978-1118808948 Chapter 13 Lecture Note Part 1

subject Type Homework Help
subject Pages 9
subject Words 2948
subject Authors William F. Samuelson

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CHAPTER THIRTEEN
THE VALUE OF INFORMATION
OBJECTIVES
1. To show how to incorporate new information into the manager's decision tree.
(The Value of Information)
2. To show how to revise probabilities using joint probability tables and Bayes
rule. (Revising Probabilities)
3. To present applications and examples of prediction pitfalls (Business Behavior
and Decision Pitfalls).
4. To examine competitive bidding strategies and explore the use of auctions.
(Auctions and Competitive Bidding).
TEACHING SUGGESTIONS
I. Introduction and Motivation
A. Information Sources. Ask students to think about the myriad kinds and
sources of information that decision makers might use. Here is a possible (partial)
list:
Types and Sources of Information
Business:
1) Macro-forecasts, Futuristic forecasts (MegaTrends), demographic data
(population data by zip-code), sociological trends, marketing surveys.
2) Budget projections, cost information of all kinds.
3) Political risk assessments, credit histories
4) Workplace: job interviews, drug testing
General:
1) Scientific evidence (medical, geological, meteorological, actuarial) Predicting
earthquakes, hurricanes, global warming, and so on
2) Legal evidence
3) Statistical information
4) Polls and political surveys
5) Educational testing (standardized tests)
6) Encyclopedia-type information
B. Guinea Pig Questions. Three intuitive prediction problems “Steve's
occupation”, “The three box problem”, and “Is it cancer?” – are discussed in the
text. These can be used in class before they are formally assigned for reading.
1. Is it Cancer? In the text example, the prior probability of the disease is 1 in
1000 and the test is 95% accurate. In light of a positive test, the revised
probability is 1 in 51 (about a 20 fold upward revision).
2. Let's Make a Deal (Problem 6). Consider the following simplified version
of the game, “Let's Make a Deal.” There is a grand prize behind one of three
curtains. The other two curtains are empty. As the contestant, you get to choose a
curtain at random. Let's say you choose curtain three. Before revealing what's
behind the curtain, the game show host offers to show you what's behind one of
the other curtains. Suppose he shows
you that curtain two is empty. In fact, he always shows you an empty curtain.
(You know that's how the game works, the audience knows it, everybody knows
it.) Now you must decide: do you stick with your original choice, curtain three,
or switch to curtain one? Which action gives you the better chance of finding the
grand prize?
3. Predicting Sales. Last year, sales at 5 department stores in a chain were:
Store A Store B Store C Store D Store E Total
Last Year 8 9 10 11 12 50
Next Year ? ? ? ? ? 55
Your job is to predict sales at each store in the coming year. The overall forecast
is for total regional sales to grow by 10 percent (from 50 to 55). What are your
predictions?
4. Odds and Urns. There are two urns – one contains 40 red balls and 20
white balls and the other contains 40 white balls and 20 red balls. The urns are
identical from the outside. After choosing an urn at random, you are allowed to
reach in and take out a ball, note its color, then return it to the urn. In ten such
picks, you observe 7 red balls and 3 white balls. What are the chances that you
originally selected the “red” urn?
5. The Cocoa Game. (We have summarized and paraphrased the original
account which appeared in Adam Smith, The Money Game.) “Cocoa,” said the
Great Winfield. “There isn't any cocoa. The world is just about out of it.” And
he proceeded to explain that the price of cocoa depends on how much cocoa
there is and that a shortage of cocoa means a sky high price. “My informant in
Ghana tells me there is no cocoa. The game then is to buy futures contracts in
cocoa, betting that the price of cocoa will go up. And when it does, we make a
killing.”
And so they bought cocoa futures and waited and watched events in Ghana.
