Let's apply the notion of pattern you've been thinking about to this point to one of the kinds of problems we all face. As we go about the business of living our lives we often would like to be able to examine a situation we find ourselves in and, after thinking about it, decide what the best thing for us to do in that situation is.
Think back with me to the set of questions about what we ought to do as we reason about those practical questions I asked about in an earlier lesson.
On the simplest level making such decisions is the same as deciding whether or not to take an umbrella or wear a raincoat after we hear a weather forecast. But we do very much the same thing when we decide whether or not to change jobs or have an operation. In all these cases we'd like to know how to make the best choice, how to bet on an outcome. We can of course just "take our chances". If the weather man says there is a 50% chance of rain and you decide to take your chances, what would you do?
Choose an action:
Whatever your choice, you made your decision based on how great you thought the risk was and which outcome, getting wet or carrying around an unneeded raincoat, was least desirable. All decision making has the same characteristics, you have to choose between risks and between outcomes.
As we noted mathematics could be used to think about making choices about questions like these. In the most straight forward application we all might ask at least two mathematical questions about each of these situations. The first is how great is the "risk" in any situation. The second is how great is the benefit or loss that comes from an outcome.
Before we deal with any numbers however we need to think about defining in some way what we mean by risk and benefit. Let's return to the weather question that heads our list.
Let's assume we got the same prediction from our weather forecaster. He says theere is a 50% chance of precipitation for the day after tomorrow.
Our hypothetical driver, Shirley, facing a 300 mile trip now has some sense of what the weather might be like.
Briefly answer each of these questions:
Assume that Shirley is driving in Minnesota in January on a shopping trip, that the amount of precipitation could be as much as ten or more inches of snow in some parts of the forecast area, and that a cold snap with sub zero temperatures is pushing through the area rapidly, producing high winds and drifting.
If Shirley was traveling to a job interview for a position that would double her current salary, and if this was her last chance for an interview before a hiring decision had to made:
In other words, the risks of traveling in mid-winter may outweigh the potential benefits of making the white sale. The potential benefits of a the mid-winter trip for a job interview may outweigh the risks of running into the storm. The chances of running into the snow are the same in either case. We have a 50/50 chance of encountering snow. What changes is our willingness to run the risks in question, for the benefits that may follow. Additional information might affect our decision further.
For example, what if Shirley was 90% sure that if she interviewed she'd get the job. How should that affect her decision?
RESPONSE
If you said Shirley ought to be more willing to go, you're saying that the more likely the benefits are, the more willing Shirley should be to run the risks in question. In fact, you're doing exactly what oil well drillers, stock brokers, farmers and other decision makers do all the time, you're trying to figure out how to minimize risks and maximize benefits.
Now that we've reasoned a bit about risks and benefits, we can turn to thinking a bit more about what mathematics does to help us consider the way chance plays a part in decisions, and in the process see again how patterns and predictions are related to each other. Suppose Shirley' gotten her job and is now facing another decision. For the moment she isn't ready to move but has to start work immediately. Should she move? Its mid-winter and her friendly forecasters have told her that the chances of heavy snow falling on any given day during the next two months are 50/50 (a bad winter indeed). Shirley wants to know, if she's driving every day, what the chances are of her getting stuck in a serious storm at least once before she moves.
Notice this isn't a question about benefits and risks. Shirley will have to consider those questions after we figure out what the chances or probability of getting stuck are. If the chance of getting stuck on any one day are 50/50 and we assume that Shirley keeps driving without getting stuck, what happens to her chances of getting stuck on her next trip. Well in one sense the answer is simple.
What are the chances of Shirley getting stuck on any one trip?
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RESPONSE
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If you said 50/50 you are absolutely right. No matter how many time Shirley makes the
trip her chances of getting stuck on any particular trip are 50/50. However if I ask a
slightly different version of the question, I'll get an answer that may help Shirley make a
better decision. Suppose I ask
QUESTION
How does the risk of getting stuck increase as the number of trip increases?
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RESPONSE
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If you said the risk increases you're right. In fact the real question is how much does it
increase? If we think of this a problem similar to what happens when we flip a fair coin,
we can develop a model that helps us to determine how likely the chances are of getting
stuck least once, given a certain number of trips. We'll begin by listing ALL the possible
outcomes of combinations of trips for small numbers of trips and then consider whether
or not we see a pattern that lets us know what happens as we increase the number of
trips. We'll identify making it back and forth with M for making it, and getting stuck will
be S for getting stuck. If we think of the coin flipping analogy, its like H for flipping
heads and T for flipping tails.
In the first trip we have only two possible outcomes making it or getting stuck. Shirley's
chances of making it are 50/50.
