Solutions to HW 3

The first 4 problems refer to the decision matrix labeled as Figure A.

****** HOW YOU ANSWER THIS PROBLEM DEPENDS ON HOW YOU INTERPRET THESE RANKING NUMBERS! THE ANSWERS BELOW ASSUME THAT #1 IS THE WORST. CREDIT WILL ALSO BE GIVEN IF YOU CONSISTENTLY ASSUMED #1 WAS THE BEST.

Figure A:

	State 1	State 2	State 3
Option 1	1	9	4
Option 2	5	3	8
Option 3	6	7	2

1. Suppose that the values shown in the decision matrix of Figure A represent a relative ranking of all the possible outcomes of the decision. (ASSUME #1 IS THE WORST.)

A. Suppose it is known that state S₁, will occur no matter which option is chosen. Which option is preferable in this case? Explain briefly.

Since we know S₁ will occur regardless of which option we choose, we need only consider the following part of our matrix:

	State 1
Option 1	1
Option 2	5
Option 3	6

Given the outcomes for our options, we should take Option 3, because it will give us our highest ranked outcome.

B. Suppose it is known that state S₂ will occur no matter which option is chosen. Which option is preferable in this case? Explain briefly.

Since we know S₂ will occur regardless of which option we choose, we need only consider the following part of our matrix:

	State 2
Option 1	9
Option 2	3
Option 3	7

Given the outcomes for our options, Option 1 is preferable, because it will give us the highest ranked outcome among the three possibilities.

C. Suppose it is known that state S₃ will occur no matter which option is chosen. Which option is preferable in this case? Explain briefly.

Since we know S₃ will occur regardless of which option we choose, we need only consider the following part of our matrix:

	State 3
Option 1	4
Option 2	8
Option 3	2

Given the outcomes for our options, Option 2 is preferable, because it will give us the highest ranked outcome among the three possibilities.

D. Suppose it is known with certainty that (1) performing the first option will lead to the occurrence of state S₃, (2) performing the second option will lead to the occurrence of state S₂, and (3) performing the third option will lead to the occurrence of state S₁. Which option is preferable in this case? Explain briefly.

With conditions (1), (2), and (3), our decision matrix can be revised to reflect the new information. We have no chance of getting State 1 or State 2 if we take Option 1, no chance of getting State 1 or State 3 if we take Option 2, and no chance of getting State 2 or State 3 if we take Option 3. Thus, our new matrix will look as follows:

	State 1	State 2	State 3
Option 1			4
Option 2		3
Option 3	6

Option 3 is preferable in this case, because it will yield the highest ranked outcome of our three possibilities.

2. Now suppose that there is complete uncertainty regarding the outcomes of the decision problem shown in Figure A.

A. Is any option better than any other? Explain briefly.

No option dominates any of the others by giving a higher value outcome no matter how the world turns out to be.

B. Is there a best option? Explain briefly.

No – same reason.

C. Suppose that a value ranking of at least 5 is regarded as a satisfactory payoff for the decision. Is there then a satisfactory option? Explain briefly.

No option has outcome of 5 or higher in every cell.

D. Suppose that a value ranking of at least 3 is regarded as a satisfactory payoff for the decision. Is there then a satisfactory option? Explain briefly.

Option 2 gives a satisfactory outcome no matter what.

E. Which Option is Preferred if one wishes to play it safe? Explain briefly.

To play it safe, one looks for the option with has the biggest number for the worst case. Option 2 in this case.

F. Which Option is preferred if one wishes to gamble? Explain briefly.

Look for the option that might give us the highest possible value. Option 1.

3. Now suppose that the values attached to the outcomes in Figure A provide not just a ranking, but also a measure of the relative value of the outcomes. Suppose further that the states have the following probabilities of occurring, regardless of which Option might be chosen: P (S₁) =0.2, P(S₂) =0.3; P(S₃) =0.5.

