There are four dice on the table with unusual numbers on their faces. Your opponent politely offers: "You choose first, and I'll take one of the ones left." Each of you rolls your own die, and the higher number wins. What is your chance if your opponent chooses wisely? Lock in your answer, then check it: play against the computer, count every pair and read the explanation.
There are four dice on the table with unusual numbers on their faces. It feels as if, choosing first, you take the best one and win at least half the rolls. But there is no best die here: die A beats B, B beats C, C beats D, and D beats A, each with chance 2/3. Whatever you take, your opponent finds a die that wins 2 rolls in 3.
The dice are ordinary, six-sided and fair: each face comes up with chance 1/6. Only the numbers on the faces are unusual. Look at them closely before you choose.
Die A has four 4s and two 0s. Die B has six 3s. Die C has two 6s and four 2s. Die D has three 5s and three 1s.
You take any die you like. Your opponent sees your choice and takes one of the three that are left.
Each of you rolls your own die, and whoever gets the higher number wins. These dice never tie: no two dice share a number.
Your opponent knows these dice as well as you do and chooses wisely. What is your chance of winning a roll?
"I choose first, so I'll take the best die and win at least half the rolls"
In fact there is no best die. Each one loses to some other die with chance 2/3, and your opponent always takes that one. Your chance is 1/3, whatever you choose.
You choose a die as well as you can, and your opponent replies as well as they can. What share of the rolls do you win? Pick a zone with a button under the slider or type a number.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Start with die B: it always shows 3. Which dice roll more than 3 more often than less?
For each die in turn, look for another that beats it in more than half the rolls. Is there a die with no such rival?
Work out D against A: how often does D show a 5, and what happens when D shows a 1?
The next rounds show your chance: a game against the computer, every pair of dice and the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
Pick a die and the computer takes its own. Roll once at a time, or 10 or 100 rolls at once. Try all four dice. Then swap roles: let the computer choose first, and you reply.
Each row is your die, each column your opponent's. A cell shows the chance that the opponent's die rolls higher: the exact fraction and the simulated share. The darker the cell, the worse for you. The outline marks your opponent's best reply to your choice.
Your chance in each row is 1 minus the number in the cell.
Let the dice play each other like teams in a league: a match is won if a die wins more than half the rolls, 3 points for a win and 1 for a draw (W–D–L: won, drawn, lost). The table is exact, from the same fractions.
The league champion, C, has the most wins and the highest average face, yet loses to B, which never rolls anything but a 3. The league table puts the dice in order, but every one of them is beaten by someone.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// Efron's four dice. You choose a die first, your opponent takes one of the rest, and the higher roll wins. What is your chance if your opponent picks the best reply?
const DICE = { A: [4, 4, 4, 4, 0, 0], B: [3, 3, 3, 3, 3, 3], C: [6, 6, 2, 2, 2, 2], D: [5, 5, 5, 1, 1, 1] };
const N = 100000; // rolls per pair
const roll = d => d[Math.floor(Math.random() * 6)];
const names = Object.keys(DICE), win = {};
for (const x of names) for (const y of names) if (x !== y) {
let w = 0;
for (let i = 0; i < N; i++) if (roll(DICE[x]) > roll(DICE[y])) w++; // these dice never tie
win[x + y] = w / N; // share of rolls where x is higher than y
}
for (const me of names) {
// the opponent takes the die that beats yours most often
const reply = names.filter(y => y !== me).reduce((a, b) => win[b + me] > win[a + me] ? b : a);
console.log('You', me, '· opponent', reply, '· your chance', (1 - win[reply + me]).toFixed(3), 'exact 1/3 = 0.333');
}
# Efron's four dice. You choose a die first, your opponent takes one of the rest, and the higher roll wins. What is your chance if your opponent picks the best reply?
import random
DICE = {"A": [4, 4, 4, 4, 0, 0], "B": [3, 3, 3, 3, 3, 3], "C": [6, 6, 2, 2, 2, 2], "D": [5, 5, 5, 1, 1, 1]}
N = 100_000 # rolls per pair
win = {}
for x in DICE:
for y in DICE:
if x != y:
# these dice never tie
w = sum(random.choice(DICE[x]) > random.choice(DICE[y]) for _ in range(N))
win[x + y] = w / N # share of rolls where x is higher than y
for me in DICE:
# the opponent takes the die that beats yours most often
reply = max((y for y in DICE if y != me), key=lambda y: win[y + me])
print("You", me, "· opponent", reply, "· your chance", f"{1 - win[reply + me]:.3f}", "exact 1/3 = 0.333")
Intuition assumes the dice can be lined up from best to worst, and whoever chooses first just takes the top one. But "wins more often" is not a number you can attach to each die; it is a relation between two dice. And such a relation can close into a circle.