Trip 1
M S
If we consider the first and second trip together we have four possible outcomes: making it on both trips;
making it on the first trip and getting stuck on the second;
getting stuck on the first trip and making it on the second;
and getting stuck on both trips.
Shirley's can only get through 2 trips without getting stuck 1 of 4 ways; her chances are
now 25/75.
Trip 1
M S
Trip 2
M S M S
If we add a third trip the outcomes map looks like this.
Trip 1
M S
Trip 2
M S M S
Trip 3
M S M S M S M S
We have 8 possible outcomes:
making it on all three trips;
making in on trips one and two getting stuck on three;
making it on one, getting stuck on two and making it on three;
making it on one, getting stuck on two and getting stuck on three;
getting stuck on one, making it on two, making it on three;
getting stuck on one, making it on two, getting stuck on three;
getting stuck on one, getting stuck on two and making it on three;
getting stuck all three times.
Shirley can get through 3 trips without getting stuck 1 of 8 ways; her chances are 12/88.
If we count these outcomes in a slightly different way a ver interesting and useful pattern
appears. Not only have we found out that chances of getting through on every trip grow
smaller, but so do the chances of getting stuck on every trip. In fact there is only one
chance that either of these things can happen however many trips we take. Another
way to map our outcomes then is as follows.
One total success, three mixed results after success, three mixed results after failure and one total failure.
QUESTION
Using this scheme for counting results, guess at what the pattern for Trip 4 would be.
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RESPONSE
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If you guessed
1 4 6 4 1
you've got the pattern and, by the way, are on your way to generating a very important
collection of numbers called the binomial distribution. If you didn't guess this time you'll
get another chance in a minute. What we can use our pattern to figure out without
going to all the trouble of listing all outcomes is what Shirley's chances of making four
trips without getting stuck are. There are 16 outcomes and only 1 way to make four
trips without getting stuck. Shirley has 1 chance in 16 of making four trips without
getting stuck; her chances are 6/94.
Lets look at our pattern again.
TRIP 1
1 1
TRIP 2
1 2 1
TRIP 3
1 3 3 1
TRIP 4
1 4 6 4 1
And now without labeling trips:
1 1
1 2 1
1 3 3 1
1 4 6 4 1
QUESTION
Guess again what the pattern would give us for trip 5. If you want a clue you can add numbers in the last row (trip) to get the next row (trip).
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RESPONSE
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If you've figured out the pattern you answered:
1 5 10 10 5 1
Lets look at this set of numbers one more time and see how any regularities you notice.
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Notice what is happening in the second and
second to last position in each row (trip).
Let's add one more row (trip).
QUESTION
What numbers complete the pattern for the next row (trip)?
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RESPONSE
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I suspect you wrote:
1 6 15 20 15 6 1
Our pattern is now:
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
If Shirley wants to take six trips without getting stuck she has only one chance in 64 of
succeeding; her chances are 2/98; or the probability of making six trips without getting
stuck is .015625 (1/64). Those aren't very good odds, so, depending on the costs of
getting stuck, the costs of moving fast and the benefits of her new job, Shirley may want
to move fairly quickly or negotiate to start her new job a little later.
Thinking about Shirley's problem has taken us a good deal beyond that problem. The symmetry of the set of numbers we generated has both practical and theoretical importance in mathematics. We used it to explore the effects of chance on courses of action and what we discovered is the binomial distribution , a general set of numbers that has many applications in mathematics. For example, this set of numbers describes
the coefficients of a whole set of algebraic functions involving polynomials.
Without bothering to get into the algebra you can see how the distribution works.
Suppose I get an algebra problem of the following sort.
Expand the following expression:
(a+b)²
Without bothering to do the actual expansion by multiplying (a+b) times (a+b) many algebra students would be able to say that:
(a+b)² = a²+2ab+b²
If we rewrite that expression including all the coefficients of each term in the expression it would read:
(a+b)² = 1a²+2ab+1b²
QUESTION
Given the binomial distribution pattern we've been studying how would you guess we would expand:
(a+b)³
You might want to think of this as (1a+1b)³, and the middle terms of the expanded expression involve a²b and ab².
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RESPONSE
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If you guessed that the coefficients of all terms followed the pattern of the binomial distributions you probably wrote:
(a+b)³ = 1a³+3a²b+3ab²+1b³
This relationship holds true for all algebraic expansions of the form (a+b).
Patterns as symmetrical and powerful as the binomial distribution are what mathematicians have been looking for and finding since the beginnings of human thought, in fact since a time before mathematicians thought of themselves as mathematicians. As it turns out such patterns have general applications in a wide variety of situations where human beings use numbers to help them understand and predict the way in which things will happen if certain conditions occur. We'll be applying such thinking to a wide variety of such situations in the remainder of this course.