With the probabilities now given to us for each state, we can complete our decision matrix:

	State 1	State 2	State 3
Option 1	P=0.2 1	P=0.3 9	P=0.5 4
Option 2	P=0.2 5	P=0.3 3	P=0.5 8
Option 3	P=0.2 6	P=0.3 7	P=0.5 2

A. Calculate the expected value for each of the options.

EV_O1 = (.2)(1) + (.3)(9) + (.5)(4) = .2 + 2.7 +2.0 = 4.9

EV_O2 = (.2)(5) + (.3)(3) + (.5)(8) = 1.0 + .9 + 4.0 = 5.9

EV_O3 = (.2)(6) + (.3)(7) + (.5)(2) = 1.2 + 2.1 + 1.0 = 4.3

B. Which is the preferred option in this case? Explain briefly.

Applying the Expected Value Strategy, Option 2 would be preferable, because it has the highest expected value.

4. Now suppose that the probability of the various states depends on which option is chosen. In particular,

P(S₁/0₁) =.2; P(S₂/0₁) =.3; P(S₃/0₁) = .5

P(S₁/0₂) = .3; P(S₂/0₂) =.5; P(S₃/0₂) =.2

P(S₁/0₃) =.5; P(S₂/0₃) =.2; P(S₃/0₃) =.3

Putting the new probabilities into our matrix we get:

	State 1	State 2	State 3
Option 1	P=0.2 1	P=0.3 9	P=0.5 4
Option 2	P=0.3 5	P=0.5 3	P=0.2 8
Option 3	P=0.5 6	P=0.2 7	P=0.3 2

A. Calculate the expected value for each of the options.

EV_O1 = (.2)(1) + (.3)(9) + (.5)(4) = .2 + 2.7 +2.0 = 4.9

EV_O2 = (.3)(5) + (.5)(3) + (.2)(8) = 1.5 + 1.5 + 1.6 = 4.6

EV_O3 = (.5)(6) + (.2)(7) + (.3)(2) = 3.0 + 1.4 + .6 = 5.0

B. Which is the preferred option in this case? Explain briefly.

Applying the Expected Value Strategy, Option 3 is preferable because of our three options it has the greatest expected value.

5. A Raffle

A friend stops you on the street saying that his fraternity is having a raffle and you now have the opportunity of buying the very last ticket for only $5. The prize, which has been donated by a rich alumnus, is worth $1000 cash. Being somewhat suspicious, you inquire as to the number of tickets being sold and are assured there are only 250. Indeed, the ticket in question has the number 250 on it. Having a spare $5, you pause to consider whether to buy the ticket.

A. Set up the decision matrix for your situation, including all the relevant information given above. Assume that your interest in buying a ticket is based solely on consideration of monetary gain or loss (and not on friendship, love of gambling, etc.).

	Win	Lose
Buy a Ticket	P=1/250 $1000-5	P=249/250 -$5
Don’t Buy a Ticket	P=1/250 0	P=249/250 0

[It might be easier to treat the $5 entry fee as a fixed cost to be deducted from the expected winnings of the lottery, in which case you could omit it from the above table.]

B. What decision strategy applies to this problem?

We have probabilities for the outcomes, so we can apply Expected Value Strategy.

C. What is the recommended choice? Exhibit any necessary calculations.

If we calculate the expected value of buying a ticket we get:

EV_BUY = (1/250)($1000 - 5) + (249/250)(-$5) = (1/250)($1000) –

(1/250)(-$5) + (249/250)(-$5) = (1/250)($1000) – (250/250)(-$5) = $4 - $5 = - $1

Our expected value for not buying a ticket will be zero, so we need to decide whether it’s worth our while to pay $5 for our ticket. Given that the overall expected value of taking the gamble is negative, we should not buy the ticket.

6. Oral Contraceptives and Breast Cancer

Rework the example of Giere Chapter 10, Section 10. 1 concerning oral contraceptives and blood clots using the information on oral contraceptives and breast cancer from a study published in Lancet. The relevant probabilities from that study are that the chances of a woman contracting breast cancer before age 36 when not using oral contraceptives are about 1/500. Using oral contraceptives for approximately eight years raises this probability to about 1/300.

A. Using the same relative values as in the blood clot example, what is the recommended choice in the breast cancer case?

