What Are the Expected Value and Standard Deviation of Dice Rolls?

For a fair six-sided die, the possible outcomes are 1, 2, 3, 4, 5, and 6, each with probability $1/6$ . The expected value of one roll is 3.5, and its standard deviation is about 1.708. These numbers describe the probability distribution, not a promise about what any particular roll will be.

For several independent rolls, the answer depends on whether you want the sum or the average. The formulas below also apply to rolling several independent fair dice at once and adding their scores.

Expected value and standard deviation for one die

The expected value, or mean, is the probability-weighted average of all possible outcomes. If $X$ is the number shown on one fair die, then

\mathbb{E}[X] = \sum_{x=1}^{6} x\,P(X=x) = \frac{1+2+3+4+5+6}{6} = 3.5.

To find the standard deviation, first calculate the variance. A useful identity is

\operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2.

For this die,

\begin{aligned} \mathbb{E}[X^2] &= \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{91}{6},\\ \operatorname{Var}(X) &= \frac{91}{6} - (3.5)^2 = \frac{35}{12},\\ \operatorname{SD}(X) &= \sqrt{\frac{35}{12}} \approx 1.708. \end{aligned}

The standard deviation is the square root of the variance. It is measured in the same units as the die score; the variance is measured in squared units.

Sum or average of several rolls

Suppose $X_1,\ldots,X_n$ are independent rolls of fair six-sided dice, and let $S=X_1+\cdots+X_n$ be their sum. Expected values add, and for independent rolls the variances add:

\mathbb{E}[S]=3.5n, \qquad \operatorname{Var}(S)=\frac{35n}{12}, \qquad \operatorname{SD}(S)=\sqrt{\frac{35n}{12}}.

If instead you calculate the average roll $\bar X=S/n$ , its expected value remains 3.5. Dividing a random variable by $n$ divides its standard deviation by $n$ , so

\operatorname{SD}(\bar X)=\frac{1}{n}\sqrt{\frac{35n}{12}} =\sqrt{\frac{35}{12n}}.

The sum becomes more variable as you add rolls, while the average becomes less variable.

Worked example: 10 rolls

For the sum of 10 independent fair-die rolls, the expected total is

\mathbb{E}[S]=10(3.5)=35.

Its variance is $10(35/12)=175/6$ , so its standard deviation is

\operatorname{SD}(S)=\sqrt{\frac{175}{6}}\approx 5.401.

For the average of those same 10 rolls, the expected value is still 3.5, and the standard deviation is about $5.401/10=0.540$ . The sum and average describe different random quantities, which is why their standard deviations differ.

If the die is not fair

The result 3.5 assumes all six faces are equally likely. For a die with outcome probabilities $p_1,\ldots,p_6$ , where $p_i$ is the probability of rolling $i$ , use

\mathbb{E}[X]=\sum_{i=1}^{6} i p_i, \qquad \operatorname{SD}(X)=\sqrt{\sum_{i=1}^{6}i^2p_i-\left(\sum_{i=1}^{6}ip_i\right)^2}.

For independent rolls of such a die, multiply the mean by the number of rolls and multiply the variance by that number. If the rolls are dependent, variances do not necessarily add this way.

Common mistakes

Using 3.5 as a guaranteed result: It is the long-run expected value, not an outcome on a standard die. A single roll is always an integer from 1 to 6.
Confusing the sum with the average: For $n$ rolls, their means and standard deviations scale differently. The sum’s standard deviation grows by $\sqrt n$ ; the average’s shrinks by $1/\sqrt n$ .
Adding standard deviations: For independent rolls, add the variances first, then take the square root. Standard deviations themselves do not add.
Assuming every die is fair: If the face probabilities are unequal, use the probability-weighted formulas rather than the fair-die values.

References

A First Course in Probability, Sheldon Ross

At a glance