}\right) + \frac{1}{3}\left(\frac{6}{6}\right) = \frac{1}{3}\left(\frac{2+2+6}{6}\right) = \frac{10}{18} \approx 55.6\% .

So A beats B more often than not. By the same style of calculation (left as the natural next check), B beats C with the same 10/1810/18 probability, and — the surprising closer — C beats A with 10/1810/18 as well. The cycle is complete: A > B > C > A, each with the same edge, and no die is uniformly best.

A B C Each arrow: "beats" with probability 10/18
A beats B, B beats C, and C beats A — a genuine cycle with no die on top, unlike a normal numeric ordering.

Why this matters beyond the puzzle

The interview point isn't the specific numbers — it's recognizing that pairwise comparison relations don't have to be transitive, and assuming they are (because "greater than" for real numbers is) is a silent, unjustified assumption. This shows up in real settings: voting systems can produce cyclic majority preferences (Condorcet's paradox — a majority prefers A to B, B to C, and C to A), and in trading, a "which strategy is better" comparison based on head-to-head backtests over different time windows can similarly fail to produce a consistent ranking if the comparisons aren't all measured on the same footing.

"Beats more than half the time" is not guaranteed to be a transitive relation — a set of dice can form a cycle (A beats B, B beats C, C beats A) with no single best die, even though each die's own face values look unremarkable in isolation. The mechanism is that each die is tuned to beat one specific opponent's distribution, not to be "better" in any absolute sense.

Don't assume that because you've shown A beats B and B beats C, you've shown A beats C — that step requires transitivity, and transitivity is exactly the property non-transitive dice are constructed to violate. Any time a comparison is defined via a match-up probability rather than a direct numeric ordering, check transitivity explicitly rather than assuming it.

Related concepts

Practice in interviews

Further reading

  • Gardner, The Colossal Book of Mathematics, ch. 27
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