Revolution in Ghana: Cocoa up 2 cents. Civil war. The Hausers are murdering
the Ibos. Who is going to harvest the crop? Cocoa up 2 more cents. The main
railway has been blown up. How will the cocoa, if there is any, get to market? It
was then that the Great Winfield decided to send his man Marvin to West Africa
to find out if it was raining (bad for the crop) and whether the dreaded Black Pod
disease was spreading. Twenty-four hours after buying out Abercrombie and
Fitch, Marvin is cabling messages from the darkest cocktail lounges of Ghanian
hotels: RAINING OFF AND ON. Three days later: BRITISHER IN HOTEL
SAYS CAPSIO FLY SPREADING. IT HAS STOPPED RAINING. All the time,
the prices of cocoa futures continued to rise.
Until strange things began to happen to the price of cocoa. The price began to
fall, first slowly, then disastrously. By the time the price of cocoa had stopped
falling, the followers of the Great Winfield were busted and Winfield himself
had lost two-thirds of his cocoa contracts.
6. Finding the Best Item Suppose you will be shown three prizes in order.
You know absolutely nothing about how valuable the prizes might be; only
after viewing all three can you determine which one you like best. You are
shown the prizes in order and are allowed to select one. However, there is no
going back. You must select a prize immediately after seeing it, before seeing
any subsequent prize.
a. Your sole objective is to obtain the best of the three prizes. (Second best
does not count.) A random selection provides a one-third chance of getting
the best prize. Find a strategy that provides a greater chance.
b. What if there are a large number of prizes (say, 10, 50, or 100)? Describe in
general terms the kind of strategy that would maximize your chances of
obtaining the best prize. (Do not try to compute an exact answer.)
7. Bidding for an Item of Unknown Value. Tell the class that you will
conduct a sealed-bid auction for a valuable item. The high bidder will pay his bid
and win the item. The item is a jar of pennies. Actually it’s easier to fill a large
plastic (pickle) jar with colored plastic Christmas ornaments (little pinwheels
about penny size) to spare the bother of dealing with pennies. Fill the jar with a
predetermined number of trinkets. (For instance, set the number at 463 by using
two packs of 250 trinkets minus 17.) Tell the class that the winning bidder will
receive the value of the jar where each trinket is redeemable for a penny. Thus,
though the students don’t know it the jar is worth $4.63. A buyer with a winning
bid of $3.50 would make a profit of $1.13. If the winning bid were $6.00, the
“winner” would lose $1.37.
Ask each student to write down 4 items: 1) an estimate of the value of the jar
(i.e. the number of trinkets translated into dollars); 2) a sealed bid for the jar; and
3) upper and lower bounds constructed to form a 90% confidence interval
around the true value. (Remind students what a confidence interval is.)
C. Guinea Pig Questions, Discussed
1. Is it Cancer? This example shows how difficult it is to estimate probabilities.
Students can get a good feel for the effect of prior probabilities by trying other
priors:
-- With prior probability 1 in 100, the revised probability is 1 in 6 (about a 17
fold upward revision).
-- With prior probability 20 in 100, the revised probability is 19 in 23 or 83
percent (representing about a 4 fold upward revision).
-- With prior probability 50 in 100, the revised probability is 47.5 in 50 or 95%.
In this case, Pr(cancer|+) = Pr(+|cancer) = .95 because of the fact that
Pr(cancer) = Pr(health) = 1/2.
2. Let's make a Deal. Almost all students (if they have not seen the problem
before) believe that either curtain offers a 50-50 chance of having the prize. This
is incorrect. In fact, the chance is one-third that the grand prize is behind your
chosen curtain and two-thirds that its behind the other curtain. After all, choosing
your original curtain at random offers a one-third winning chance. The fact that
you are shown an empty curtain does not change this prior probability (although
it does eliminate one curtain from consideration). Since your winning chances
are 1/3 if you “stick”, you should switch and gain a 2/3 winning chance.
Many students will still be unconvinced by this argument. One response is to
challenge students (in pairs) to simulate the setup for themselves: shuffling 2
black cards and one red card, let's say, and playing 60 times). (Always “sticking”
after a black card has been revealed will produce only 20 wins on average.) A
second argument is to suppose their are 100 curtains and one prize. After a
curtain is chosen (say, curtain 1), the host reveals 98 empty curtains (all but
curtain 71). Most students will easily see that the odds are 99 to 1 in favor of
curtain 71 over curtain 1.
