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
Best Answer
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.