1
00:00:00,292 --> 00:00:01,750
PROFESSOR: In this
segment, we will

2
00:00:01,750 --> 00:00:04,000
look at the famous
example, which

3
00:00:04,000 --> 00:00:06,530
was posed by Comte de Buffon--

4
00:00:06,530 --> 00:00:11,720
a French naturalist--
back in the 18th century.

5
00:00:11,720 --> 00:00:13,750
And it marks the
beginning of a subject

6
00:00:13,750 --> 00:00:18,280
that is known as the subject
of geometric probability.

7
00:00:18,280 --> 00:00:20,600
The problem is pretty simple.

8
00:00:20,600 --> 00:00:22,990
We have the infinite
plane, and we

9
00:00:22,990 --> 00:00:26,830
draw lines that are
parallel to each other.

10
00:00:26,830 --> 00:00:31,190
And they're spaced
apart d units.

11
00:00:31,190 --> 00:00:33,790
So this distance here is d.

12
00:00:33,790 --> 00:00:37,270
And the same for
all the other lines.

13
00:00:37,270 --> 00:00:41,110
We take a needle that has
a certain length-- l--

14
00:00:41,110 --> 00:00:44,290
and we throw it at
random on the plane.

15
00:00:44,290 --> 00:00:47,890
So the needle might
fall this way, so

16
00:00:47,890 --> 00:00:51,220
that it doesn't cross any line,
or it might fall this way,

17
00:00:51,220 --> 00:00:55,880
so that it ends up
crossing one of the lines.

18
00:00:55,880 --> 00:00:59,620
If the needle is long enough,
it might actually even end up

19
00:00:59,620 --> 00:01:01,870
crossing two of the lines.

20
00:01:01,870 --> 00:01:05,830
But we will make the assumption
that the length of the needle

21
00:01:05,830 --> 00:01:08,350
is less than the distance
between the two--

22
00:01:08,350 --> 00:01:11,620
between two adjacent
lines, so that we're

23
00:01:11,620 --> 00:01:13,720
going to have either
this configuration,

24
00:01:13,720 --> 00:01:16,130
or that configuration.

25
00:01:16,130 --> 00:01:19,240
So in this setting, we're
interested in the question

26
00:01:19,240 --> 00:01:22,720
of how likely is it that the
needle is going to intersect

27
00:01:22,720 --> 00:01:28,070
one of the lines if the needle
is thrown completely at random?

28
00:01:28,070 --> 00:01:31,670
We will answer this question,
and we will proceed as follows.

29
00:01:31,670 --> 00:01:34,510
First, we need to
model the experiment--

30
00:01:34,510 --> 00:01:37,450
the probabilistic
experiment-- mathematically.

31
00:01:37,450 --> 00:01:40,930
That is, we need to define
an appropriate sample space,

32
00:01:40,930 --> 00:01:43,780
define some relevant
random variables,

33
00:01:43,780 --> 00:01:46,240
choose an appropriate
probability law,

34
00:01:46,240 --> 00:01:50,740
identify the event of
interest, and then calculate.

35
00:01:50,740 --> 00:01:54,100
Let us see what it takes to
describe a typical outcome

36
00:01:54,100 --> 00:01:55,670
of the experiment.

37
00:01:55,670 --> 00:02:01,240
Suppose that the needle fell
this way, so that the nearest

38
00:02:01,240 --> 00:02:03,790
line is the one above.

39
00:02:03,790 --> 00:02:07,650
And let us mark here the
center of the needle.

40
00:02:07,650 --> 00:02:12,340
One quantity of interest
is this vertical distance

41
00:02:12,340 --> 00:02:15,920
between the needle
and the nearest line.

42
00:02:15,920 --> 00:02:18,265
Let us call this quantity x.

43
00:02:21,030 --> 00:02:23,820
We're using here a
lowercase x, because we're

44
00:02:23,820 --> 00:02:27,990
dealing with a numerical value
in one particular outcome

45
00:02:27,990 --> 00:02:29,560
of the experiment.

46
00:02:29,560 --> 00:02:32,490
But we think of this x
as being the realization

47
00:02:32,490 --> 00:02:34,830
of a certain random
variable that we

48
00:02:34,830 --> 00:02:37,560
will denote by capital X.