“In plain words” is hidden on all puzzles.
Each arrow points from the winner to the loser. Die A beats B, B beats C, C beats D, and D beats A, 2 rolls in 3 each time. As in rock, paper, scissors, there is no best gesture, only a reply to your opponent's. So whoever chooses second always takes the die that comes before yours in the circle.
This card shows what happens but not yet why it is possible. The next card works out each arrow by hand.
A vs B · 2/3B always shows 3, so A wins when it shows 4: 4 faces out of 6.B vs C · 2/3A 3 loses only to a 6, and C has two 6s out of six faces. The other 4 times in 6 it shows 2, and the 3 wins.C vs D · 2/3A 6 on C (chance 1/3) beats everything. A 2 (chance 2/3) beats only a 1 on D, half the time: 2/3 × 1/2 = 1/3. Together 1/3 + 1/3 = 2/3.D vs A · 2/3A 5 on D (half the time) beats anything A rolls. A 1 (the other half) beats only a 0, and a third of A's faces are 0: 1/2 × 1/3 = 1/6. Together 1/2 + 1/6 = 2/3.Each die is built to beat one neighbour and pays for it with weakness against the other. B holds a steady 3: enough against C's rare 6s, too little against A's frequent 4s.
| Die | Faces | Average | Beats with chance 2/3 | Loses with chance 2/3 to |
|---|---|---|---|---|
A | 4 4 4 4 0 0 | 2.67 | B | D |
B | 3 3 3 3 3 3 | 3 | C | A |
C | 6 6 2 2 2 2 | 3.33 | D | B |
D | 5 5 5 1 1 1 | 3 | A | C |
The average measures how many points a die collects over many rolls. But the game asks something else: whose number is higher in a single roll. C's big 6 adds a lot to its average, yet wins a duel just as a 4 would: a win is a win, by 1 or by 4. So the averages can be lined up (C, then B and D, then A), but the duels cannot.
“Rigorous” is hidden on all puzzles.
Two dice together give 6 × 6 = 36 equally likely pairs of faces. The row is the first die's face, the column the second's, and the cell's colour shows who won. On the four arrows of the cycle the winner takes 24 cells out of 36, that is 2/3. Across the circle, in the pairs A–C and B–D, the edge is smaller or gone: 20 of 36 and 18 of 36.
For each of your dice the opponent compares three options and takes the one that wins most often. Against A the best is D (2/3), since C gives only 5/9 and B 1/3. Against B the best is A (2/3). Against C the best is B (2/3). Against D the best is C (2/3).
The best reply gives exactly 2/3 every time, so your choice changes nothing: your chance is 1/3 with any die. If you chose at random and your opponent also picked at random from the rest, you would win exactly half: every pair would come up both ways round.
The first die wins if it shows x and the second shows less than x.
The four links of the cycle:
Your chance when the opponent takes the best reply:
“Deeper” is hidden on all puzzles.
A relation is transitive if, whenever A is better than B and B is better than C, A is also better than C. For numbers this always holds: 5 > 3 and 3 > 2, so 5 > 2. Intuition carries this over to dice, as if each die were a single "strength" number. But a die is six numbers, and "rolls higher more often" compares all 36 pairs of faces. Such a comparison does not reduce to one number, and nothing guarantees transitivity.
The shortest example hides in the 3 × 3 magic square. Take its rows {4, 9, 2}, {3, 5, 7} and {8, 1, 6} as the faces of three dice (each number twice). Then {8, 1, 6} beats {3, 5, 7}, {3, 5, 7} beats {4, 9, 2}, and {4, 9, 2} beats {8, 1, 6}, each with chance 5/9. It is the same as the three voters of the Condorcet paradox: every option loses to a majority, and there is no winner.
When for every pair we know which is stronger, mathematicians call it a tournament: a graph with exactly one arrow between any two vertices. A tournament can be turned into a ranking only if it has no cycles. Efron's dice give a cycle of four, and no ordering agrees with every duel: every way of lining up A, B, C, D contradicts at least one arrow.
The Marquis de Condorcet spotted such cycles as early as 1785, in elections where candidates are compared in pairs. The coin version of the same idea is Penney's game: for any triple of heads and tails there is a triple that comes up sooner, and the second player again has the edge. And Parrondo's paradox shows another side of the same trap: intuition about "worse" and "better" breaks when games are combined.
A cycle can flip when each player rolls two dice. The edge in a cycle has a ceiling. And rock, paper, scissors is played by elections, bacteria and lizards alike.
