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Chapter 8 Expected Value

Decisions need more than probability.

Most of the time, when we make a decision, there are several different kinds of considerations. The chance of rain may be countered by my unwillingness to spend money on an umbrella (It’s only water, right?). There are simply some things, like money, that I value more.

If the chance of rain is below a certain percent, then I may feel that there is really no need to buy an umbrella. I decide that if that chance of rain is <60% then I will risk getting rained on, but if it is ≥60% then I will buy an umbrella.

Fair's Fair p. 81

Your aunt bought a lottery ticket for $1.

What is the fair price for you to buy it from her?

You might think $1 but, according to expected value theory...

A fair price for the ticket would make the expected value of buying the ticket $0.

So, if your expected value of buying it for $1 is -$0.10, to make it fair your aunt should sell the ticket to you for $0.90.

Let C be buying the ticket for $0.90

Act B = Buying the ticket for $1

C1 = the ticket is a winner

C2 = the ticket is not a winner

U(C1) = $90-$1 (your cost)

U(C2) = -$1

Pr(C1/B) = 1/100

Pr(C2/B) = 99/100

Exp(C) = U(C1)Pr(C1/C) + U(C2)Pr(C2/C)

= $89.10 x (1/100) + (-$0.90) x (99/100)

= $0.891 + -$0.891

= $0

Exp(B) = U(C1)Pr(C1/B) + U(C2)Pr(C2/B)

= $89 x (1/100)

+ -$1 x (99/100)

= $0.89 + (-0.99)

That's not fair!

= -$0.10

Notation

U: Utility: The value of possible outcomes for our actions

We make decisions like this everyday, and what we’re doing is calculating the expected value of our decision.

We are balancing the risks versus the rewards.

A: Act: the act of making the choice based on the expected value calculation

A Raffle p. 83

C: Consequence: the potential result of making choices

There is a raffle in which two numbers are drawn. The first winner gets $90 and the second winner gets $9. There are 100 tickets total.

Act A: Take a free ticket

U(C): The Utility of a Consequence

Pr(C1/A) = 1/100

Pr(C2/A) = 1/100

Pr(C3/A) = 98/100

C1 = Winner #1

C2 = Winner #2

C3 = Lose

U(C1) = $90

U(C2) = $9

U(C3) = $0

Pr(C/A): The probability of C happening, if act A is taken

Exp(A) = U(C1)Pr(C1/A) + U(C2)Pr(C2/A) + U(C3)Pr(C3/A)

= $90 x (1/100) + $9 x (1/100) + $0 x (98/100)

For two consequences:

Act: A

Consequences: C1, C2

Utilities: U(C1), U(C2)

Probabilities Pr(C1/A), Pr(C2/A)

= $0.90 + $0.09 + $0

Exp(A) = U(C1)Pr(C1/A) + U(C2)Pr(C2/A)

Exp(A): The expected value of A

Let's use the example of the umbrella

= $.99

A = You don't buy an umbrella

C1 = W = You get wet (It rains)

Your expected value is positive, so take the ticket!

C2 = D = You stay dry (It doesn't rain)

U(C1) = U(W) = -$25 (You have to buy a new shirt)

U(C2) = U(D) = $0

Exp(A) = U(W)Pr(W/A) + U(D)Pr(D/A)

Pr(C1/A) = Pr(W/A) = .5

= (-$25) x .5 + ($0) x .5

Pr(C2/B) = Pr(D/A) = .5

= -$12.50 + $0 = -$12.50

Is it worth it to you? It depends on how much an umbrella costs. If it costs less than $12.50, then it's smarter to go ahead and buy it, according to expected value theory.

