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I'm considering rolling six fair dice. For each die, we have outcomes ${1, 2, 3, 4, 5, 6}$ that each occur with a probability of $\dfrac{1}{6}$.

The expectation, or mean value, of each die is $\dfrac{1+2+3+4+5+6}{6} = 3.5$

Now, if we take 10 of these dice, we have a range of outcomes $10, 11, 12, ..,58,59,60$.

Can we calculate the expectation of rolling the dice by calculating: $\dfrac{ \sum^{60}_{i=10} i } {50}$ ? For some reason I doubt this is right, as each outcome does not have an equal probability of occurrence.. (14 can be obtained by rolling 1,1,1,1,1,1,2,2,2,2 or 1,1,1,1,1,1,1,1,3,3 or many other rolls). Even if this is correct, it is tedious to calculate.

How can we calculate the expectation?

Thank you

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up vote 2 down vote accepted

Let $X_1, X_2,\dots,X_n$ be the results on the dice. (Suppose the dice have little numbers $1$ to $n$ written on them, to make them distinct).

Then by the linearity of expectation, $$E(X_1+X_2+\cdots +X_n)=E(X_1)+E(X_2)+\cdots +E(X_n).$$

Thus for the $6$ dice the expectation of the sum is $(6)(3.5)$.

Remark: In principle, we could compute the probabilities of the various sums, and use these to calculate the mean. The sums range from $6$ to $36$. If $p_k$ is the probability that the sum is $k$, then the mean is $\sum_{k=6}^{36}kp_k$. Although we can use symmetry to cut the work in half, the calculations of the $p_k$ are still unpleasantly lengthy.

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When would I use $E(x_1 * x_2 * .. * x_n) = E(x_1) * E(x_2) * .. * E(x_n)$ ? I'm asking because I was trying to find the answer to my problem and found this: – connorbode Nov 16 '12 at 3:30
This is true under the assumption of independence. (The linearity of expectation does not require independence, which is very useful, though in this case we do have independence.) You would use your formula if you wanted the expected value of the product of the numbers on the dice. But your question asks for the mean of the sum. – André Nicolas Nov 16 '12 at 3:33
Is the same true of variance? Does $Var(x_1 + x_2 + .. + x_n) = Var(x_1) + Var(x_2) + .. + Var(x_n)$ ? – connorbode Nov 16 '12 at 3:53
The variance of a sum of independent random variables is the sum of the variances. More generally, if the $X_i$ are independent, the variance of $a_1X_1+\cdots +a_nX_n$ is $a_1^2\text{Var}(X_1)+\cdots +a_n^2\text{Var}(X_n)$. – André Nicolas Nov 16 '12 at 4:01
Responding to the remark: these calculations are not too tough. I did it for this question with a description of how in a comment. It is correct that it is not needed here. – Ross Millikan Nov 16 '12 at 4:53

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