Putting our new probabilities into the matrix from the example of Section 10.1 we get:

	Contract Breast Cancer	Don’t Contract Breast Cancer
Use Oral Contraceptives (OC)	P=1/300 1	P=299/300 100
Don’t Use Oral Contraceptives (DOC)	P=1/500 0	P=499/500 99

We can now calculate the two expected values:

EV_OC = (1/300)(1) + (299/300)(100) = 1/300 + 299/3 = 99.67

EV_DOC = (1/500)(0) + (499/500)(99) = 98.80

B. Using the information from the study on oral contraceptives and breast cancer, represent the choice whether or not to use Oral contraceptives as a choice between two jars from which one marble is to be selected at random.

Using the approach in Chapter 10 of Giere (see Figure 10.5), we can set up the following jars with the appropriate number of marbles. Since our probabilities are so small (1/300 = .0033 and 1/500 = .002) and it would be difficult to imagine having.33 or .2 marbles, we can multiply our decimal values by 100 to get whole numbers. Let the red marbles represent getting breast cancer and the blue marbles represent not getting breast cancer:

Use Oral Contraceptive

33 Red

(.33 * 100)

9967 Blue

(99.67 *100)

Don’t Use Oral Contraceptive

20 Red

(.2 * 100)

9980 Blue

(99.8 *100)

Since the choice of a marble will be random from either jar, is taking the oral contraceptive worth facing the additional 13 red marbles in the “Use Oral Contraceptive” jar? Note that this is just a method of making the choice vivid to our intuitions. Giere calls it “confronting our values”.

7. Generating Electricity: An Economic Analysis

Imagine that the officials of your state's electric company are trying to decide whether a needed new generator should be powered by coal or by a nuclear reactor. Suppose they have reached the point where the only unknown is the nature of new federal regulations on air pollution. If the regulations stay as they are now, a coal generator will involve a net cost of $50 million. On the other hand, if the regulations are tightened, equipment to "clean" the smoke will cost another $10 million. The officials are completely in the dark as to whether the regulations will be tightened. The cost of the nuclear plant is $55 million.

A. Set up the officials' decision problem as specified above. (Express the values in negative millions of dollars.)

We know that the cost for the nuclear plant will be -$55 million regardless of any changes in the federal regulations. For the coal plant, we can set up the following matrix:

	Regulations Stay As Are	Regulations Change
Coal-powered Generator	-$50 million	-$50 - $10 million

B. Does the problem exhibit a best option? Explain briefly.

There is no best option, because if we choose the coal plant, there is a chance that we would incur less cost than choosing the nuclear plant (if regulations stay as are) and there is also a chance that we could pay more (if regulations change).

C. What decision would the officials make if they were to play it safe? Explain briefly.

If the officials wanted to play it safe, they would choose the nuclear plant, because they know they won’t incur the most possible cost (i.e., the -$60 million if regulations change).

D. What decision would the officials make if they were to gamble? Explain briefly.

If the officials wanted to gamble, they would choose the coal-powered plant, because there is a chance they would incur the least possible cost (i.e., the -$50 million if regulations stay as are).

8. Generating Electricity: A Worst Case Analysis

Another way of looking at the decision between coal- and nuclear-powered, electrical generating plants is to focus on the possible environmental consequences of an accident. For a coal-fired plant, the worst possible accident could at most kill a small number of workers on the site and leave a city without electricity for some days. With a nuclear plant, the worst possible case is an accident such as the one that occurred in Chernobyl, USSR, or the one that almost occurred at Three Mile Island in Pennsylvania. Thousands of people may be exposed to damaging radiation. A meltdown could contaminate underground water supplies for several states, perhaps even the whole Mississippi River basin from St. Louis to New Orleans.

A. Set up the decision problem as specified above, assuming that a coal-fired plant with no accidents is better than a nuclear plant with no accidents. Rank the outcomes beginning with 1 for the worst outcome and ending with 4 for the best.

	Accidents	No Accidents
Coal-fired Plant	2	4 (best)
Nuclear Plant	1 (worst)	3

B. What would be the recommended option? Explain briefly.

Here we do have a best option. No matter whether accidents occur or not, the outcome of the coal plant option is higher in rank.

C. Now set up the decision problem assuming that a nuclear plant with no accidents is better than a coal-fired plant with no accidents. Rank the outcomes beginning with 1 for the worst outcome and ending with 4 for the best.

