Dice Probability Calculator

Set the number of dice and sides, choose the question, and get the exact probability with the full distribution of sums.

6 for ordinary dice; 4, 8, 10, 12 or 20 for polyhedral dice.
The three sum questions use the target sum; the face question uses the face value.
Must lie between the smallest and largest possible sum.
Used only by the "at least one die shows a face" question.

Probability

16.6667%

Exact fraction

6/36 = 1/6

Favourable outcomes

6

Total outcomes

36

One chance in

6

Expected (average) sum

7

How it works

Every die is equally likely to land on any of its faces, so with n dice of s sides there are s^n equally likely outcomes. The probability of an event is the number of outcomes that produce it divided by s^n, which is why every answer here is an exact fraction before it is a percentage.

For the sum questions the calculator counts how many of those outcomes add up to each possible total. It does this one die at a time: starting from one die (one way to make each face), adding a die spreads every existing count across the next s sums. The table shows the result for every sum, with the running at-most and at-least totals, so you can read off any threshold without re-running the calculation.

For the face question it is quicker to count the complement. The chance that one die avoids the face is (s − 1)/s, so the chance that all n dice avoid it is ((s − 1)/s)^n, and at least one showing it is one minus that.

Formula

total outcomes        = s^n
P(sum = k)            = ways(k) / s^n     ways(k) by convolving n uniform dice
P(sum ≥ k)            = Σ ways(j) / s^n  for j = k … n·s
P(sum ≤ k)            = Σ ways(j) / s^n  for j = n … k
P(at least one face)  = 1 − ((s − 1) / s)^n
expected sum          = n (s + 1) / 2

Example

Two six-sided dice have 36 outcomes. Six of them add to 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), so P(sum = 7) = 6/36 = 1/6 = 16.6667%, or one chance in 6. The expected sum is 2 × 3.5 = 7.

Rolling four six-sided dice, the chance of at least one 6 is 1 − (5/6)^4 = 671/1296 = 51.7747%: slightly better than even, the result de Méré famously confirmed at the gaming table.

Assumptions and limitations

  • Dice are fair: every face is equally likely, and the dice do not influence one another.
  • All dice in a roll have the same number of sides. Mixed sets (a d6 and a d8 together) are not modelled.
  • Limited to 10 dice of up to 20 sides so that every count stays an exact integer and the table stays readable; the face formula works for any n and s if you need more.
  • The face question counts dice showing exactly that face, not "that face or higher". Use the sum questions with one die for a "roll k or higher" check.

Frequently asked questions

Why is 7 the most likely sum with two dice?

There are more ways to make 7 than any other total: six of the 36 outcomes. Sums near the middle can be formed from many pairs of faces; 2 and 12 can each be formed only one way. With more dice the distribution gets narrower around the expected sum n(s + 1)/2.

Is the chance of at least one 6 in four rolls really over 50%?

Yes, 671/1296 ≈ 51.8%. The common mistake is to add 1/6 four times and get 66.7%, but that double-counts rolls with more than one 6. Taking one minus the chance of no sixes at all, (5/6)^4, gives the right answer.