Dice Probability Calculator

What is a dice probability calculator?

A dice probability calculator works out how likely a particular total is when you roll several dice at once and add the faces together. Rolling a single die is easy — every face is equally likely, so a six-sided die shows any given number one time in six. Adding dice together is not: totals in the middle of the range can be reached in many different ways, while the extremes can be reached in only one. That is why 7 is the most common result on two six-sided dice and 2 is the rarest.

The calculator answers questions such as “What is the chance of rolling a total of exactly 7 with two dice?” or “What is the chance of rolling 10 or more?” for dice with 4, 6, 8, 10, 12, 20, or 100 faces, and for anywhere from one to twelve dice at a time.

Why the sum of dice is not a coin-flip problem

Counting heads in a series of coin tosses is a binomial problem: each toss has two outcomes, and every head contributes exactly one to the count. Dice sums are different. Each die contributes anything from 1 to $s$, so the distribution of the total is built by convolution — repeatedly combining the distribution you already have with the flat 1-to-$s$ distribution of one more die.

The result is a hump-shaped distribution that gets smoother and narrower (relative to its range) as you add dice, but it is not binomial and no single binomial coefficient describes it. This is why a coin-flip formula gives the wrong answer for dice totals, and why the calculator builds the whole distribution instead.

How does the calculator work?

You provide four inputs:

• Number of dice ($n$) — how many dice are rolled and added together, from 1 to 12.
• Sides per die ($s$) — the number of faces on each die: 4, 6, 8, 10, 12, 20, or 100. All dice are assumed identical and fair.
• Condition on the sum — whether you want the total to be exactly, at least, at most, more than, or less than your target.
• Target sum ($t$) — the total you are interested in.

The calculator then builds the full distribution of possible totals, counts the outcomes that satisfy your condition, and reports:

• the probability as a percentage,
• the probability as a decimal between 0 and 1,
• the odds expressed as “one in …”,
• the number of favorable outcomes (the ways to satisfy the condition),
• the number of total outcomes ($s^{n}$).

A target outside the possible range is not an error — it simply has probability 0. The smallest total you can roll is $n$ (every die shows 1) and the largest is $n \times s$ (every die shows its top face), so asking for 13 on two six-sided dice correctly returns 0.

Formulas

Let $N(n, s, t)$ be the number of ways to roll a total of $t$ with $n$ dice of $s$ sides. Since every one of the $s^{n}$ ordered outcomes is equally likely, the probability of an exact total is

The counts themselves come from a convolution. Starting from $N(0, s, 0) = 1$, adding one more die spreads each existing count across the $s$ faces:

The cumulative cases add up the individual counts:

The strict comparisons are the same sums shifted by one, because a total is always a whole number:

There is also a closed form for the count, obtained by inclusion-exclusion on the dice that “overflow” past $s$:

Finally, the odds are the reciprocal of the probability:

Worked examples

1. Exactly 7 on two six-sided dice. With $n = 2$, $s = 6$, $t = 7$, the total 7 can be made as 1+6, 2+5, 3+4, 4+3, 5+2 and 6+1 — 6 favorable outcomes out of $6^{2} = 36$. So $P = 6/36 \approx 0.166667$, or about 16.6667%, which is odds of one in 6. This is the most likely total on two dice.

2. Exactly 2 on two six-sided dice. Only 1+1 works, so there is 1 favorable outcome out of 36: $P = 1/36 \approx 0.027778$, about 2.7778%, or one in 36.

3. Exactly 10 on three six-sided dice. With $n = 3$, $s = 6$, $t = 10$ there are 27 favorable outcomes out of $6^{3} = 216$, so $P = 27/216 = 0.125$ — exactly 12.5%, or one in 8. Together with 11, this is the peak of the three-dice distribution.

4. At least 10 on two six-sided dice. The qualifying totals are 10, 11 and 12, reachable in 3, 2 and 1 ways respectively — 6 outcomes out of 36, so $P = 6/36 \approx 16.6667\%$.

5. More than 10 on two six-sided dice. “More than” excludes 10 itself, leaving totals 11 and 12: $2 + 1 = 3$ outcomes out of 36, so $P = 3/36 \approx 8.3333\%$. The mirror case, “less than 4”, leaves totals 2 and 3 and gives the same $3/36$.

6. Exactly 13 on two six-sided dice. The largest possible total is $2 \times 6 = 12$, so there are 0 favorable outcomes and $P = 0$.

7. Exactly 10 on four four-sided dice. With $n = 4$, $s = 4$, $t = 10$ there are 44 favorable outcomes out of $4^{4} = 256$, so $P = 44/256 \approx 0.171875$, about 17.1875%.

Practical notes

• The distribution of a dice total is always symmetric around $n(s+1)/2$. On two six-sided dice that centre is 7; on three it is 10.5, which is why 10 and 11 tie for the most likely total.
• Asking for “at least” the minimum total $n$, or “at most” the maximum total $n \times s$, always returns a probability of 1, because every roll satisfies the condition.
• Odds are only shown when the probability is greater than zero — an impossible total has no meaningful “one in” figure.
• The counts of favorable and total outcomes are shown only while $s^{n}$ is small enough to be an exact whole number. With twelve hundred-sided dice, $100^{12} = 10^{24}$ outcomes exceeds that limit, so the counts are hidden while the probabilities, which are ratios, remain accurate.
• Every calculation assumes fair, independent dice. Loaded dice, rerolls, “drop the lowest” rules and exploding dice all change the distribution and are not covered here.
• For a two-outcome experiment repeated many times, the coin flip probability calculator uses the binomial model instead. To combine the chances of independent events, see the probability calculator, and to count selections rather than sums, see the combinations calculator.