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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/   
    
Associated Topics:
High School Analysis
High School Probability
High School Sets

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