Independence, sums, and convolutions — what 2d6 = 1d6 + 1d6 actually means.

Independence in plain language

Two random variables X and Y are independent when knowing the value of X tells you nothing about the value of Y. Two physical dice rolled at the same time are independent: the first die can't see the second, can't communicate with it, can't bias it. The second die is going to come up uniform on the integers 1 through 6 whether the first one came up 1 or 6.

Formally, independence means

P(X = a and Y = b) = P(X = a) · P(Y = b)

for every pair of outcomes (a, b). The probability of a joint outcome is the product of the individual probabilities. That factoring is the entire mathematical content of independence.

Counter-examples (non-independent variables) come up in dice math whenever a mechanic looks at a die before deciding what to do next: 2d20kh1 (advantage — keep the higher of two) is the higher of the two rolls, which depends on both of them, so the kept value isn't independent of either single roll. Reroll-on-1 mechanics (2d6r1+5 for Great Weapon Fighting) similarly condition the second look on the first. Those mechanics need the full conditional structure, not just sums of independent dice — and the engine handles that, but the closed-form mean-and-variance shortcuts here only apply to the truly-independent case.

Sum of two dice = convolution

What is the distribution of X + Y when you know the distributions of X and Y separately and they are independent? You enumerate every joint outcome and group the ones that produce the same sum.

For 2d6 = 1d6 + 1d6, there are 6 × 6 = 36 joint outcomes, each with probability 1/36. Group by sum:

• Sum 2: only (1,1). One outcome. P = 1/36.
• Sum 3: (1,2) and (2,1). Two outcomes. P = 2/36 = 1/18.
• Sum 4: (1,3), (2,2), (3,1). Three outcomes. P = 3/36 = 1/12.
• … and so on, peaking at sum 7 with six outcomes.

The general rule, written out: the probability of the sum equalling k is

P(X + Y = k) = Σⱼ P(X = j) · P(Y = k − j)

summed over every j in the support of X. That is the definition of convolution of two probability mass functions.

Read the chart against the convolution: the bar at 7 is 6/36 because there are 6 ways to make 7. Tap any bar to see its exact rational and the number of joint outcomes that produce it.

What the engine actually does

For an expression like 2d6 + 1d4 + 3, the engine:

1. Builds the PMF of each base die (1d6 uniform on 6, 1d4 uniform on 4).
2. Convolves the PMF of 1d6 with itself to get 2d6.
3. Convolves 2d6 with the PMF of 1d4.
4. Shifts the resulting PMF right by 3 (the constant offset).

Every step is in exact rationals. There is no floating point in the convolution, so every probability you see on the site is exact, not an approximation that's been rounded to display.

Mean and variance under independent sums

The two operations you use most after computing a distribution:

Linearity of expectation (no independence needed): E[X + Y] = E[X] + E[Y]

Additivity of variance (independence needed): Var(X + Y) = Var(X) + Var(Y)

Linearity holds always. Additivity of variance holds only when the variables are independent. That asymmetry is the most common quiet error in back-of-napkin probability: people add variances on dependent rolls, and the answers come out wrong by exactly the covariance term they ignored.

For independent dice, you get a mechanical computation of mean and variance for any sum:

• E[NdM] = N · (M+1)/2
• Var(NdM) = N · (M² − 1)/12

So E[8d6] = 8 · 7/2 = 28 and Var(8d6) = 8 · 35/12 = 70/3 ≈ 23.33. The engine gives the same results from the convolved distribution, but these closed forms are how you read the answers off a character sheet without a calculator.

Try it yourself

↦ /strike/1d6 — base case ↦ /strike/2d6 — one convolution step ↦ /strike/3d6 — two convolution steps ↦ /strike/8d6 — seven convolutions, visibly bell-shaped ↦ /strike/1d8+1d6+3 — heterogeneous sum + constant