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Chapter 1
Expected Value
12-5-1
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Expected Value
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Expected Value
Games and Gambling
Investments
Business and Insurance
12-5-2
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Expected Value
Children in third grade were surveyed and told to pick
the number of hours that they play electronic games
each day. The probability distribution is given below.
# of Hours x
Probability P(x)
0
.3
1
.4
2
.2
3
.1
12-5-3
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Expected Value
Compute a “weighted average” by multiplying
each possible time value by its probability and then
adding the products.
0(.3)  1(.4)  2(.2)  3(.1)  1.1
1.1 hours is the expected value (or the mathematical
expectation) of the quantity of time spent playing
electronic games.
12-5-4
© 2008 Pearson Addison-Wesley. All rights reserved
Expected Value
If a random variable x can have any of the
values x1, x2 , x3 ,…, xn, and the corresponding
probabilities of these values occurring are
P(x1), P(x2), P(x3), …, P(xn), then the
expected value of x is given by
E ( x)  x1  P( x1 )  x2  P( x2 ) 
 xn  P( xn ).
12-5-5
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Finding Expected Value
Find the expected number of boys for a three-child
family. Assume girls and boys are equally likely.
Solution
S = {ggg, ggb, gbg,
bgg, gbb, bgb, bbg,
bbb}
The probability
distribution is on
the right.
# Boys Probability
x
P(x)
0
1/8
1
3/8
2
3/8
3
1/8
Product
x  P( x)
0
3/8
6/8
3/8
12-5-6
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Finding Expected Value
Solution (continued)
The expected value is the sum of the third column:
3 6 3 12
0   
8 8 8 8
3
  1.5.
2
So the expected number of boys is 1.5.
12-5-7
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Example: Finding Expected Winnings
A player pays $3 to play the following game: He rolls
a die and receives $7 if he tosses a 6 and $1 for
anything else. Find the player’s expected net winnings
for the game.
12-5-8
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Example: Finding Expected Winnings
Solution
The information for the game is displayed below.
Die Outcome Payoff Net P(x) x  P ( x )
1, 2, 3, 4, or 5
$1
–$2 5/6
6
$7
$4
–$10/6
1/6
$4/6
Expected value: E(x) = –$6/6 = –$1.00
12-5-9
© 2008 Pearson Addison-Wesley. All rights reserved
Games and Gambling
A game in which the expected net winnings
are zero is called a fair game. A game with
negative expected winnings is unfair against
the player. A game with positive expected net
winnings is unfair in favor of the player.
12-5-10
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Example: Finding the Cost for a Fair
Game
What should the game in the previous example
cost so that it is a fair game?
Solution
Because the cost of $3 resulted in a net loss of $1,
we can conclude that the $3 cost was $1 too high. A
fair cost to play the game would be $3 – $1 = $2.
12-5-11
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Investments
Expected value can be a useful tool for
evaluating investment opportunities.
12-5-12
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Example: Expected Investment Profits
Mark is going to invest in the stock of one of the two
companies below. Based on his research, a $6000
investment could give the following returns.
Company ABC
Company PDQ
Profit or Probability Profit or Probability
Loss x
P(x)
Loss x
P(x)
–$400
.2
$600
.8
$800
.5
1000
.2
$1300
.3
12-5-13
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Expected Investment Profits
Find the expected profit (or loss) for each of the
two stocks.
Solution
ABC: –$400(.2) + $800(.5) + $1300(.3) = $710
PDQ: $600(.8) + $1000(.2) = $680
12-5-14
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Business and Insurance
Expected value can be used to help make
decisions in various areas of business,
including insurance.
12-5-15
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Example: Expected Lumber Revenue
A lumber wholesaler is planning on purchasing a
load of lumber. He calculates that the probabilities
of reselling the load for $9500, $9000, or $8500 are
.25, .60, and .15, respectfully. In order to ensure an
expected profit of at least $2500, how much can he
afford to pay for the load?
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Example: Expected Lumber Revenue
Solution
The expected revenue from sales can be found below.
Income x
P(x)
x  P( x)
$9500
.25
$2375
$9000
.60
$5400
$8500
.15
$1275
Expected revenue: E(x) = $9050
12-5-17
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Expected Lumber Revenue
Solution (continued)
profit = revenue – cost or cost = profit – revenue
To have an expected profit of $2500, he can pay up to
$9050 – $2500 = $6550.
12-5-18
© 2008 Pearson Addison-Wesley. All rights reserved