Act B: Pay $1 for a ticket

What about the expected value of Act B, buying an umbrella? That's easy, it's just the cost of the umbrella! (Let's say $8)

C1 = Winner #1

C2 = Winner #2

C3 = Lose

Pr(C1/A) = 1/100

Pr(C2/A) = 1/100

Pr(C3/A) = 98/100

U(C1) = $89

U(C2) = $8

U(C3) = -$1

Street Life p. 83

Exp(B) = U(W)Pr(W/B) + U(D)Pr(D/B)

Martin sells stuff illegally for a gross profit of $300 per day

If you buy the umbrella and still manage to get wet, the utility is the cost of the new shirt plus the cost of the umbrella = $37.50

If you buy the umbrella, the probability of staying dry is 100%

= (-$37.50) x 0 + (-$8) x 1

Expected Time of Travel pp. 84-5

The fine for illegal vending is $100

Exp(B) = U(C1)Pr(C1/A) + U(C2)Pr(C2/A) + U(C3)Pr(C3/A)

(Time and money are both valued, so they are both considered utilities)

= -$8

He works 5 days a week and is ticketed twice per week

But the probability of that happening is zero because if you have the umbrella with you, you won't get wet!

= $89 x (1/100) + $8 x (1/100) + (-$1) x (98/100)

His cost for the stuff he sells is $100

= $0.89 + $0.08 + (-$0.98)

You are going to a job interview and can either take a plane or a train. But you are worried about the possibility of a bad storm, so you find the expected value of taking a plane and the expected value of taking a train and compare them.

What is the expected value of a day's work, Exp(W)?

= -$0.01

U(C1) = $300-$100-$100 = $100

A bad storm is predicted with probability 0.2

C1 = Gets fined

C2 = Doesn't get fined

Not quite fair. The price of the ticket should be

$0.99

U(C2) = $300-$100 = $200

If there is a storm (C2):

If there is no storm (C1):

Pr(C1/W) = 2/5 = .4

Pr(C2/W) = 3/5 = .6

U(~S/T) = Time by train = 5 hours

U(~S/P) = Time by plane = 1 hour

U(S/T) = 7 hours

U(S/P) = 11 hours

Exp(W) = U(C1)Pr(C1/W) + U(C2)Pr(C2/W)

Pr(S) = 0.2

Pr(~S) = 0.8

= $100 x .4 + $200 x .6

Exp(T) = U(~S/T)Pr(~S) + U(S/T)Pr(S)

Exp(P) = U(~S/P)Pr(~S) + U(S/P)Pr(S)

= $40 + $120 = $160

We can think of this differently

= 5 x .8 + 7 x .2

= 1 x .8 + 11 x .2

= 4 + 1.4

= .8 + 2.2

We could say that he is sure to get the $200, but that 40% of the time, he has to pay the $100 fine:

= 3 hours

= 5.4 hours

= $200 - (.4 x $100) = $200 - $20 = $160

Exercise 2 Gimmicks

What is the expected value of act M, mailing in a coupon for a chance to win $10,000 with a chance of winning of 1/40,000? The only cost is the postage stamp: $0.45

Exercise 6 Insurance

William has a car and watch that he is considering getting insured by one policy. To find what they will charge him, the insurance company takes the value of the items and the probability that they will be stolen into account.

Probability of a watch being stolen: 1/30

Value of the watch: $600

Value of the car: $5,400

Probability of a car being stolen: 1/900

Because he is insuring them with one policy, there are four scenarios: both stolen, only watch stolen, only car stolen, and neither stolen.

Two consequences: Winning and Losing

W = Watch is stolen

C = Car is stolen

C1: Pr(W&C) = 1/30 x 1/900

= 1/27,000

U(W&C) = -$5,400 + -$600

U(W) = $10,000-$0.45

U(L) = -$0.45

U(W&~C) = -$600

= 899/27,000

C2: Pr(W&~C) = 1/30 x 899/900

Exercise 3 Street vendor

U(~W&C) = -$5,400

= 29/27,000

C3: Pr(~W&C) = 29/30 x 1/900

C4: Pr(~W&~C) = 29/30 x 899/900)

U(~W&~C) = $0

= 26,071/27,000

Pr(L/M) = 39,999/40,000

Pr(W/M) = 1/40,000

First, we have to figure out the expected value of declining to buy insurance.

