SOLVETUTORMATH SOLVER

Instrument MI-07-059 · Statistics

Dice Probability Calculator

The classic 2d6 sum-of-7 odds are only one special case. Enter any number of dice, any number of sides per die, and any target sum, and this returns the exact probability for that specific roll.

Instrument MI-07-059
Sheet 1 OF 1
Rev A
Verified
Type 07 — Probability SER. 2026-07059

P(sum = target)

0.16666667

P(sum = target) = (ways to reach target with n dice) / sides^n

6 Number of ways to reach the target sum
36 Total possible outcomes
The working Every figure verified twice
  1. n=2, sides=6, target=7 -> 6/36 = 0.16666667
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Rolling n dice with s sides each and asking for the probability of a specific total sum is a classic probability question with a genuinely non-trivial answer once n grows beyond 1 or 2 — there's no simple closed-form shortcut the way there is for a single die. The number of ways to reach a given sum with multiple dice has to be counted directly, and this calculator does that counting with a dynamic-programming convolution: it builds up, one die at a time, the full distribution of possible sums, tracking exactly how many combinations of individual die results produce each possible total.

The classic 2d6 case is the most famous instance of this question, and it's why 7 is the most commonly rolled sum in games like craps: there are 6 distinct ways to roll a 7 with two six-sided dice (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) out of 36 equally likely combinations, giving P=6/36=1/6, noticeably more likely than any other single sum. This calculator generalizes that exact same logic to any number of dice, any number of sides per die, and any target sum — 3d6 summing to 10, 4d8 summing to 20, or any other combination you can think up.

Every individual outcome — each specific sequence of results across all the dice — is equally likely, at (1/sides)ⁿ each; what varies is how many of those equally likely sequences add up to your specific target sum. Sums near the middle of the possible range (n to n×sides) have far more ways to be reached than sums near either extreme, which is exactly why the middle sums dominate the probability distribution and the extremes — all-ones or all-max-value — are always the rarest possible outcomes, each with only a single way to occur.

P(sum=t)=ways(t)sidesnP(\text{sum}=t) = \frac{\text{ways}(t)}{\text{sides}^{\,n}}
n (diceCount) — number of dice rolled · sides — number of sides on each die · target — the specific sum being asked about · waysCount — the number of distinct dice-result combinations that add up to target, found by a dynamic-programming convolution across all n dice · totalOutcomes — sidesⁿ, every equally likely combination of results.
  • Enter how many dice are being rolled into Number of dice.
  • Enter how many sides each die has into Sides per die (6 for a standard die).
  • Enter the total you want the probability for into Target sum.
  • Read P(sum = target) for the exact probability, Number of ways to reach the target sum for the raw combinatorial count, and Total possible outcomes for the denominator that count is drawn from.

Worked example — the classic two-dice sum of 7

Entering Number of dice = 2, Sides per die = 6, and Target sum = 7 — the single most famous dice-probability question there is — Number of ways to reach the target sum reads 6 (the pairs 1+6, 2+5, 3+4, 4+3, 5+2, and 6+1), and Total possible outcomes reads 36 (every one of the 6×6 equally likely combinations of two six-sided dice).

P(sum = target) then reads 0.16666667, exactly 6/36 = 1/6 — noticeably higher than the probability of any other single sum with two six-sided dice, which is precisely why 7 is the most commonly rolled total in games built around two standard dice, from craps to countless board games.

Questions

Why is 7 the most likely sum with two standard dice?

Because more combinations of two six-sided dice add up to 7 than to any other possible sum. The possible totals range from 2 (only 1+1) up to 12 (only 6+6), and the number of ways to reach each total increases steadily toward the middle of that range and back down symmetrically — 7 sits exactly in the middle, with 6 different combinations reaching it, more than any sum on either side.

Does this work for dice with more or fewer than 6 sides?

Yes — Sides per die accepts any number of sides from 2 upward, covering coins framed as 2-sided 'dice,' standard 6-sided dice, and less common shapes like 4-sided, 8-sided, 10-sided, 12-sided, or 20-sided dice from tabletop games. The same dynamic-programming convolution handles any of these the same way, just building up the sum distribution across however many sides each die actually has.

How does the calculation work with more than 2 dice, where there's no simple table to look up?

It builds up the answer one die at a time rather than trying to enumerate every possible roll directly, which would get unmanageably large fast (6 dice already has over 46,000 possible outcomes). The instrument tracks, after each additional die, how many ways exist to reach every possible running total so far, then combines that with the next die's possible results — a standard technique called convolution — until all the dice have been accounted for, at which point the count for your specific target sum is read straight off.

What's the probability of the lowest or highest possible sum?

Always the smallest possible probability in the whole distribution, and always exactly 1/sidesⁿ — there's only ever one way to roll the minimum sum (every single die shows its lowest face) and one way to roll the maximum (every die shows its highest face). For 2 standard six-sided dice, that means both a sum of 2 and a sum of 12 have probability exactly 1/36, six times less likely than the sum of 7.

Can the target sum be impossible for the dice entered?

Yes — any target below the minimum possible sum (equal to the number of dice, since each die shows at least 1) or above the maximum possible sum (the number of dice times the number of sides) simply has zero ways to reach it, so P(sum = target) comes out to exactly 0. Entering a target of 25 with 3 standard six-sided dice, for instance, is asking for a sum above the maximum possible 18, so it will always return zero.

References