Every roll is favourable outcomes over total outcomes
A fair die with s sides has each face equally likely. Roll n of them and there are sn equally likely results — 6×6 = 36 for two ordinary dice. The probability of any total is just how many of those results reach it, divided by the whole count.
Totals bunch in the middle
The extremes need one exact pairing each; the middle totals can be reached many ways. That is why 7 sits at the peak for two dice and the edges are rare.
Two six-sided dice: the chance climbs steadily to a peak at 7, then falls away symmetrically. Add more dice and the shape tightens into a bell curve.
Two six-sided dice — every total
| Total | Ways | Probability |
|---|---|---|
| 2 | 1 | 1/36 ≈ 2.8% |
| 3 | 2 | 2/36 ≈ 5.6% |
| 4 | 3 | 3/36 ≈ 8.3% |
| 5 | 4 | 4/36 ≈ 11.1% |
| 6 | 5 | 5/36 ≈ 13.9% |
| 7 | 6 | 6/36 ≈ 16.7% |
| 8 | 5 | 5/36 ≈ 13.9% |
| 9 | 4 | 4/36 ≈ 11.1% |
| 10 | 3 | 3/36 ≈ 8.3% |
| 11 | 2 | 2/36 ≈ 5.6% |
| 12 | 1 | 1/36 ≈ 2.8% |
"At least" and "at most" just sum the rows from that total outward — for example "at most 4" is 1 + 2 + 3 = 6/36.
Averages and spread by die
- Expected value. One fair die averages (s + 1) ÷ 2 — that is 3.5 for a d6. For n dice the mean total is n times that, so two d6 average 7 and three d6 average 10.5.
- Spread. A d20 (standard deviation about 5.77) scatters far more than a d6 (about 1.71), which is why a d20 roll feels so much swingier.
- More dice, tighter curve. Sums of many dice pile up near the mean, so extreme totals get rare fast — the central limit theorem at work.
Common questions
What is the probability of rolling a 7 with two dice?
Six of the 36 equally likely outcomes add up to 7, so the chance is 6 in 36, which is 1 in 6, or about 16.7%. Seven is the single most likely total on two six-sided dice.
Why is 7 more likely than 2 or 12?
A total of 2 needs both dice to show 1, and 12 needs both to show 6 — one combination each. A 7 can happen six different ways (1-6, 2-5, 3-4 and their reverses), so it comes up six times as often.
How do I find the chance of rolling at least a certain total?
Add up the outcomes for that total and every higher one, then divide by the total number of outcomes. For "at least 10" on two dice: 3 + 2 + 1 = 6 outcomes out of 36, which is 1 in 6.