After 20 years, this classic problem finally made it to the front page of The
New York Times: “Behind Monty Hall's Doors: Puzzle, Debate, and Answer,”
NYT, July 21, 1991, p. 1. The discussion and analysis are recommended.
Eighteen years later, the conundrum is still going strong.
3. Predicting Sales. Most students do what comes naturally making the
predictions: 8.8, 9.9, 11, 12.1, and 13.2, i.e. raising last year's sales for each
location by 10%. This prediction would be correct only if differences in sales
between locations were due entirely to “real” and persistent factors. The contrary
hypothesis is that last year's differences were completely random (i.e. due to
chance elements). In this case, the same prediction, namely 11, would be
suitable for all locations in the coming year. What can one say about reasonable
forecasts? Only that they lie somewhere between the two extreme hypotheses
(since sales differences are neither entirely deterministic nor entirely random). If
these components were equally important, the best forecasts would be 9.9, 10.45,
11, 11.55, and 12.1 (half way between the extreme forecasts). This
example illustrates regression toward the mean. A component of the best store's
sales (or the worst store's) is due to luck. Next period's results will not be as
good (nor as bad) but rather will move toward the overall mean.
4. Odds and Urns. Most students guess probabilities between 60 and 80
percent that the draw of 7 red and 3 white is from the “red” urn. The actual
probability is 16/17 = .94, much higher than most people expect. Students
recognize that the sample is a likely one from the red urn. What they tend to
overlook is how unusual the sample would be from the white urn (where white
balls outnumber red ones two to one). The calculation via Bayes theorem is:
=
)5(.21)5(.12
)5(.12
3737
37
=
37
7
22
2
= .94.
In the calculation, we have used the fact: Pr(Red) = Pr(White) = .5. After
cancellation, these equal priors disappear from numerator and denominator. (We've
also canceled all factors of 3.)
Remark. Here, the revised probability depends on the difference between the
number of reds and whites, not the ratio. Thus, 15r&10w is stronger evidence for
the red urn than 7r&3w, which in turn is stronger than 4r&1w. To see this
intuitively, note that 15r&10w is equivalent to 10r&10w (which calls for no
revision) followed by 5r.
Here are some other results:
=12
8
12
8
+ 8
Pr(Red|7r
&3w) =
Pr(7r&3w|Re
d)Pr(Red)
Pr(7r&3w|R)Pr(R) +
Pr(7r&3w|W)Pr(W)
= (2/
3)
7
(1/
3)
3
(0.
5)
(2/
3)
7
(1/
3)
3
(0.5)
+
(2/
3)
3
(1/
3)
7
(0.
5)
Pr(Red|1r) = (2/3)(.5)/[(2/3)(.5) + (1/3)(.5)] = 2/3.
Pr(Red|2r) = (4/9)(.5)/[(4/9)(.5) + (1/9)(.5)] = 4/5.
This last probability can also be computed one ball at a time. After 1 red, the red
urn is two-thirds likely. Use this as the new prior to be revised after drawing a
second red:
Pr(Red|2nd r) = (2/3)(2/3)/[(2/3)(2/3) + (1/3)(1/3)]
= 4/5 = .8 (the same answer as above).
5. The Cocoa Game. What went wrong? There were revolutions in Nigeria and
Ghana and outbreaks of Black Pod disease and torrential rain and railways blown
up. But something like this happens almost every year in Ghana, and there is still a
cocoa crop. Just a large enough crop to satisfy demand at “normal” prices. The
lesson here is to know which information is meaningful and which should be
“discounted” or ignored.
A Related Question: Yesterday, Company A reported that its earnings were up 14
percent over last year. Today, the price of the company's common stock fell by two
points. The explanation? Stock watchers had expected earnings to be up by 15 to
20 percent and the stock's previous price reflected this. Thus, announced earnings
(though solid) were a disappointment relative to expectations; this prompted the
price drop.