49
00:02:37,560 --> 00:02:40,740
What else does it take
to describe the needle?

50
00:02:40,740 --> 00:02:42,990
Suppose that the
needle had fallen

51
00:02:42,990 --> 00:02:46,140
somewhere so that it is at
the same vertical distance

52
00:02:46,140 --> 00:02:50,340
from the nearest
line, but it has

53
00:02:50,340 --> 00:02:53,910
an orientation of this kind.

54
00:02:53,910 --> 00:02:57,390
This orientation
compared to that one

55
00:02:57,390 --> 00:02:59,220
should make a difference.

56
00:02:59,220 --> 00:03:02,340
Because when it falls
that way, it's more likely

57
00:03:02,340 --> 00:03:07,050
that it's going to cut the next
line as opposed to this case.

58
00:03:07,050 --> 00:03:12,660
So the angle that the needle is
making with the parallel lines

59
00:03:12,660 --> 00:03:14,500
should also be relevant.

60
00:03:14,500 --> 00:03:17,950
So let us give a name to
that particular angle.

61
00:03:17,950 --> 00:03:24,660
So let's extend that line until
it crosses one of the lines.

62
00:03:24,660 --> 00:03:30,720
And let us give a name to
this angle, and call it theta.

63
00:03:30,720 --> 00:03:32,850
So if I tell you
x and theta, you

64
00:03:32,850 --> 00:03:36,840
know how far away the needle
is from the nearest line,

65
00:03:36,840 --> 00:03:39,510
and at what angle it is.

66
00:03:39,510 --> 00:03:42,240
It looks like these are
two useful variables

67
00:03:42,240 --> 00:03:44,520
to describe the outcome
of the experiment, so let

68
00:03:44,520 --> 00:03:46,620
us try working with these.

69
00:03:46,620 --> 00:03:50,610
So our model is going to involve
two random variables defined

70
00:03:50,610 --> 00:03:54,690
the way we discussed
it just now.

71
00:03:54,690 --> 00:03:59,340
What is the range of
these random variables?

72
00:03:59,340 --> 00:04:06,570
Since we took x to be the
distance from the nearest line,

73
00:04:06,570 --> 00:04:09,300
and the lines are
d units apart, this

74
00:04:09,300 --> 00:04:14,250
means that x is going to
be somewhere between 0

75
00:04:14,250 --> 00:04:18,260
and d over 2.

76
00:04:18,260 --> 00:04:20,500
How about theta?

77
00:04:20,500 --> 00:04:24,640
So the needle makes two angles
with the part of the line.

78
00:04:24,640 --> 00:04:27,790
It's this angle, and
the complimentary one.

79
00:04:27,790 --> 00:04:29,150
Which one do we take?

80
00:04:29,150 --> 00:04:30,940
Well, we use a
convention that theta

81
00:04:30,940 --> 00:04:34,690
is defined as the acute
angle that the direction

82
00:04:34,690 --> 00:04:37,210
of the needle is
making with the lines,

83
00:04:37,210 --> 00:04:44,740
so that theta will vary over
a range from 0 to pi over 2.

84
00:04:48,080 --> 00:04:50,630
And our sample space
for the experiments

85
00:04:50,630 --> 00:04:55,370
who will be the set of
all pairs of x and theta,

86
00:04:55,370 --> 00:04:58,220
that satisfy these
two conditions.

87
00:04:58,220 --> 00:05:01,820
These will be the
possible x's and thetas.

88
00:05:01,820 --> 00:05:05,060
Having defined the
sample space, next we

89
00:05:05,060 --> 00:05:08,240
need to define a
probability law.

90
00:05:08,240 --> 00:05:11,960
At this point, we do not want to
make any arbitrary assumptions.

91
00:05:11,960 --> 00:05:16,970
We only have the words
completely at random to go by.

92
00:05:16,970 --> 00:05:19,310
But what do these words mean?

93
00:05:19,310 --> 00:05:24,080
We will interpret them to mean
that there are no preferred x

94
00:05:24,080 --> 00:05:27,660
values, so that all x
values are- in some sense--

95
00:05:27,660 --> 00:05:28,950
equally likely.

96
00:05:28,950 --> 00:05:33,800
So we're going to assume that
x is a uniform random variable.

