Nontransitive Dice, Explained
After reading this you will understand why "beats" can fail to be a ranking, how to build Efron's four dice, and how to compute the exact 2/3 win rate that lets the second player win against any choice the first player makes.
What nontransitive dice are
You expect "beats" to work like "taller than". If A is taller than B and B is taller than C, then A is taller than C. That property is called transitivity. Most comparisons obey it.
Dice do not have to. Efron's dice are four six-sided dice labeled A, B, C, D whose faces are chosen so that A beats B, B beats C, C beats D, and D beats A, each about two rolls out of three. There is no best die. Whichever die your opponent picks, one of the others beats it more often than not.
The hook: let your opponent pick first. Offer them any of the four dice, then pick the one that beats theirs. Over many rolls you win roughly 67 games out of 100. The trick works every time because the "beats" relation runs in a loop, not a line.
The standard face values
Efron's classic set uses these faces. Each die has six faces, and every value is repeated to force the win rates.
| Die | Faces | Sum | Mean |
|---|---|---|---|
| A | 4, 4, 4, 4, 0, 0 | 16 | 2.667 |
| B | 3, 3, 3, 3, 3, 3 | 18 | 3.000 |
| C | 6, 6, 2, 2, 2, 2 | 20 | 3.333 |
| D | 5, 5, 5, 1, 1, 1 | 18 | 3.000 |
Notice the means do not line up with the cycle. C has the highest mean (3.333) yet C loses to B. A has the lowest mean (2.667) yet A beats B. The average face value tells you almost nothing about who wins a head-to-head roll.
Ranking dice by their average face value is the mistake the design is built to punish. Two dice with equal means (B and D both average 3) still have a lopsided matchup: D beats B two times out of three. What matters is the face-by-face count, not the sum.
The formula and the intuition
To compare two dice you count, over all pairs of faces, how often the first die shows a higher number. Each die has six faces, so there are 6 \times 6 = 36 equally likely face pairs when neither die can tie.
Here x_i is face i of die X, y_j is face j of die Y, and the bracket [\, x_i \gt y_j \,] equals 1 when the first value is larger and 0 otherwise. You add up all 36 comparisons and divide by 36. Ties, when they can happen, are counted separately and split or replayed.
The intuition: a die wins by covering the opponent's common values with a slightly larger number, while accepting a rare large loss elsewhere. Efron's dice each trade one big weakness for many small wins. That is why the cycle can close on itself instead of forming a chain.
Reproducing the demo numbers
Counting A versus B, then the whole cycle
Start with A (faces 4,4,4,4,0,0) against B (all faces 3). Because B always shows 3, only A's face matters.
- A shows 4 on four faces. Every 4 beats a 3. That is 4 \times 6 = 24 winning pairs out of 36.
- A shows 0 on two faces. Every 0 loses to a 3. That is 2 \times 6 = 12 losing pairs.
- No ties are possible here, so P(A \gt B) = 24/36 = 0.6667.
Now B (all 3) against C (faces 6,6,2,2,2,2). Only C's face matters.
- C shows 6 on two faces. Those beat B's 3: 2 \times 6 = 12 pairs where C wins, so B loses.
- C shows 2 on four faces. Those lose to B's 3: 4 \times 6 = 24 pairs where B wins.
- So P(B \gt C) = 24/36 = 0.6667.
Next C against D (faces 5,5,5,1,1,1). Now both dice vary, so count all 36 pairs.
- C shows 6 (2 faces): beats every D face (5 or 1), so 2 \times 6 = 12 wins.
- C shows 2 (4 faces): beats D's 1 (3 faces) but loses to D's 5 (3 faces). That gives 4 \times 3 = 12 more wins.
- Total C wins: 12 + 12 = 24, so P(C \gt D) = 24/36 = 0.6667.
Finally D against A (faces 4,4,4,4,0,0).
- D shows 5 (3 faces): beats every A face (4 or 0), so 3 \times 6 = 18 wins.
- D shows 1 (3 faces): beats A's 0 (2 faces) but loses to A's 4 (4 faces). That adds 3 \times 2 = 6 wins.
- Total D wins: 18 + 6 = 24, so P(D \gt A) = 24/36 = 0.6667.
