Dice Probability Calculator

Find the exact probability that a roll of several dice reaches a target sum. The calculator enumerates every possible outcome by dynamic programming — no simulation, no rounding shortcuts — and reports the chance as a decimal, a percentage, and "1 in X" odds, together with the count of favorable outcomes behind it.

How to Use This Calculator

  1. Enter how many dice you roll (1 to 10).
  2. Pick the die type: 4, 6, 8, 10, 12, or 20 faces, matching standard gaming dice.
  3. Enter the target sum and choose exactly, at least, or at most.
  4. Press Calculate to get the exact probability and the outcome counts it comes from.

Reading the Output

Probability and percentage: the same value on two scales, exact to the digits shown.

Odds: the "1 in X" form — X is total outcomes divided by favorable ones.

Favorable / total outcomes: the raw counts, where the total is always n^k ordered rolls.

How the Outcomes Are Counted

Each die is independent and each face equally likely, so a roll of k n-sided dice has n^k equally likely ordered outcomes. The calculator counts how many of them hit each possible sum by adding one die at a time — a convolution, computed by dynamic programming:

W₁(s) = 1 for s = 1 … n

Wₖ(s) = Wₖ₋₁(s − 1) + Wₖ₋₁(s − 2) + … + Wₖ₋₁(s − n)

P(sum = s) = Wₖ(s) / n^k

In words: to end on sum s after k dice, the first k − 1 dice must have summed to s minus whatever the last die shows, so the counts for k dice are built from the counts for k − 1 dice. Every intermediate value is an exact integer — with at most 10 dice of 20 faces the grand total is 20^10 = 10,240,000,000,000 outcomes, still comfortably below the 2^53 boundary where JavaScript integers stay exact.

The Shape of Dice-Sum Distributions

One die is flat: every face has probability 1/n. Two dice produce a triangle, rising by one count per step to a single peak and falling symmetrically:

Two d6 — ways per sum (2 through 12):

1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 (36 total)

From three dice on, the corners smooth out and the distribution turns bell-shaped — a small-scale preview of the central limit theorem, which says sums of independent random quantities tend toward a normal curve. With three d6 the peak splits across 10 and 11, each achievable in 27 of the 216 outcomes (probability 0.125). The symmetry is general: sums s and k(n + 1) − s always have identical counts, because replacing every face f with n + 1 − f pairs the outcomes up exactly.

Exactly, At Least, and At Most

The three comparison modes answer different questions from the same table of counts. "Exactly" reads one entry; the other two accumulate a tail:

P(sum ≥ s) = Σ P(sum = t) for t = s … k×n

P(sum ≤ s) = Σ P(sum = t) for t = k … s

P(sum ≥ s) + P(sum ≤ s − 1) = 1

Tail probabilities are what most game decisions need — "do I hit at least 10 to land this attack?" With two d6, rolling at least 10 succeeds in 3 + 2 + 1 = 6 of 36 outcomes (16.67%), while rolling at least 7 succeeds in 21 of 36 (58.33%). The complement identity is a handy shortcut: the chance of at least 5 is 1 minus the chance of at most 4.

From Counts to Probability and Odds

Because all n^k ordered rolls are equally likely, probability is simply favorable count divided by total count — the classical definition our probability calculator applies to generic events. The "1 in X" odds form inverts that fraction: X = total ÷ favorable. Rolling a 20 on one d20 is 1 in 20 (5%); rolling at least 17 is 4 in 20, or 1 in 5 (20%); snake eyes on two d6 is 1 in 36 (about 2.78%).

Note that the counting is over ordered outcomes: a 3 on the red die with a 4 on the blue die is a different outcome from 4-then-3. Treating (3, 4) and (4, 3) as one case is the classic mistake that makes dice math come out wrong — the 11 "unordered" sums of two d6 are not equally likely, but the 36 ordered pairs are.

Worked Example: Sum of 7 on Two Dice

What is the probability that two six-sided dice sum to exactly 7? Enter 2 dice, 6 faces, target 7, mode "exactly". By hand:

  1. Total outcomes: 6² = 36 ordered pairs, all equally likely.
  2. Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 pairs.
  3. Probability: 6 ÷ 36 = 1/6 = 0.166667, i.e. 16.6667%.
  4. Odds: 36 ÷ 6 = 6, so the odds are 1 in 6.

Seven is the unique most likely sum on two d6 — every other total has fewer ordered pairs, dropping one at a time down to a single pair each for 2 (1,1) and 12 (6,6), which is why 7 anchors the dice mechanics of so many board games. Switching the mode to "at least 7" adds the counts for 8 through 12 (5 + 4 + 3 + 2 + 1 = 15) to the 6 ways of rolling exactly 7, giving 21 of 36 outcomes = 0.583333.

Frequently Asked Questions

Why is 7 the most likely sum for two six-sided dice?

Because more ordered pairs reach 7 than any other total: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) make 6 of the 36 equally likely outcomes, a 16.67% chance. Totals further from 7 have fewer pairs, down to just one each for 2 and 12. In general the most likely sum of k n-sided dice sits at the center k(n+1)/2, or at the two whole numbers on either side when that center is not an integer, as with the 10-and-11 double peak of three d6.

Does the calculator simulate dice rolls?

No. It counts every possible outcome exactly using dynamic programming, building the sum distribution one die at a time. A simulation would carry sampling error that shrinks only slowly with more trials; exact enumeration gives the true probability every time, with no randomness involved.

What happens if my target sum is impossible?

The calculator still answers honestly. The sum of k n-sided dice always lies between k and k times n, so asking for exactly 13 on two d6 returns probability 0, and asking for at least 1 returns probability 1. A note under the results points out when the target lies outside the reachable range.

How are 'at least' and 'at most' related to 'exactly'?

They are running totals of the exact counts: at least s adds up every sum from s to the maximum, and at most s adds up every sum from the minimum to s. The two tails overlap only at s itself, which gives the useful identity P(at least s) = 1 - P(at most s-1). With two d6, at least 10 collects 3 + 2 + 1 = 6 outcomes of 36.

Are results still exact with 10 twenty-sided dice?

Yes. That worst case has 20^10 = 10,240,000,000,000 total outcomes, and every count in the calculation is an exact integer well below 2^53, the limit where JavaScript numbers stop representing integers exactly. No floating-point rounding enters until the final division that turns counts into a probability.