Line Segments and Size of Infinites
Date: 03/19/97 at 23:02:54 From: ALLEN Subject: Line Segments and Size of Infinites Take a line segment. Divide it into three segments such that one is half the length of the line and the other two are a quarter of the line segment. Choose a point at random along this line segment. What is that probability that this point lands in the 1/2 segment? 50 percent because that segment is half the line, or 33 percent because each segment has an infinite number of points, so any segment is equally likely? Thank you!
Date: 03/20/97 at 12:23:08 From: Doctor Rob Subject: Re: Line Segments and Size of Infinites The answer to your question lies in the phrase "AT RANDOM". Implicit in the statement is that the point is chosen from the uniform distribution on the points of the line segment (although it is not explicitly stated). This means that the probability of picking a point in any subset of total length ("measure") x is given by x divided by the total length ("measure") of the line segment. In your case, the probabilities of the point lying in the three subsets is 1/2, 1/4, and 1/4, respectively, since the probability is proportional to the lengths. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 03/20/97 at 23:11:04 From: ALLEN Subject: Line Segments and Size of Infinites Dear Dr. Rob, Thank you for your response. However, doesn't each line segment contain an infinite number of points, and therefore the probability of picking a point in any line segment would be equal, since all infinities are equal?
Date: 03/21/97 at 10:54:00 From: Doctor Rob Subject: Re: Line Segements and Size of Infinites Yes, every line segment contains an infinite number of points. This, however, does not affect the probability. Perhaps looking at it this way would help. Split the 1/2-long segment in two equal parts, so that the entire segment is divided into four parts, each 1/4 as long as the whole. Now I think that by any analysis, the chance of picking a point at random in any of the four parts should be equal, so all must have probability 1/4. Now undo your splitting, and you should be able to see why the probability of the long segment should be 1/2, not 1/3, and the smaller segments each 1/4, not 1/3. In essence, all line segments have the same number of points, but they cannot all have the same probability, or else the probabilities cannot add properly. Remember the rule that if events are mutually exclusive, the probability that one or the other happens must be the sum of the probabilities that the two events happen individually. The correct substitute for the number of points to describe the size of an infinite set is something called "measure", and the measure which is appropriate for the uniform distribution on a line segment is the length of the segment. Of course, in the case of finite sets of points, the measure which is appropriate is the number of points in the set, but this breaks down in the case of infinite sets. By the way, not all infinities are equal. In the case at hand, however, all the sets in question have the same cardinality C, the same as the cardinality of the real numbers. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 03/21/97 at 17:35:33 From: ALLEN Subject: Line Segments and Size of Infinites Dear Dr. Rob, You made your point that if you split the line segment 1/2 into two sections of 1/4 so that you have four sections of the same size, then the probability that the point lies in any section is 1/4. If you add probabilities for the two 1/4 segments that make up the 1/2 line segment, you get a probability of 1/2. However, infinity + infinity = infinity, and when you deal with infinite sets, the rules change. Also, by the same point you make, I could split one of the 1/4 sections in two 1/8 sections, the add the probability of these together. Then I would get the probability of 1/2 for one of the 1/4 sections!
Date: 03/21/97 at 19:43:00 From: ALLEN Subject: Line Segments and Size of Infinites Dear Dr. Rob, After further review of your letter, I came across the last paragraph, where you said that the cardinality of all the sets is equal. So, doesn't that mean that there is an equal chance of choosing at random a point from any of the line segments?
Date: 03/28/97 at 12:06:08 From: Doctor Ken Subject: Re: Line Segments and Size of Infinites Hi Allen - I saw this question-answer exchange and couldn't resist jumping in to offer another perspective. I think it's great that you're wondering about these kinds of things! First of all, when you are trying to figure out probabilities that involve making choices from infinite sets, it's seldom a simple matter of counting up the number of things in the various sets and comparing them. See, when we compare the sizes of these infinite sets, there are certain kinds of statements that don't really make sense. For instance, you can't really say that the infinite set A is twice as big (has twice as may elements) as the infinite set B, because someone could just turn around and reply that B is twice as big as A. Each of these claims could be made by matching up each element of set A with 2 elements of B, and then vice versa. So if we want to deal with these kinds of probability questions in a meaningful way, it's not going to be by counting up the size of the sets and comparing them. The key to this question is in the phrase "at random." How do we choose a point at random on the line segment? Well, to answer this question I'm going to play the part of Your Worst Enemy. Your Worst Enemy always tries to give you what you don't want. For instance, if you tell me to start picking random numbers between 0 and 1, I'll give you these: .5, .5, .5, .5, .5, .5, ..., continuing in the obvious pattern, and you'll got upset (hey, I'm Your Worst Enemy). So then you might say something like "don't make all your numbers the same." At which point I give you .5, .55, .555, .5555, ..., continuing in the obvious pattern. So it becomes apparent that you're going to have to be more clever. What if you say this: "choose your numbers so that if I pick any interval and it has length L, eventually about 1/L of your points are in that interval." Then I have essentially no choice but to give you what you really want, a string of random-looking numbers. And you're happy. So let's re-examine what you told me to do. You phrased your conditions in terms of the length of line segments, so it's no surprise that the probability (of a point chosen at random being in a certain region) will rely on the length of the interval and not the number of points in the interval. Here's another explanation from a fellow Dr. Math (Steven) that you might be interested in. It's from yot another different point of view: Don't look at the point, but look at the distance of the point from the start of the line. The probability that a point lies in the first half of the section is the probability that the length of the line segment from the start of line to the picked point is less than or equal to 1/2. The probability that the point lies in the second section is the probability that the length of the line segment from the start to the point is between 1/2 and 3/4 inclusive. The probability that the point is in the last fourth is the probability that the line segment is between 3/4 and 1 inclusive. Another thing: what if you told me to pick a point in the segment but you didn't tell me how you had divided it up? -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 04/01/97 at 19:57:23 From: ALLEN Subject: Re: Line Segments and Size of Infinites Dr. Math, Thank you for your response about infinite sets. I am still confused about how to determine the probability of infinite sets, uncountable and countable. How do the rules change when dealing with infinities where change in involved? For example, the set of even numbers is infinite, the set of whole numbers is infinite, but the chance of selecting an even number from the set of whole numbers is 1/2. Why?