Every edge of the cycle is exactly 24/36, which is 2/3 \approx 0.6667. The loop closes: A beats B beats C beats D beats A.
Reading the round-robin matrix
The full picture is a 4-by-4 table where the cell in row X, column Y holds P(X \gt Y). Cells on the diagonal are blank because a die does not play itself.
| vs A | vs B | vs C | vs D | |
|---|---|---|---|---|
| A | 0.6667 | 0.4444 | 0.4444 | |
| B | 0.3333 | 0.6667 | 0.3333 | |
| C | 0.5556 | 0.3333 | 0.6667 | |
| D | 0.6667 | 0.6667 | 0.3333 |
Read across each row to see what a die beats and loses to. No row is all above 0.5, so no die dominates. Read down a column and you always find at least one entry above 0.5: the counter-die. That guarantee is the whole game. If your opponent shows you die C, you answer with B (P(B \gt C) = 0.6667). If they show D, you answer with C. The matrix is your cheat sheet.
When to use this, and when not
Use nontransitive dice as a clean example of a preference cycle. The same structure appears in voting (Condorcet cycles), in ecology (rock, paper, scissors among competing species), and in any situation where pairwise comparison hides a loop. If a single number seems to rank options but the pairwise contests disagree, suspect nontransitivity.
Do not use these dice to claim that one die is "stronger" in general. Strength is only defined per matchup. Also do not assume the 2/3 rate survives changes to the game. If both players roll several dice and sum them, the cycle can weaken, hold, or even reverse depending on the count. Efron's dice reverse direction when you roll two of each and compare sums, which is a genuinely surprising second layer.
There are stronger sets. Efron built a different four-dice set with faces that push the win rate up toward 2/3 exactly on every edge, and other designers have found three-dice cycles and sets that stay nontransitive even under summing. The 2/3 value here is not a universal limit, just the rate of this particular set.
Common mistakes
- Ranking by mean
- C's mean of
3.333is the highest, yet C loses to B. Averages do not predict head-to-head outcomes. - Assuming a best die
- People look for the winner. There is none. Every die loses two of its three matchups by rate, and wins the third by rate.
- Ignoring who chooses first
- The advantage belongs entirely to whoever picks second. Insist on choosing first and you hand the edge away.
- Treating one game as proof
- At
0.6667per game, the favored die still loses about 1 game in 3. Over 10 games it can lose 4 or 5 by chance. The edge only shows clearly over hundreds of rolls.
Related tools
If cycles and preference paradoxes interest you, the Monty Hall Simulator shows another result that fights intuition, settling on 2/3 for switching. To watch a running average settle onto its true value the way the win rate above does, try the Law of Large Numbers demo. For a trend that reverses when groups merge, a cousin of nontransitivity, see the Simpson's Paradox Visualizer. And to build intuition for how random samples estimate a fixed probability, the Monte Carlo Playground throws random points at the problem directly.
Frequently asked questions
Why does A beat B if A's average is lower?
A wins on 4 of its 6 faces (the four 4s beat B's constant 3) and loses on 2 (the two 0s). That is a 4:2 face ratio, a win rate of 0.6667. The two 0s drag A's mean down to 2.667 without hurting its win count against B, because a 0 and a large loss cost the same single game.
What is the best strategy playing these dice?
Always let your opponent choose first, then pick the die that beats theirs from the matrix: answer A with D, B with A, C with B, D with C. Each answer wins at 0.6667.
Does the trick still work if we roll twice and add?
Not the same way. For Efron's set, comparing the sum of two rolls of each die can shift or even reverse some matchups. Always fix the exact rules before betting, because the cycle depends on them.
How many rolls do I need to see the edge clearly?
At a true rate of 0.6667, roughly 1000 rolls pin the observed rate to within about \pm 0.03. Under 20 rolls the noise can swamp the edge entirely, so short sessions prove nothing.
Are there dice where three of them form a cycle?
Yes. Three-dice nontransitive sets exist, such as the Miwin dice, where die 1 beats die 2 beats die 3 beats die 1. Four dice simply make the closed cycle easy to display and easy to verify by hand.
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