Probability
Probability of Rolling 12 With Two Dice: All Ordered Outcomes
Two fair six-sided dice sum to 12 in 1 of 36 ordered outcomes, giving probability 1/36. See every successful pair and the assumptions.
Number JD Editorial Team · Published 2026-10-13
The answer under a fair-dice model
For two independent fair six-sided dice, the probability of a total of 12 is 1/36 = 1/36, or approximately 2.7778%. Each die has six equally likely faces, giving 6 × 6 = 36 equally likely ordered pairs.
This calculation assumes faces numbered 1 through 6 and independent results. Loaded dice, a different number of faces or a rule that rerolls certain outcomes creates a different model.
List the successful outcomes
The complete list is (6, 6). There are 1 successful ordered pairs.
The first coordinate records die A and the second records die B. For a nonmatching pair, reversing the coordinates represents another outcome. Matching pairs such as (3, 3) occur only once. Enumerating ordered pairs prevents inconsistent counting.
Why the eleven possible totals are not equally likely
Totals run from 2 to 12, but the paths to each total differ. A total of 7 has 6 ordered outcomes, giving 6/36. Compare that count with the 1 outcomes for 12 rather than assigning every total a probability of 1/11.
The sample space consists of pairs of faces, not just the labels of their sums. Whenever several equally likely outcomes collapse into one label, count how many outcomes belong to each label.
Expected counts are not a schedule
Across 360 independent rolls, the expected count of sum 12 is 360 × 1/36 = 10. Actual counts can differ. “Expected” describes the average across many imagined repetitions, not a promise for the next block of rolls.
If the last ten rolls did not total 12, the probability on the next roll remains 1/36 under the independent model. Previous misses do not create a debt that the dice must repay.
A related question: two successes in a row
The probability that two consecutive rolls both total 12 is (1/36) × (1/36) = 1/1296, approximately 0.0772%. Multiply because both events must happen and the rolls are independent.
The probability of not getting 12 on one roll is 1 − 1/36 = 35/36. Those two complementary probabilities add to 1. Always distinguish “both,” “at least one” and “exactly one” before selecting an operation.
Explore the examples
Check the mathematical definitions
For fractions, percentages and worked arithmetic, see OpenStax Prealgebra. For number definitions, see OpenStax on prime and composite numbers and Wolfram MathWorld on digital roots. Worked examples on this page are calculated for this guide.
Continue learning
Exactly 0 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 0 heads has probability 1/8 = 1/8. Compare one sequence with all possible arrangements.
Exactly 1 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 1 heads has probability 3/8 = 3/8. Compare one sequence with all possible arrangements.
Exactly 2 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 2 heads has probability 3/8 = 3/8. Compare one sequence with all possible arrangements.
Exactly 3 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 3 heads has probability 1/8 = 1/8. Compare one sequence with all possible arrangements.