you can think of expected number as the â€śaverageâ€ť number you would get if you repeated the experiment many, many times. as s2k pointed out, you need to know if the die is fair (i.e. all outcomes are equally likely) or not (in which case, at least one of the outcomes will be more likely than the others).

in general, expected value for discrete events (such as rolling of a die) is defined as follows:

E(X) = sum from x = min(k) to x = max(k) of x*p(x = k)

(mathjax doesnt seem to work for me)

E(X) = \sum_{x=min(k)}^{max(k)} x * p(x = k)

for a fair, 6-sided die numbered 1 through 6, the expeted value is, by definition,

1 * (1/6) + 2 * (1/6) + â€¦ + 6 * (1/6) = 21/6 = 3.5

that is, if you roll a die many times, on average you would get 3.5

the answer will be different if the die is not fair. for example, if 1 had 0 probability of occurring and 2 had twice the probability, then the expected value would be

1 * (0/6) + 2 * (2/6) + â€¦ + 6 * (1/6) = 3.67

use the same steps to find the expected value of the 8-sided die.

edit: if the question is asking for the expected value of the sum, there is a neat formula from probability theory:

E (X + Y) = E(X) + E (Y)

that is, the expected value of the sum of two random variables is the sum of the expected values of the individual random variables.

the expected value for a fair, 6-sided die is 3.5

the expected value for a fair, 8-sided die with the extra 2 numbers being 6 is 4.125

thus, the expected value of the sum is 3.5 + 4.125 = 7.625, which agrees with bread.

edit: mathjax works now, thanks magician