Rolling two identical dice and adding them means playing with a new 36-faced "die". Its chances depend not only on how often a die wins, but also on how spread out its numbers are. So the cycle can break, or even reverse.
Efron's dice rolled twice no longer form a circle: C wins more often than it loses against each of the other three. James Grime's five dice are designed so that one of their two cycles flips completely with two dice. The player who chooses second also decides whether to play one die or two, and wins again.
How strong can each link of a cycle be? For three independent random variables (dice are a special case) the smallest chance around the circle cannot exceed (√5 − 1)/2 ≈ 0.618, the inverse golden ratio (Stanisław Trybuła, 1961). In 1964 Zalman Usiskin showed how to find this ceiling for any number of them: for four it is exactly 2/3, so Efron's dice are the best possible, and as the number grows the ceiling approaches 3/4.
The same magic square gives an election too. Let its rows 4 9 2, 3 5 7 and 8 1 6 be the scores of three candidates, and each column one voter's ballot: a higher score means a higher place. Check it yourself: in every pair of candidates, two voters out of three prefer one and only one prefers the other.
| Voter | 1st choice | 2nd | 3rd |
|---|---|---|---|
| 1 (column 4 3 8) | Mark | Iris | Oscar |
| 2 (column 9 5 1) | Iris | Oscar | Mark |
| 3 (column 2 7 6) | Oscar | Mark | Iris |
Iris (row 4 9 2) beats Oscar (3 5 7) 2 : 1 (voters 1 and 2), Oscar beats Mark (8 1 6) 2 : 1 (voters 2 and 3), and Mark beats Iris 2 : 1 (voters 1 and 3). Whoever is declared the winner, a majority would prefer someone else. The same goes for sport: a team that beat the leader can lose to the bottom side, and a points table only hides the cycle. And in A/B tests where variants are compared in pairs, "B beats A" and "C beats B" do not yet mean that C beats A: that has to be tested on its own. More on elections in the Condorcet paradox.
A cycle with no strongest player lets nobody win for good. That is why in nature such cycles keep diversity alive: each species or strategy survives because it has its own "prey".
| Where | The cycle | What follows |
|---|---|---|
| E. coli bacteria: Kerr et al., Nature, 2002 | A strain that makes the poison colicin kills a sensitive strain. The sensitive strain outgrows a resistant one. The resistant strain outgrows the producer, since it spends nothing on poison. | On a static Petri dish, where neighbours meet only locally, all three strains survived. In a constantly shaken flask, only one was left. |
| Side-blotched lizards, Uta stansburiana: Sinervo and Lively, Nature, 1996 | Orange-throated males with large territories beat blue-throated males guarding one female. Blue-throated males beat yellow-throated "sneakers". Yellow-throated males beat orange ones by slipping into their large territories. | The shares of the three kinds of males cycle over the years: as soon as one kind becomes common, the one that beats it thrives. |
| Tournaments and ratings | Player A usually beats B, B beats C, and C beats A: playing styles that catch one another. | A single rating number always puts players in order, so a cycle cannot show in it. A forecast for a specific pair is better built on their head-to-head record. |
The wider world learned of these dice from the "Mathematical Games" column in the December 1970 Scientific American: "The paradox of the nontransitive dice and the elusive principle of indifference". Gardner described the dice of the Stanford statistician Bradley Efron there.
The story often told is that Warren Buffett offered Bill Gates a game with such dice and politely let him choose first. Gates looked the dice over, grew suspicious and insisted that Buffett choose first. Scientific American, among others, retells it (2023).
In 1964 Zalman Usiskin proved that in a cycle of four independent random variables the weakest link cannot be stronger than 2/3. Efron's dice have exactly 2/3 on every link, so no four dice can make the trap for the first player any deeper.
For three dice the ceiling of the edge is the inverse golden ratio, (√5 − 1)/2. Stanisław Trybuła proved it in 1961. Dice with whole-number faces can only approach this irrational number: the magic square gives 5/9 ≈ 0.556.
The mathematician James Grime devised five dice with two different cycles of wins: one follows the alphabetical order of the colour names, the other their length. If each player rolls two dice, the length cycle flips while the alphabetical one holds, with one exception: the arrow between red and olive.
In 1975 the physicist Michael Winkelmann made three dice with faces 1, 2, 5, 6, 7, 9; 1, 3, 4, 5, 8, 9; and 2, 3, 4, 6, 7, 8. Each has the same face total, 30, and an average of 5. Yet each beats the next one around the circle: 17 rolls in 36 against 16, with the other 3 tied.
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