Exp(D) = U(W&C)Pr(W&C) + U(W&~C)Pr(W&~C) + U(~W&C)Pr(~W&C) + U(~W&~C)Pr(~W&~C)

Conditions are the same as the example we did before, except that the fine is rarely enforced. So even though Martin gets a ticket, he only ends up paying the fine 20% of the time.

= (1/27,000)(-$5,400 +-$600) + (899/27,000)(-$600) + (29/27,000)(-$5,400) + (26,071/27,000)($0)

= -$6000/27,000 + (899 x -$600)/27,000 + (29 x -$5,400)/27,000 + $0

Exp(M) = U(W)Pr(W/M) + U(L)Pr(L/M)

= (1/27,000)(-$6,000 + -$539,000 + -$156,600)

= (1/27,000)(-$701,600)

Exp(W) = U(C1)Pr(C1/W) + U(C2)Pr(C2/W)

Exp(D) = -$26

= $9,999.55 x (1/40,000) + -$0.45 x 39,999/40,000

New Pr(C1) = .4 x .2 = .08, so new Pr(C2) = .92

The insurance company will charge $60 plus 10% of the value of his lost property.

The value of his lost property is -$26, or the cost for him if he does not have insurance.

= $100 x .08 + $200 x .92

=$0.25 + (-$0.45)

The consequences for buying insurance are the same as for not buying insurance: whether one, the other, both, or neither of the things are stolen.

= $8 + $184 = $192

We could figure it up again using all four consequences. The probabilities would be the same but the utilities would be different. But we'll just do the shorter version now, using Exp(D).

Or thinking about it the other way, the money he'll make minus what he'll pay if he is cited:

Exp(I) = -$60 + (.1 x -$26)

= -$0.20

Exercise 4 Street Vendor with Bad Boss

= -$60 + -$2.60

If we were to change the value of the goods, for example if they were appraised at twice the value, would this make more of a difference for Exp(D) or Exp(I)?

"Big Fish" offers Martin a deal. For the cost of $50 a day, he will pay any fines that Martin gets. But it turns out that "Big Fish" is not reliable. He only pays the fine 75% of the time! (So Martin must pay the fine 25% of the time.)

= -$62.60

What is the expected value of a day's work under these conditions?

Remember:

= $200 - ($100 x .08)

U(C1) = Money he has left after paying fine

= his profit minus the fine minus what he pays "Big Fish"

C1 = pays fine

C2 = no fine

U(C2) = Money he makes with no fine

= his profit minus what he pays "Big Fish"

Exp(D) = -$52

Pr(C1/W) = Probability he pays the fine (given he's working that day)

= the probability he's caught, the fine is enforced, and that "Big Fish" does not pay the fine.

Pr(C2/W) = Probability he does not pay the fine

Exp(I) = -$60 + (.1 x -$52) = -$60 - $5.20 = -$65.20

= $200 - $8 = $192

Exp(W) = U(C1)Pr(C1/W) + U(C2)Pr(C2/W)

even though Exp(D) doubled, Exp(I) only increased by -$2.60

= ($200 - $100 - $50)(.4 x .2 x .25) + ($200-$50)(1-.4x.2x.25)

Or if we think about it as the cost of the postage stamp plus the chance of winning:

= $50 x .02 + $150 x .98

= $1 + $147

So, difference in value of the watch and car makes a bigger difference for the act of declining insurance.

= $148

Or thinking about it in terms of what he makes minus the chance of the fine:

= $150 - ($50 x .02)

= $148

= $150 - $2

Exercise 5 The Best of Possible Worlds

= -$0.45 + $10,000(1/40,000)

Out of these four scenarios, which is the best and which the worst for Martin?

= -$0.45 + $0.25

1. (Fined or Not) Exp(W) = $160

= -$0.20

2. ("Big Fish" pays fine 100% of the time) Exp(W) = $150

3. (Fine only enforced 20% of the time) Exp(W)= $192

3. is the best because it is the highest net profit.

4. ("Big Fish" only pays fine 75% of the time) Exp (W) = $148

4. is the worst because the lowest net profit.