WEBVTT
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PROFESSOR: In this
segment, we will
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look at the famous
example, which
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was posed by Comte de Buffon--
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a French naturalist--
back in the 18th century.
00:00:11.720 --> 00:00:13.750
And it marks the
beginning of a subject
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that is known as the subject
of geometric probability.
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The problem is pretty simple.
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We have the infinite
plane, and we
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draw lines that are
parallel to each other.
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And they're spaced
apart d units.
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So this distance here is d.
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And the same for
all the other lines.
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We take a needle that has
a certain length-- l--
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and we throw it at
random on the plane.
00:00:44.290 --> 00:00:47.890
So the needle might
fall this way, so
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that it doesn't cross any line,
or it might fall this way,
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so that it ends up
crossing one of the lines.
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If the needle is long enough,
it might actually even end up
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crossing two of the lines.
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But we will make the assumption
that the length of the needle
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is less than the distance
between the two--
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between two adjacent
lines, so that we're
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going to have either
this configuration,
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or that configuration.
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So in this setting, we're
interested in the question
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of how likely is it that the
needle is going to intersect
00:01:22.720 --> 00:01:28.070
one of the lines if the needle
is thrown completely at random?
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We will answer this question,
and we will proceed as follows.
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First, we need to
model the experiment--
00:01:34.510 --> 00:01:37.450
the probabilistic
experiment-- mathematically.
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That is, we need to define
an appropriate sample space,
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define some relevant
random variables,
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choose an appropriate
probability law,
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identify the event of
interest, and then calculate.
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Let us see what it takes to
describe a typical outcome
00:01:54.100 --> 00:01:55.670
of the experiment.
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Suppose that the needle fell
this way, so that the nearest
00:02:01.240 --> 00:02:03.790
line is the one above.
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And let us mark here the
center of the needle.
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One quantity of interest
is this vertical distance
00:02:12.340 --> 00:02:15.920
between the needle
and the nearest line.
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Let us call this quantity x.
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We're using here a
lowercase x, because we're
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dealing with a numerical value
in one particular outcome
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of the experiment.
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But we think of this x
as being the realization
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of a certain random
variable that we
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will denote by capital X.
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What else does it take
to describe the needle?
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Suppose that the
needle had fallen
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somewhere so that it is at
the same vertical distance
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from the nearest
line, but it has
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an orientation of this kind.
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This orientation
compared to that one
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should make a difference.
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Because when it falls
that way, it's more likely
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that it's going to cut the next
line as opposed to this case.
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So the angle that the needle is
making with the parallel lines
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should also be relevant.
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So let us give a name to
that particular angle.
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So let's extend that line until
it crosses one of the lines.
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And let us give a name to
this angle, and call it theta.
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So if I tell you
x and theta, you
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know how far away the needle
is from the nearest line,
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and at what angle it is.
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It looks like these are
two useful variables
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to describe the outcome
of the experiment, so let
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us try working with these.
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So our model is going to involve
two random variables defined
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the way we discussed
it just now.
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What is the range of
these random variables?
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Since we took x to be the
distance from the nearest line,
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and the lines are
d units apart, this
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means that x is going to
be somewhere between 0
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and d over 2.
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How about theta?
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So the needle makes two angles
with the part of the line.
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It's this angle, and
the complimentary one.
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Which one do we take?
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Well, we use a
convention that theta
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is defined as the acute
angle that the direction
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of the needle is
making with the lines,
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so that theta will vary over
a range from 0 to pi over 2.
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And our sample space
for the experiments
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who will be the set of
all pairs of x and theta,
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that satisfy these
two conditions.
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These will be the
possible x's and thetas.
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Having defined the
sample space, next we
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need to define a
probability law.
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At this point, we do not want to
make any arbitrary assumptions.
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We only have the words
completely at random to go by.
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But what do these words mean?
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We will interpret them to mean
that there are no preferred x
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values, so that all x
values are- in some sense--
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equally likely.
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So we're going to assume that
x is a uniform random variable.
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Since it is uniform,
it's going to be
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a constant over this range.
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And in order to integrate
to 1, that constant
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will have to be 2 over d.
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And we understand that the PDF
of x is 0 outside that range.
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Similarly for theta,
we do not want
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to assume that some
orientations are more
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likely than other orientations.
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So we will again assume
a uniform probability
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distribution.
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And therefore, that PDF
must be equal to 2 over pi
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for theta's over this
particular range.
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So far, we have specified
the marginal PDFs of each one
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of the two random variables.
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How about the adjoined PDF?
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In order to have
a complete model,
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we need to have a
joint PDF in our hands.
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Here, we're going to make the
assumption that x and theta are
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independent of each other.
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And in that case,
the joint PDF is
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determined by just taking the
product of the marginal PDFs.
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So the joint PDF is
going to be equal to 4
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divided by pi times d.
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By this point, we
have completely
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specified a probabilistic model.
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We have made some assumptions,
which you might even
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consider arbitrary.
