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From:http://forum.swarthmore.edu/dr.math/problems/nesser12.9.97.html

Expected value

It means your average expected payoff in the long run.

For example, if we play a game where we flip a fair coin, and if it's heads, you pay me a dollar, but if it's tails, I pay you two dollars (a good game for you, right?), then on average, out of every two times we play, you will win once and I will win once.

But when you win, you get $2 and when I win, you get -$1 (in other words, you lose a dollar). So in the long run, for every two times we play, you will gain $2 + (-$1) = $1.

On average, for every two games, you will win $1. So on average, you will win half that, or 50 cents, every time you play.

This is useful, because now it's easy to estimate how much you will win if we play 1000 games. At 50 cents per game, on average, you "expect" to win about 1000 times 50 cents, or $500 over the course of our 1000 game match.

To calculate expected value, just add up the sum of the probabilities of each event times the payoff for each event.

In the example above, the probability of heads or tails is 1/2 each, so the expected value is:

E = 1/2($2) + 1/2(-$1) = 1 - 1/2 = 1/2, or 1/2 dollar.

To show how this works in another situation, suppose that again we agree to play a game where we roll a single die, and here's what happens for each result:

 

1: you win $3

2: nobody wins or loses

3: I win $5 4: you win $3

5: I win $4

6: you win $2

Is this a good game for you? To find out, work out the expected value (to you).

Remember that when you win, the payoff is positive, and when I win, your payoff is negative, since you pay me.

If the die is fair, every one of the possibilities above has an equal probability of 1/6: E = 1/6(3) + 1/6(0) - 1/6(5) + 1/6(3) - 1/6(4) + 1/6(2) E = 3/6 - 5/6 + 3/6 - 4/6 + 2/6 = -1/6

So on average, you lose 1/6 dollar every time we play, so it's a bad game for you. Every 600 times we play, you will lose, on average, 600(1/6) = 100 dollars. Your expected loss on each game is 1/6 dollar.

How should the expected value be interpreted?

1) it is the average value of the random variable over a LARGE number of trials

2) it is analogous to the sample mean, but applies to probabilities rather than data

3) the expected value does not have to be a particular value of the random variable

4) beware of fallacy of law of small numbers when using expected values (exaggeration of the degree to which a small sample resembles the population from which it is drawn, also called ‘gamblers fallacy’)

next from: http://garnet.acns.fsu.edu/~ppeters/fin3403/work/expval.htm

Calculating the Expected Value and Standard Deviation

Problem:

Consider the following probability distribution:

 

Probability

Possible Outcome

30%

$100

50%

$50

20%

-$10

Calculate the expected value and standard deviation of this distribution.

Solution:

Step 1: Calculate the expected value

Probability (pi)

Possible Outcome (xi)

Weighted Outcome  pixi

30%

$100

$30

50%

$50

$25

20%

-$10

-$2

 

E(x) = $53

 

The expected value is $53.

 

Step 2: Calculate the variance of the distribution

Probability(pi)

Possible Outcome (xi)

Deviation

xi E(x)

Deviation squared

(xi-E(x))2

Weighted squared deviation

pi (xi-E(x))2

30%

$100

$47

2,209

662.7

 

50%

$50

-$3

9

4.5

20%

-$10

-$63

3,969

793.8

 

s2= 1,461

 

Note: the variance is in squared units; in this example, the variance is in terms of squared dollars. 

Step 3: Calculate the standard deviation of the distribution Standard deviation = s = Ö1,461 = $38.22

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