Date: 04/02/97 at 22:50:28 From: Doctor Rob Subject: Re: Line Segements and Size of Infinites You are not alone in your confusion about infinite sets. It was not until about 100 years ago that a mathematician named Georg Kantor put infinite sets on a firm mathematical footing. Until then, all sorts of strange errors were made using infinite sets. For infinite sets, the number of points is not the correct way to try to figure probabilities. Every line segment of any length has the same number of points, but not every one can be assigned the same probability, if the points are chosen "at random". And that is the place where things get complicated: what do we mean by "at random"? If we pick a person "at random" from all living human beings, and measure his height, we will not get any numbers less that 12 inches or more than 120 inches, and the ones near 66 inches will be much more likely than much smaller or much larger ones. This is a way of picking a number from a "non-uniform distribution." (By the way, we lump all heights within 1/16" together into one value, so there are only about 8000 possible "heights" for people.) We can talk about picking a height "at random" from these 8000 using this method, but the chance of picking each height varies, and is not a constant. If we pick one number from a set of "size" n with the "uniform distribution", we mean that the chance of the number being in any set of "size" k will be k/n, and in particular the chance of it being a particular number in the set is 1/n. Here the number of points in the set works fine for "size". This is precisely what we mean by the "uniform distribution": all size-k subsets have the same probability of k/n, so the probability is uniform (the same everywhere). If we pick one point from a line segment of "size" n with the "uniform distribution", we mean that the chance of the point being in any subset of "size" k will be k/n. Since all line segments have the same number of points, number of points will *not* work for "size", and instead we have to use something called a "measure" of a set. In the case of line segments, length works fine for "size". Here are some rules that "size" has to satisfy. Let S be any set, T be any subset of S, and R disjoint from S (R and S have nothing in common). Then: 1) The size of S is never negative. 2) The size of T cannot be greater than the size of S. 3) The size of the set obtained from joining S and R together is exactly equal to the size of S plus the size of R. 4) If S and R are congruent, they have the same size. You can check that what I said above is true about using "number of points" for "size" for finite sets. A few examples should convince you that using "length" for "size" for line segments also works. It is rule number 4 that rules out the 1/3 probabilities of your original problem. Probabilities also have rules, as you probably know. The probability that an event doesn't happen is 1 minus the probability that it does. Probabilities cannot be negative or greater than 1. If two events cannot both happen, then the probability that one *or* the other happens is the sum of the probabilities that each happens. These are some of those rules. You want to compute probabilities as the quotient of the size of the set of good outcomes over the size of the set of all possible outcomes. If you used "cardinality" for "size" in line segments, then all sets would have the same "size" C, except for the line segment of a single point, which would have "size" zero. This fits the above rules for "size", but these "sizes" would not allow you to compute probabilities by division, since infinity over infinity does not have a fixed value. Now about your example of even integers. The way to think of this is to just consider all the integers in an interval [-n, n+1], where n > 0 is an integer. There are 2*(n+1) of them. No matter whether n is even or odd, n+1 integers are even and n+1 are odd. The probability of picking an even integer from this interval is (n+1)/(2*(n+1)) = 1/2. But this is true no matter what the size of n, so in some sense, the same will be true if we let n approach infinity, and let the interval include all the integers. This kind of argument, where a variable wanders off to infinity, is something you will encounter when you study limits in calculus, probably late in your high school career. I hope this helps. Actually, I rather expect I have just confused you even more. You are thinking very deep thoughts! Explanations for these ideas are not easy, and involve very hard ideas. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 04/03/97 at 19:25:58 From: ALLEN Subject: Line Segments and Size of Infinites Dear Dr. Rob, You said that when dealing with infinite sets, you cannot compute probabilities by division since infinity/infinity does not have a fixed value. Why not?
Date: 04/04/97 at 16:41:02 From: Doctor Rob Subject: Re: Line Segments and Size of Infinites The definition of a/b is that number c such that b*c = a. When both a and b are infinite, you get infinity*c = infinity for this last equation. But infinity*1 = infinity*2 = infinity*3 = ... = infinity, so c = 1, 2, 3, or any other number works fine. As a result, you cannot say that infinity/infinity = 1, 2, 3, or any other number, since all of them are possible (along with 7/3, 8.346, and any other number). Infinity/infinity is one of the so-called "indeterminate forms." Similarly 0/0 is one, too. For a discussion of the indeterminate form 0^0, see the URL http://mathforum.org/dr.math/faq/faq.0.to.0.power.html -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum