The expectation value is a value applied to a probability distribution that roughly
corresponds to what we would think of as the average. Given that you have some process with
a probability distribution, if you were to do that process many times and average the
outcomes that mean value would tend toward the expectation value of the probability
distribution as the number of trials increased toward infinity. Formally,
given a distribution of probabilities for a set of outcomes `x`(`i`) with probability
`P`(`x`(`i`)), the expectation value of a function `f`(`x`) for that probability
distribution, denoted <`f`>, would be the sum of the values `f`(`x`(`i`))
weighted by the probabilities `P`(`x`(`i`)).

The expectation value may also be denoted in others ways including f with a bar above or
exp(`f`) (which I know is like the way people express `e`, sometimes, but people use
it) and others. There are a few important things to note about the expectation value. For
a finite distribution there will be a finite set of possible values of a given function
`f`, but the expectation value does not have to be the value of `f` for one of the outcomes
actually possible for an individual trial, since it represents the average of multiple
outcomes. Also, if all outcomes are equally likely (for `n` possible outcomes, `P`=1/`n`)
then the expectation value simply becomes the arithmetic mean of `f` over all
outcomes. Finally, I think an example is in order.

#### Example

Let's take the example of rolling a pair of dice. If you roll a pair of 6 sided dice,
the possible outcomes for the total amount showing are: 2,3,4,5,6,7,8,9,10,11,12. Not too
interesting so far. The number of ways, N, you can get each outcome, s, is as follows:

s N
-----------
2 | 1
3 | 2
4 | 3
5 | 4
6 | 5
7 | 6
8 | 5
9 | 4
10 | 3
11 | 2
12 | 1

The total number of possible combinations of the values on each die (not
the sum) is 36, so the probability of each outcome for the sum, s, is N/36. Now we can
look at some expectation values. Let's calculate <`s`>. We'll leave the factor of 36 to
the end and divide through by it.

<`s`> = (2*1+3*2+4*3+5*4+6*5+7*6+8*5+9*4+10*3+11*2+12*1)/36=252/36=7

So, if you roll two properly weighted dice lots of times and average the values you get for
the total amount showing, the resulting mean would be close to 7. The more you roll the
dice, the less likely that your value will differ much from 7. Like I said, all this
counts on you having properly weighted dice, which many are not (have you ever noticed that
about the dice that come with Risk?). If, however, you calculate the expectation value
of the function `s^2`, then:

<`s^2`>=Sum of (`s^2`)*`P`(`s`)=1974/36=54.8333... Notice, that that is not the value of
`s^2` for any of the possible values of `s`.

The concept of the expectation value is important in many fields that use statistics, including statistical mechanics and quantum mechanics.

Note: The above example, while simple, was just pulled out of my ass
as I write this, so let me know if I messed up the math. Also, please /msg me
and let me know if I made any other mistakes or if you find something unclear.