Dice pool success probability calculator.

Answer

With 6 six-sided dice where each die succeeds on a 5 or 6 (a 1 in 3 chance), you roll an average of 2.00 successes, and the chance of at least 2 successes is 64.88% (1 in 1.5).

Worked example

Six d6, a die succeeds on 5 or 6, so p = 2/6 = 1/3. The chance of no successes is (2/3)^6 = 8.78%. The chance of exactly one is 6 x (1/3) x (2/3)^5 = 26.34%. At least two is 1 - 8.78% - 26.34% = 64.88%.

How it works

Binomial model. Each die succeeds independently with probability p = (sides - target + 1) / sides. The number of successes in a pool of n dice then follows a binomial distribution: P(k) = C(n, k) x p^k x (1 - p)^(n - k).

At least. The chance of at least k successes is the sum of P(j) for j from k up to n. The mean is n x p.

What this covers. Plain success counting only. Rules that re-roll or explode dice, subtract failures, add automatic successes or let you keep a variable number of dice need a different calculation.

Sources and basis (opened 2026-10-08): Binomial distribution (Wikipedia) for the number of successes in n independent trials; arithmetic otherwise, checked against full enumeration in the test suite.

FAQ

How do I calculate the chance of at least one success on a dice pool?

Compute 1 minus the chance that every die fails: 1 - (1 - p)^n, where p is the chance one die succeeds. Enter 1 as the successes needed and the calculator does this.

What is p for a d10 that succeeds on 8 or higher?

Three faces (8, 9, 10) succeed out of ten, so p = 0.3.

How many successes should I expect on average?

Dice in the pool times p. Six d6 that succeed on 5 or 6 average 2 successes.

Why is the most likely result not always the average?

The most likely count sits next to the mean, but results spread out on both sides. In the example above the single most likely outcome, exactly 2 successes, is only 32.92%.

Does this handle exploding or re-rolled dice?

No. It assumes each die is rolled once and succeeds or fails independently.