	Accidents	No Accidents
Coal-fired Plant	2	3
Nuclear Plant	1 (worst)	4 (best)

D. Is there now a recommended option? Explain briefly.

Now we have neither a best option or a satisfactory option, so playing it safe, we would choose the coal-powered plant. If we wanted to gamble, we would choose the nuclear plant.

9. Vacation Plans

Vacation time is coming and you are trying to decide whether to go to Florida with your friends or to go home. Your main worry is the weather, about which there is total uncertainty. If the weather in Florida turns out to be good, it would be worth the money to go. If, however, the weather turns out to be bad, you would rather be home with your family. Besides, it is free. On the other hand, if you decide to go home, you prefer that the weather in Florida be bad so you will not feel that you are missing something. You do not reveal that preference to your friends.

A. Using the numbers 1, 2, 3, and 4 to rank the four outcomes, construct the decision matrix for this problem. THIS TIME WE’LL ADOPT THE CONVENTION OF MAKING 1 REPRESENT THE BEST OUTCOME. [IF YOU CONSISTENTLY ADOPT THE OPPOSITE CONVENTION YOUR ANSWERS WILL STILL BE OK.]

	Good Weather	Bad Weather
Go to Florida	1 st (best)	4 th (worst)
Go Home	3rd	2 nd

B. Does the matrix exhibit a best option? Explain briefly.

There is no best option in this case – which option is best turns on the weather.

C. What option would you choose if you were to play it safe? Explain briefly.

If we wanted to play it safe we would choose to go home, because then we are guaranteed not to get the worst possible outcome.

D. What option would you choose if you were to gamble? Explain briefly.

If we wanted to gamble we would choose to go to Florida, because there is a possibility of getting our highest-ranked outcome.

10. Flu Epidemic

Imagine that the Centers for Disease Control in Atlanta has issued a bulletin that there is an impending flu epidemic. The flu is serious but only rarely fatal – typically, leaving one incapacitated for thirty days. Fortunately, there is a very effective vaccine ready to be used, and it is being made available free to anyone who requests it. Unfortunately, this flu vaccine has the side effect that it may cause temporary paralysis, which, though again rarely fatal, also has the effect of leaving one incapacitated for thirty days. If one were so unlucky as to get both, one would be incapacitated for forty days. The officials are wary of making recommendations and are leaving the decision whether to take the vaccine up to individuals.

A. Set up the decision problem giving the possible options, the relevant, possible states of the world, and values measured in negative days of incapacitation. Note that there will be four exclusive and exhaustive, possible states of the world.

	Get the Flu Alone	Get neither the Flu nor the Side-effect	Get the Side-Effect Alone	Get the Side-Effect and the Flu
Vaccinated	-30	0	-30	-40
Not Vaccinated	-30	0	-30	-40

[WE FIND OUT LATER IN THIS PROBLEM THAT ONE CAN GET THE PARALYSIS EVEN THOUGH ONE WAS NOT VACCINATED SO WE MUST INCLUDE THAT POSSIBILITY.]

B. If you have set up the problem correctly, there is no applicable decision strategy using only the information given above. Explain briefly why this is so.

The two options have exactly the same values for the same outcomes.

11. Flu Epidemic, Cont. In the whole population, the estimated risk of getting flu without a shot is 1/10, but with a shot only 1/10,000 (1000 times less). On the other hand, the probability of getting the paralysis without the shot is only 1/100,000; with the shot, it jumps to 1/100 (1000 times more).

A. Determine the approximate expected values of the two options. Treat very small probabilities as 0 and very large probabilities as 1. Which option is preferable? Explain briefly.

	Get the Flu Alone	Get Neither the Flu nor the Side-effect	Get the Side-Effect Alone	Get the Side-Effect and the Flu
Vaccinated	P=0 (1/10,000) -30	P=99/100 0	P=1/100 -30	P=0 (1/10,000 x 1/100) -40
Not Vaccinated	P=1/10 -30	P=9/10 0	P=0 (1/100,00) -30	P=0 (1/100,000 x 1/10) -40

EV(Vaccination) = (1/100) x ( - 30) = - 0.3

EV(not-Vaccination) = (1/10) x ( - 30) = - 3

We should choose the option that maximizes expected value and since – 0.3 is bigger that – 3, we should get vaccinated. (Note that the expectations are both negative numbers because they represent days sick.)