# Expected sum if a fair die is rolled continuously [closed]

The problem statement is :

What sum can I expect in rolling a fair die n times?

How to determine the mean in a convenient way? Consider the fair die is 6 sided

-

## closed as off-topic by Did, Claude Leibovici, USER91500, Jonas, WatsonJun 9 at 9:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Claude Leibovici, USER91500, Jonas, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

If you've used dice a lot, you may have noticed that the pips on opposite sides always add up to $7$. Since this is the same for all three pairs of opposite sides and each pair has the same chance of being rolled and both sides of each pair have the same chance of being rolled, the expected value of a single six-sided die is $7/2$. Then by linearity of expectation the expected value of $n$ six-sided dice is $n\cdot7/2$.

-
Excellent!! Never thought this way. – Diptarag Oct 13 '12 at 21:23

The sum $Y$ is given by $Y=X_1+X_2+\cdots +X_n$, where $X_i$ is the number rolled on the $i$th toss. We have $E(X_i)=\dfrac{21}{6}$.

Since the expectation of a sum is the sum of the expectations, $E(Y)=\dfrac{21n}{6}$.

-
The answer I was looking for. Thanks a lot. – Diptarag Oct 13 '12 at 21:22

The problem can be easily solved by using the concept of indicator variables. Let $X_i$ be the random variable which counts the number of the times $i$ comes up in $n$ rolls of dice. So, we have events $X_1, X_2, X_3, X_4, X_5$ & $X_6$ representing all possible scenario (for e.g. $X_1$ counts the number of times the face $1$ came up in $n$ rolls of dice).

Note that: $$E[X_i] = P[X_i] = \frac n{6}$$

Also let's assume that random variable $X$ correspond to the sum of $n$ dice rolls. Then, $E[X]$ is given by:

$$E[X] = \sum_{i=1}^{i=6} i*E[X_i] = \sum_{i=1}^{i=6} i*P[X_i] = \sum_{i=1}^{i=6} i*\frac n{6} = 3.5n$$

-