97
00:05:33,800 --> 00:05:35,780
Since it is uniform,
it's going to be

98
00:05:35,780 --> 00:05:37,640
a constant over this range.

99
00:05:37,640 --> 00:05:40,280
And in order to integrate
to 1, that constant

100
00:05:40,280 --> 00:05:42,620
will have to be 2 over d.

101
00:05:42,620 --> 00:05:47,180
And we understand that the PDF
of x is 0 outside that range.

102
00:05:47,180 --> 00:05:49,280
Similarly for theta,
we do not want

103
00:05:49,280 --> 00:05:51,560
to assume that some
orientations are more

104
00:05:51,560 --> 00:05:54,090
likely than other orientations.

105
00:05:54,090 --> 00:05:57,320
So we will again assume
a uniform probability

106
00:05:57,320 --> 00:05:59,630
distribution.

107
00:05:59,630 --> 00:06:04,850
And therefore, that PDF
must be equal to 2 over pi

108
00:06:04,850 --> 00:06:08,460
for theta's over this
particular range.

109
00:06:08,460 --> 00:06:11,750
So far, we have specified
the marginal PDFs of each one

110
00:06:11,750 --> 00:06:13,710
of the two random variables.

111
00:06:13,710 --> 00:06:15,800
How about the adjoined PDF?

112
00:06:15,800 --> 00:06:17,540
In order to have
a complete model,

113
00:06:17,540 --> 00:06:20,540
we need to have a
joint PDF in our hands.

114
00:06:20,540 --> 00:06:24,620
Here, we're going to make the
assumption that x and theta are

115
00:06:24,620 --> 00:06:26,820
independent of each other.

116
00:06:26,820 --> 00:06:29,510
And in that case,
the joint PDF is

117
00:06:29,510 --> 00:06:34,460
determined by just taking the
product of the marginal PDFs.

118
00:06:34,460 --> 00:06:38,120
So the joint PDF is
going to be equal to 4

119
00:06:38,120 --> 00:06:42,110
divided by pi times d.

120
00:06:42,110 --> 00:06:43,700
By this point, we
have completely

121
00:06:43,700 --> 00:06:47,740
specified a probabilistic model.

122
00:06:47,740 --> 00:06:51,370
We have made some assumptions,
which you might even

123
00:06:51,370 --> 00:06:52,990
consider arbitrary.

124
00:06:52,990 --> 00:06:56,860
But these assumptions
are a reasonable attempt

125
00:06:56,860 --> 00:06:59,350
at capturing the idea
that the needle is

126
00:06:59,350 --> 00:07:02,840
thrown completely at random.

127
00:07:02,840 --> 00:07:05,170
This completes the
subjective part--

128
00:07:05,170 --> 00:07:07,270
the modeling part.

129
00:07:07,270 --> 00:07:11,440
The next step is much
more streamlined.

130
00:07:11,440 --> 00:07:14,050
There's not going
to be any choices.

131
00:07:14,050 --> 00:07:17,700
We just need to consider
the event of interest,

132
00:07:17,700 --> 00:07:20,140
express it in terms of
the random variables

133
00:07:20,140 --> 00:07:24,610
that we have in our hands, and
then use the probability model

134
00:07:24,610 --> 00:07:26,950
that we have to
calculate the probability

135
00:07:26,950 --> 00:07:29,000
of this particular event.

136
00:07:29,000 --> 00:07:32,920
So let us identify
the event of interest.

137
00:07:32,920 --> 00:07:39,040
When will the needle
intersect the nearest line?

138
00:07:39,040 --> 00:07:43,010
This will depend
on the following.

139
00:07:43,010 --> 00:07:48,190
We can look at the vertical
extent of the needle.

140
00:07:48,190 --> 00:07:51,770
By vertical extent,
I mean the following.

141
00:07:51,770 --> 00:07:54,700
Let's see how far
the needle goes

142
00:07:54,700 --> 00:07:59,680
in the vertical direction,
which is the length

143
00:07:59,680 --> 00:08:04,860
of this green segment here.

144
00:08:04,860 --> 00:08:08,520
In this example, the
vertical extent of the needle

145
00:08:08,520 --> 00:08:13,740
is less than the distance
from the next line.

146
00:08:13,740 --> 00:08:16,840
And we do not have
an intersection.