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But these assumptions
are a reasonable attempt
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at capturing the idea
that the needle is
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thrown completely at random.
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This completes the
subjective part--
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the modeling part.
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The next step is much
more streamlined.
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There's not going
to be any choices.
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We just need to consider
the event of interest,
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express it in terms of
the random variables
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that we have in our hands, and
then use the probability model
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that we have to
calculate the probability
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of this particular event.
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So let us identify
the event of interest.
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When will the needle
intersect the nearest line?
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This will depend
on the following.
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We can look at the vertical
extent of the needle.
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By vertical extent,
I mean the following.
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Let's see how far
the needle goes
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in the vertical direction,
which is the length
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of this green segment here.
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In this example, the
vertical extent of the needle
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is less than the distance
from the next line.
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And we do not have
an intersection.
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If the figure was
something like this,
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the vertical extent of the
needle would have been that,
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but x would have been
just this little segment.
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The vertical extent is
bigger than x and the needle
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intersects the line.
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So we have an
intersection if and only
00:08:42.070 --> 00:08:44.500
if the vertical extent--
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which is this vertical
green segment--
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is larger than the distance x.
00:08:51.520 --> 00:08:56.120
Or equivalently, if x is less
than the vertical extent.
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So we will have an
intersection if x is less than
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or equal to the vertical
extent of the needle.
00:09:06.850 --> 00:09:10.210
Now, how big is this
vertical extent?
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Let's use some
trigonometry here.
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This angle here is theta, so
this angle here is also theta.
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Here, we have a right
triangle and the hypotenuse
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of this triangle is l over 2.
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This angle is theta, therefore
this vertical segment
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is equal to l over
2 times sine theta.
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So this is the
geometrical condition
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that describes the event
that the needle intersects
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the nearest line.
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And all we need to do
now is to calculate
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the probability of this event.
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So here is what we have so far.
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This is the picture that
we had before, but drawn
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in a somewhat nicer way.
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This is the joint PDF
that we decided upon.
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And we wish to calculate
the probability
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of this particular event--
00:10:15.460 --> 00:10:20.460
that x is less than or equal
to l over 2 sine theta.
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How do we calculate
the probability
00:10:26.400 --> 00:10:29.740
of an event that has to do
with two random variables?
00:10:29.740 --> 00:10:34.980
What we do is we
take the joint PDF--
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which in our case
is four over pi d--
00:10:39.390 --> 00:10:49.140
and integrate it over the set
of x's and theta's for which
00:10:49.140 --> 00:10:51.180
the PDF is non-zero.
00:10:51.180 --> 00:10:55.380
So it's only going to be over
x's and theta's in those ranges
00:10:55.380 --> 00:10:58.920
and also, only for
those x theta pairs
00:10:58.920 --> 00:11:01.540
for which the event occurs.
00:11:01.540 --> 00:11:03.840
So what are these pairs?
00:11:03.840 --> 00:11:08.410
This event can occur
with any choice of theta.
00:11:08.410 --> 00:11:16.050
So theta is free to vary
from 0 up to pi over 2.
00:11:16.050 --> 00:11:17.610
How about x?
00:11:17.610 --> 00:11:22.380
For this event to
occur, x can be
00:11:22.380 --> 00:11:26.970
anything that is non-negative
as long as it is less than
00:11:26.970 --> 00:11:29.490
or equal to this number.
00:11:29.490 --> 00:11:32.220
So the upper limit
of this integration
00:11:32.220 --> 00:11:38.630
is going to be l over
2 times sine theta.
00:11:38.630 --> 00:11:42.710
And all we need to do now is to
evaluate this double integral.
00:11:42.710 --> 00:11:45.560
Let's start with
the inner integral.
00:11:45.560 --> 00:11:48.440
Because we're just
integrating a constant,
00:11:48.440 --> 00:11:56.520
the inner integral
evaluates to the quantity
00:11:56.520 --> 00:11:58.920
that we're integrating--
the constant that we're
00:11:58.920 --> 00:12:02.580
integrating-- which is 4
times pi d times the length
00:12:02.580 --> 00:12:05.430
of the interval over
which we're integrating,
00:12:05.430 --> 00:12:07.855
which is l over 2 sine theta.
00:12:12.520 --> 00:12:15.860
And now we need to carry
out the outer integral.
00:12:15.860 --> 00:12:20.490
Let us pull out the constants,
which is this 4 with this 2
00:12:20.490 --> 00:12:22.290
give us a 2.
00:12:22.290 --> 00:12:27.220
We have 2l over pi d.
00:12:27.220 --> 00:12:31.280
And then the integral from 0
to pi over 2 of sine theta.
00:12:31.280 --> 00:12:35.500
Now the integral of sine
theta is minus cosine theta.
00:12:35.500 --> 00:12:39.250
And we need to evaluate
this at 0 and pi over 2.
00:12:39.250 --> 00:12:41.660
This turns out to be equal to 1.
00:12:41.660 --> 00:12:47.400
So the final result
is 2 l over pi d.
