Dice Probability Calculator

Using the dice probability calculator

Three choices define the question. First pick the die: 4, 6, 8, 10, 12 or 20 sides. Second pick how many dice are rolled, from 1 to 100; this also works for one die rolled several times in a row. Third choose the outcome in the last row: a number and one of nine conditions.

• dice sum =, ≥, ≤: the total of all dice equals, reaches or stays under your number.
• at least one die =, ≥, ≤: one or more of the dice shows your number, or a number at or above or at or below it.
• all dice =, ≥, ≤: every single die meets the condition.

The result shows the probability as a decimal and in brackets as a percentage, and then the odds in the form "1 in N", which is simply 1 divided by the probability. The target must be a whole number inside the possible range: for a sum that is from the number of dice up to dice × sides, otherwise from 1 up to the number of sides.

Probabilities for two six-sided dice

Two fair dice have 6 × 6 = 36 equally likely ordered outcomes, listed by the pair (first die, second die). A total of 7 can be formed six ways: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1. A total of 2 can only be 1+1. Dividing the number of ways by 36 gives each probability:

The distribution is a triangle with its peak at 7, which is why 7 is the most common total. Cumulative questions add rows: a sum of 10 or more is (3 + 2 + 1) ÷ 36 = 1/6, or 16.67%, and the calculator gives 1 in 6 for it.

The formula for the sum of n dice

Counting pairs by hand does not scale to ten dice, so the number of ways to reach a total x with n dice of s sides is counted with inclusion-exclusion over the binomial coefficients:

ways(x) = Σ_k (−1)^k × C(n, k) × C(x − s×k − 1, n − 1), k = 0 … ⌊(x − n) ÷ s⌋

P(sum = x) = ways(x) ÷ sⁿ

Example: the chance that three six-sided dice add up to 10. For k = 0 the term is C(9, 2) = 36, for k = 1 it is −3 × C(3, 2) = −9, and k = 2 needs a negative count so the sum stops. That is 27 ways out of 6³ = 216, so P = 27/216 = 0.125 or 12.5%, which matches the calculator. Probabilities for "at least" or "at most" a total are sums of this expression over the totals included. See the combination calculator for the binomial coefficients.

At least one die and all dice

These two families need no sums, only the chance of one die meeting the target, raised to a power.

All dice meet a condition with probability (m ÷ s)ⁿ, where m is how many of the s faces satisfy it. All five dice showing a 6 is (1/6)⁵ = 1/7776, about 0.0129%. All dice at 15 or below on a twenty-sided die is (15/20)ⁿ.

At least one die meets it with probability 1 − (1 − m ÷ s)ⁿ, which is the complement of "none does". The classic example is at least one 6 in four rolls of one die: 1 − (5/6)⁴ = 0.5177, a little better than even. Rolling a d20 twice and wanting at least one natural 20 gives 1 − (19/20)² = 0.0975, or 9.75%.

at least one = 1 − (1 − m/s)^n all = (m/s)^n

For repeated yes-or-no trials with other success counts, such as exactly two sixes in ten rolls, use the binomial probability calculator.

Worked examples

Rolling three dice for an ability score of 15 or more. The totals 15, 16, 17 and 18 can be made 10, 6, 3 and 1 ways with three six-sided dice. That is 20 ways out of 216, so the probability is 20/216 ≈ 0.0926, or about 9.26%, or 1 in 10.8.

Ten dice reaching a total of 40 or more. The expected total is 10 × 3.5 = 35. Counting every total from 40 to 60 exactly gives a probability of about 0.2050, or 20.5%; you will see it as 1 in 4.9. A bell-curve estimate with a continuity correction lands near 0.20, which shows why the normal distribution is a good shortcut once many dice are involved. See the normal distribution calculator.

To actually roll dice instead of computing the odds, use the random dice roller.

Limits and things to watch

• Fair, independent dice. Every face is assumed equally likely and each die is unaffected by the others. Loaded dice or rerolls change the answer.
• Order does not matter for the sum. The probabilities count every ordered pair, so 1+6 and 6+1 are separate outcomes, as they should be.
• Big counts are exact. The ways to reach a total are counted with whole-number arithmetic of unlimited size, so 100 twenty-sided dice, with 20¹⁰⁰ (about 1.27 × 10¹³⁰) equally likely outcomes, give the true probability and not an overflow. The only rounding happens once, when the exact fraction is written out in decimals.
• Tiny probabilities. A value smaller than 5 × 10⁻¹⁵ is displayed as 0 in the probability, though the "1 in N" odds still carry the size of the number. Those odds are exact: below a quadrillion they are rounded to one decimal, and beyond that they are rounded to a whole number and every digit is shown.
• A normal shortcut for many dice. For a quick estimate with many dice, the total of n dice has mean n(s + 1) ÷ 2 and variance n(s² − 1) ÷ 12, and is close to a bell curve.