What Is the Standard Deviation of a Dice Roll?

A fair die is a discrete random variable $X$ that takes each value $1, 2, \dots, 6$ with probability $\tfrac16$. Its standard deviation is a fixed property of that distribution, so there is one exact answer and nothing needs to be estimated from data.

Derivation for a six-sided die

Mean:

$$E[X] = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$

Mean of the squares:

$$E[X^2] = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$$

Variance, using $\operatorname{Var}(X) = E[X^2] - (E[X])^2$:

$$\operatorname{Var}(X) = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.9167$$

Standard deviation:

$$\sigma = \sqrt{\frac{35}{12}} \approx 1.7078$$

Formula for any fair die

For a die with faces $1, \dots, s$, the same steps (using $\sum k = \tfrac{s(s+1)}{2}$ and $\sum k^2 = \tfrac{s(s+1)(2s+1)}{6}$) give:

$$E[X] = \frac{s+1}{2}, \qquad \operatorname{Var}(X) = \frac{s^2 - 1}{12}$$

Die	Mean	Variance	Standard deviation
d4	2.5	1.25	1.118
d6	3.5	2.917	1.708
d8	4.5	5.25	2.291
d10	5.5	8.25	2.872
d12	6.5	11.917	3.452
d20	10.5	33.25	5.766

Rolling several dice

Sum of $n$ dice. Independent variances add, so

$$\operatorname{Var}(X_1 + \dots + X_n) = n \cdot \frac{35}{12}, \qquad \sigma_{\text{sum}} = \sqrt{\frac{35n}{12}}$$

For two dice: mean 7, variance $70/12 \approx 5.833$, standard deviation $\approx 2.415$. Doubling the number of dice does not double the spread. It grows by a factor of $\sqrt{2}$.

Average of $n$ dice. Dividing by $n$ divides the variance by $n^2$:

$$\sigma_{\bar{X}} = \frac{1.7078}{\sqrt{n}}$$

This is why averages of many rolls cluster tightly around 3.5, which is the law of large numbers in action.

Why not divide by n − 1?

The $n-1$ (“sample”) formula is for estimating a standard deviation from observed data. Here we know the exact distribution, so we compute the population value $\sqrt{35/12}$ directly. If you rolled a die 60 times and computed the sample standard deviation of your results, you’d get a number near 1.708 that varies from experiment to experiment.

Common mistakes

• Treating the six faces as a data set and using $n - 1 = 5$, which gives $\sqrt{17.5/5} \approx 1.871$. That estimates the spread of an unknown population from a sample, which is a different question.
• Adding standard deviations instead of variances for sums: $1.708 + 1.708 = 3.416$ is wrong. The correct value for two dice is $\sqrt{2} \times 1.708 \approx 2.415$.
• Assuming the sum of dice is uniform. It isn’t: two dice make a triangle peaking at 7, and many dice approach a normal curve. Explore it with the calculator above.