00:12:47.400 --> 00:12:49.890
And this is the final
answer to the problem
00:12:49.890 --> 00:12:53.710
that we have been considering.
00:12:53.710 --> 00:12:57.700
And now, a curious thought.
00:12:57.700 --> 00:13:03.310
Suppose that you do not
know what the number pi is
00:13:03.310 --> 00:13:05.950
and all you have
in your hands is
00:13:05.950 --> 00:13:11.500
your floor, lines drawn on
your floor, and the needle.
00:13:11.500 --> 00:13:14.320
And you do know the length
between adjacent lines
00:13:14.320 --> 00:13:15.490
on your floor.
00:13:15.490 --> 00:13:18.140
And you do know your
length of your needle.
00:13:18.140 --> 00:13:21.070
How can you figure
out the number pi?
00:13:21.070 --> 00:13:25.120
Take your needle, throw it
at random a million times,
00:13:25.120 --> 00:13:29.080
and count the frequency with
which the needle ends up
00:13:29.080 --> 00:13:30.730
crossing the line.
00:13:30.730 --> 00:13:32.230
If you believe
that probabilities
00:13:32.230 --> 00:13:35.200
can be interpreted
as frequencies,
00:13:35.200 --> 00:13:39.250
the frequency that you observe
gives you a good estimate
00:13:39.250 --> 00:13:40.600
of this probability.
00:13:40.600 --> 00:13:44.560
So it gives you a good estimate
of this particular number.
00:13:44.560 --> 00:13:47.800
And if you know the length of
your needle and of the distance
00:13:47.800 --> 00:13:50.530
between the different
lines, you can
00:13:50.530 --> 00:13:55.660
use the estimate of that number
to determine the value of pi.
00:13:55.660 --> 00:13:59.530
This is a so-called
Monte Carlo method,
00:13:59.530 --> 00:14:03.430
which uses simulation to
evaluate experimentally
00:14:03.430 --> 00:14:07.690
the value, in this case,
of the constant pi.
00:14:07.690 --> 00:14:12.250
Of course, for pi, we have much
better ways of calculating it.
00:14:12.250 --> 00:14:15.660
But there are many applications
in engineering and in physics
00:14:15.660 --> 00:14:18.430
where certain quantities
are hard to calculate,
00:14:18.430 --> 00:14:22.960
but they can be calculated
using a trick of this kind
00:14:22.960 --> 00:14:24.700
by simulation.
00:14:24.700 --> 00:14:26.990
Here's a typical situation.
00:14:26.990 --> 00:14:29.500
Consider the unit cube.
00:14:29.500 --> 00:14:31.810
And for simplicity,
I'm only taking
00:14:31.810 --> 00:14:33.710
a cube in two dimensions.
00:14:33.710 --> 00:14:35.710
But in general, think
of the unit cube
00:14:35.710 --> 00:14:40.390
in n dimensions, which is an
object that has unit volume.
00:14:40.390 --> 00:14:46.590
Inside that unit cube, there
is a complicated subset
00:14:46.590 --> 00:14:51.000
which is described maybe by
some very complicated formulas.
00:14:51.000 --> 00:14:52.920
And you want to
calculate the volume
00:14:52.920 --> 00:14:56.070
of this complicated subset.
00:14:56.070 --> 00:14:58.630
The description of the
subset is so complicated
00:14:58.630 --> 00:15:03.280
that using integration,
multiple integrals, and calculus
00:15:03.280 --> 00:15:05.290
is practically impossible.
00:15:05.290 --> 00:15:06.700
What can you do?
00:15:06.700 --> 00:15:11.080
What you can do is to start
throwing at random points
00:15:11.080 --> 00:15:13.330
inside that unit cube.
00:15:13.330 --> 00:15:14.470
So you throw points.
00:15:14.470 --> 00:15:15.640
Some fault inside.
00:15:15.640 --> 00:15:18.070
Some fall outside.
00:15:18.070 --> 00:15:21.070
You count the frequency
with which the points
00:15:21.070 --> 00:15:24.520
happen to be inside your set.
00:15:24.520 --> 00:15:29.620
And as long as you're
throwing the points uniformly
00:15:29.620 --> 00:15:34.150
over the cube, then
the probability
00:15:34.150 --> 00:15:39.970
of your complicated set is going
to be the volume of that set.
00:15:39.970 --> 00:15:42.190
You estimate the
probability by counting
00:15:42.190 --> 00:15:46.930
the frequency with which
you get points in that set.
00:15:46.930 --> 00:15:50.830
And so, by using these
observed frequencies,
00:15:50.830 --> 00:15:55.210
you can estimate the
volume of a set--
00:15:55.210 --> 00:15:57.670
something that might
be very difficult to do
00:15:57.670 --> 00:16:00.100
through other numerical methods.
00:16:00.100 --> 00:16:04.840
It turns out that these days,
physicists and many engineers
00:16:04.840 --> 00:16:07.885
use methods of this
kind quite often
00:16:07.885 --> 00:16:11.610
and in many important
applications.