Download Section 11.8 - Expected Value

Section 11.8 Expected Value
Objectives
1. Compute expected value.
2. Use expected value to solve applied
problems
3. Use expected value to determine
the average payoff or loss in a game of
chance.
Expected value
Expected value is a mathematical way
to use probabilities to determine what
to expect in various situations in the
long run.
Probability distribution
Expected value - example
Definition: A probability distribution is a
table containing all possible outcomes and
their probabilities.
Example: A probability for the number of
heads in an experiment of 3 coin tosses
is:
Number of
A coin is tossed three times. To find the expected
value for the number of heads, we multiply the
probability for each outcome times that outcome,
then sum up those products:
Expected
Number of
Value times
value =
Probability probability
heads
expected
0
1/8
0 . 1/8 =0
number of
1
3/8
1 . 3/8 = 3/8
heads =
2
3/8
2 . 3/8 = 6/8 0 + 3/8 +
6/8 + 3/8 =
3 . 1/8 = 3/8
3
1/8
3/2
heads
Probability
0
1/8
1
3/8
2
3/8
3
1/8
Expected value
Of course, we can never get 3/2 (1 ½
heads).
We say that if we performed this experiment
over and over, the number of heads in all
of these experiments would (theoretically)
average out to 1 ½ per 3 coin tosses in
the long run.
Expected Value - Definition
For a random occurrence that can take on n
different values, say
x1, x2, . . . , xn
with the probabilities related to these values
P(x1), P(x2), . . . , P(xn),
then we say that the expected value, E, of that
occurrence is
E = P1(x1) . x1 + P2(x2) . x2 + . . . + Pn(xn) . xn
1
Expected value - Example
A basketball player gets 2 free throws. Each
free throw is worth one point. The
probability of making none, 1, or both
baskets is given in the table. What is the
expected value of the points that will be
scored?
Number of
Probability
shots made
0
.16
1
.48
2
.36
Expected value – special case
When an expected value problem involves
1. A number n trials of the same experiment.
2. Only 2 possible outcomes.
3. Same probability for the outcomes every time the
experiment is performed.
Expected value - Example
Solution:
Number of
Value .
Probability
shots made
Probability
0
.16
0
1
.48
.48
2
.36
.72
The expected value is:
0 + .48 + .72 = 1.2 points
This means that if this basketball goes up for 2 free
throws many times, on average he will score 1.2
points each time.
Example
Back to our coin toss problem. Since a coin
toss has only 2 outcomes (heads or tails),
and the probability of either is ½, then in
the experiment that involves 3 trials of a
coin toss, the expected value for heads is
then the expected value of one of those outcomes,
equals:
n
.
3 . ½ = 3/2.
P(outcome)
= number of trials times probability of the outcome.
Expected value - application
An investor has $1000 to invest. After
researching, the investor decides to buy
stock in either company A, company B, or
company C. The following are projected
profits (or losses) for each company, and
a reasonable expectation of the
probability for each.
Calculate the expected values of the $1000
investment in each company. Which
company is the best choice?
Expected value - application
Company A
Company B
Company C
Profit
or Loss
Probability
Profit or
Profit
Probability
Loss
or Loss
-$400
.2
$500
.8
$0
.4
800
.5
1000
.2
700
.3
1500
.3
1200
.1
2000
.2
Probability
2
Expected value - application
Example
Expected return – company A:
A company is planning to bid on a contract.
The bidding process will cost $1500.
.
.
.
EA = -400 .2 + 800 .5 + 1500 .3 = $770
Expected return – company B:
.
The probability of their bid being accepted is
.2. If the bid is accepted, they will earn
$40,000.
.
EB = 500 .8 + 1000 .2 + = $600
Expected return – company C:
EC = 0 . .4 + 700 . .3 + 1200 . .1 + 2000 . .2
= $730
Go with company A.
Example
What is their expected profit?
Expected value – game of chance
Solution: You need to set up your own probability
distribution. Remember to subtract the amount
spent on the bidding process from any possible
gains. Set up a table:
Outcome: Profit/Loss (Value) Prob.
Value . Prob.
Win bid
40,000 – 1500 =
38,500
.2
7700
Loose bid
0 – 1500 =
-1500
.8
-1200
A charity is holding a raffle. It is selling
3000 tickets for $5 each. One grand prize
winner will get a new car worth $18,000,
and 2 second prize winners get $100 gift
certificates.
If you buy one $5 ticket, what is the
expected value of its return?
Expected value:
7700 + -1200 = $6,500
Expected value – game of chance
Solution: Set up a table:
Outcome:
Profit/Loss
Prob.
Value . Prob.
Win car
18,000 – 5 =
17,995
1/3000
17,995/3000
Win gift
certificate
100 – 5 = 95
2/3000
190/3000
Loose
0 – 5 = -5
2997/3000
-14,985/3000
E=
17,995 190 − 14,985 3200
+
+
=
≈ $1.07
3000 3000
3000
3000
3