147
00:08:16,840 --> 00:08:22,390
If the figure was
something like this,

148
00:08:22,390 --> 00:08:27,700
the vertical extent of the
needle would have been that,

149
00:08:27,700 --> 00:08:32,770
but x would have been
just this little segment.

150
00:08:32,770 --> 00:08:36,669
The vertical extent is
bigger than x and the needle

151
00:08:36,669 --> 00:08:38,559
intersects the line.

152
00:08:38,559 --> 00:08:42,070
So we have an
intersection if and only

153
00:08:42,070 --> 00:08:44,500
if the vertical extent--

154
00:08:44,500 --> 00:08:47,950
which is this vertical
green segment--

155
00:08:47,950 --> 00:08:51,520
is larger than the distance x.

156
00:08:51,520 --> 00:08:56,120
Or equivalently, if x is less
than the vertical extent.

157
00:08:56,120 --> 00:09:02,650
So we will have an
intersection if x is less than

158
00:09:02,650 --> 00:09:06,850
or equal to the vertical
extent of the needle.

159
00:09:06,850 --> 00:09:10,210
Now, how big is this
vertical extent?

160
00:09:10,210 --> 00:09:12,400
Let's use some
trigonometry here.

161
00:09:12,400 --> 00:09:17,530
This angle here is theta, so
this angle here is also theta.

162
00:09:20,090 --> 00:09:23,840
Here, we have a right
triangle and the hypotenuse

163
00:09:23,840 --> 00:09:29,330
of this triangle is l over 2.

164
00:09:29,330 --> 00:09:34,100
This angle is theta, therefore
this vertical segment

165
00:09:34,100 --> 00:09:41,880
is equal to l over
2 times sine theta.

166
00:09:41,880 --> 00:09:43,820
So this is the
geometrical condition

167
00:09:43,820 --> 00:09:49,380
that describes the event
that the needle intersects

168
00:09:49,380 --> 00:09:50,820
the nearest line.

169
00:09:50,820 --> 00:09:53,190
And all we need to do
now is to calculate

170
00:09:53,190 --> 00:09:54,700
the probability of this event.

171
00:09:59,330 --> 00:10:01,930
So here is what we have so far.

172
00:10:01,930 --> 00:10:04,990
This is the picture that
we had before, but drawn

173
00:10:04,990 --> 00:10:07,120
in a somewhat nicer way.

174
00:10:07,120 --> 00:10:11,020
This is the joint PDF
that we decided upon.

175
00:10:11,020 --> 00:10:13,270
And we wish to calculate
the probability

176
00:10:13,270 --> 00:10:15,460
of this particular event--

177
00:10:15,460 --> 00:10:20,460
that x is less than or equal
to l over 2 sine theta.

178
00:10:23,470 --> 00:10:26,400
How do we calculate
the probability

179
00:10:26,400 --> 00:10:29,740
of an event that has to do
with two random variables?

180
00:10:29,740 --> 00:10:34,980
What we do is we
take the joint PDF--

181
00:10:34,980 --> 00:10:39,390
which in our case
is four over pi d--

182
00:10:39,390 --> 00:10:49,140
and integrate it over the set
of x's and theta's for which

183
00:10:49,140 --> 00:10:51,180
the PDF is non-zero.

184
00:10:51,180 --> 00:10:55,380
So it's only going to be over
x's and theta's in those ranges

185
00:10:55,380 --> 00:10:58,920
and also, only for
those x theta pairs

186
00:10:58,920 --> 00:11:01,540
for which the event occurs.

187
00:11:01,540 --> 00:11:03,840
So what are these pairs?

188
00:11:03,840 --> 00:11:08,410
This event can occur
with any choice of theta.

189
00:11:08,410 --> 00:11:16,050
So theta is free to vary
from 0 up to pi over 2.

190
00:11:16,050 --> 00:11:17,610
How about x?

191
00:11:17,610 --> 00:11:22,380
For this event to
occur, x can be

192
00:11:22,380 --> 00:11:26,970
anything that is non-negative
as long as it is less than

193
00:11:26,970 --> 00:11:29,490
or equal to this number.

194
00:11:29,490 --> 00:11:32,220
So the upper limit
of this integration

195
00:11:32,220 --> 00:11:38,630
is going to be l over
2 times sine theta.