6. Finding the Best Item
a. Here is a strategy that improves upon random choice. Observe but bypass the
first prize. Select the second prize only if it is better than the first; otherwise go on
and select the third prize. We list below the six distinct (equally likely) orderings
of the prizes:
First Prize best best 2nd best 2nd best worst worst
Second Prize 2nd best worst best worst best 2nd best
Third Prize worst 2nd best worst best 2nd best best
The italicized items show the prizes selected in each ordering following the
suggested strategy. In three cases out of six, the individual obtains the best prize,
in two cases the second-best prize, and in only one case the worst prize. Observing
the relative merits of the prizes and making a contingent choice (even though there
is no going back) improves your outcomes.
b. For a large number of items, the optimal strategy is to observe but bypass a
certain fraction of the total, then select a subsequent item if and only if it’s the best
item of all you have seen.
Remarks. For an arbitrary number of items (not only three), this is variously
called the “secretary problem” (interviewing and selecting the best secretary), the
“hotel problem” (deciding to stop at one of a number of hotels situated hours apart
along a cross-country route), or the “art gallery problem” (in the gallery you are
allowed to view paintings one at a time and choose only one without going back to
a previous painting).
For a large number of items, a remarkable result has been proven. The optimal
strategy is to bypass about 37% of the items before choosing the best item to date.
The chances of obtaining the best item using this strategy is also 37%! (The
optimal proportion and resulting chance are 1/e, where e = 2.71828 is Euler's
constant. For a derivation, see M. DeGroot, Optimal Statistical Decisions,
McGraw-Hill, 1970.) This is a truly remarkable result. Even if there are one
million items, this strategy will select the best of the million 37% of the time. (By
contrast, a random selection would provide only a one-in-a-million chance.) Here
is one way to see why the sequential method does so well. Suppose you were to
observe and bypass half the items. Consider the case that the second-best item
appears in the first half of the total number while the best item appears in the
second half. This occurs with probability .25 (since there is a .5 chance that each
item is in the specified half). In this case, the sequential strategy must stop at the
best item. (Make sure you understand why.) Thus, the winning chances are at least
.25 regardless of the number of items. By bypassing only 37% of the items and
accounting for other ways of winning, one can raise the overall winning chances to
37%.
7. Bidding for an Item of Unknown Value. This example introduces
complications posed by uncertainty. Each bidder is uncertain about the value of the
jar and also about the level of competitors’ bids. Before revealing the true value of
the jar, ask a number of students to reveal their estimates. Students will be
surprised by the wide dispersion of estimates. Ask the class what the distribution
might look like if it were plotted. A number of students are sure to suggest a
bell-shaped normal curve. By shows of hands, find roughly the median of the class
estimates and sketch a normal curve with this mean. Then ask students how they
bid. Compare bids to estimates to emphasize the logic of placing bids below
estimates. Note that the optimal degree of shading involves a tradeoff between the
chance of winning and the profit from winning. (Finding an optimal solution is a
complex problem.) Superimpose the bid distribution next to the estimate
distribution. Your graph should resemble Figure 13.3.
Now is the time to suggest the possibility of the winner’s curse by raising two
hypotheses. First, suppose estimates are unbiased, i.e. centered around the true
value of the item. (This seems reasonable.) Second, suppose that the estimates and
bids are quite disperse. (Also reasonable.) These two points imply that the highest
bids (the shaded tail in Figure 13.3) are above the true value of the item. On
average, the winning bidder loses money on the acquisition. At this point, you may
ask some of the high bidders in the class, whether this argument worries them. (Do
they still want to play for real or only for fun?) Find the highest bid. Then, reveal
the value of the item. With 20 or more students in the class, nine times out of ten,
the winning bidder will fall prey to the winner’s curse, i.e. pay more than the item’s
value.
Having made the main point, you can raise some other questions. What factors
increase the likelihood and magnitude of the winner’s curse? As discussed in the
text, the winner’s curse increases with the degree of uncertainty and the number of
bidders. Does the winner’s curse happen in real life? Yes, bidding for off-shore oil
leases, competitive tender-offers, and free-agent bidding for athletes are just three
examples. How well did you (the student) gauge your degree of uncertainty? Most
individuals are overconfident of their estimation abilities. Ask students if the true
estimate fell between their lower and upper bounds. Typically, less than half the
students answer in the affirmative by a show of hands. By construction, the
intervals should bracket the true value 90 percent of the time. Students are
surprised that their knowledge is so uncertain.

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