196
00:11:38,630 --> 00:11:42,710
And all we need to do now is to
evaluate this double integral.

197
00:11:42,710 --> 00:11:45,560
Let's start with
the inner integral.

198
00:11:45,560 --> 00:11:48,440
Because we're just
integrating a constant,

199
00:11:48,440 --> 00:11:56,520
the inner integral
evaluates to the quantity

200
00:11:56,520 --> 00:11:58,920
that we're integrating--
the constant that we're

201
00:11:58,920 --> 00:12:02,580
integrating-- which is 4
times pi d times the length

202
00:12:02,580 --> 00:12:05,430
of the interval over
which we're integrating,

203
00:12:05,430 --> 00:12:07,855
which is l over 2 sine theta.

204
00:12:12,520 --> 00:12:15,860
And now we need to carry
out the outer integral.

205
00:12:15,860 --> 00:12:20,490
Let us pull out the constants,
which is this 4 with this 2

206
00:12:20,490 --> 00:12:22,290
give us a 2.

207
00:12:22,290 --> 00:12:27,220
We have 2l over pi d.

208
00:12:27,220 --> 00:12:31,280
And then the integral from 0
to pi over 2 of sine theta.

209
00:12:31,280 --> 00:12:35,500
Now the integral of sine
theta is minus cosine theta.

210
00:12:35,500 --> 00:12:39,250
And we need to evaluate
this at 0 and pi over 2.

211
00:12:39,250 --> 00:12:41,660
This turns out to be equal to 1.

212
00:12:41,660 --> 00:12:47,400
So the final result
is 2 l over pi d.

213
00:12:47,400 --> 00:12:49,890
And this is the final
answer to the problem

214
00:12:49,890 --> 00:12:53,710
that we have been considering.

215
00:12:53,710 --> 00:12:57,700
And now, a curious thought.

216
00:12:57,700 --> 00:13:03,310
Suppose that you do not
know what the number pi is

217
00:13:03,310 --> 00:13:05,950
and all you have
in your hands is

218
00:13:05,950 --> 00:13:11,500
your floor, lines drawn on
your floor, and the needle.

219
00:13:11,500 --> 00:13:14,320
And you do know the length
between adjacent lines

220
00:13:14,320 --> 00:13:15,490
on your floor.

221
00:13:15,490 --> 00:13:18,140
And you do know your
length of your needle.

222
00:13:18,140 --> 00:13:21,070
How can you figure
out the number pi?

223
00:13:21,070 --> 00:13:25,120
Take your needle, throw it
at random a million times,

224
00:13:25,120 --> 00:13:29,080
and count the frequency with
which the needle ends up

225
00:13:29,080 --> 00:13:30,730
crossing the line.

226
00:13:30,730 --> 00:13:32,230
If you believe
that probabilities

227
00:13:32,230 --> 00:13:35,200
can be interpreted
as frequencies,

228
00:13:35,200 --> 00:13:39,250
the frequency that you observe
gives you a good estimate

229
00:13:39,250 --> 00:13:40,600
of this probability.

230
00:13:40,600 --> 00:13:44,560
So it gives you a good estimate
of this particular number.

231
00:13:44,560 --> 00:13:47,800
And if you know the length of
your needle and of the distance

232
00:13:47,800 --> 00:13:50,530
between the different
lines, you can

233
00:13:50,530 --> 00:13:55,660
use the estimate of that number
to determine the value of pi.

234
00:13:55,660 --> 00:13:59,530
This is a so-called
Monte Carlo method,

235
00:13:59,530 --> 00:14:03,430
which uses simulation to
evaluate experimentally

236
00:14:03,430 --> 00:14:07,690
the value, in this case,
of the constant pi.

237
00:14:07,690 --> 00:14:12,250
Of course, for pi, we have much
better ways of calculating it.

238
00:14:12,250 --> 00:14:15,660
But there are many applications
in engineering and in physics

239
00:14:15,660 --> 00:14:18,430
where certain quantities
are hard to calculate,

240
00:14:18,430 --> 00:14:22,960
but they can be calculated
using a trick of this kind

241
00:14:22,960 --> 00:14:24,700
by simulation.

242
00:14:24,700 --> 00:14:26,990
Here's a typical situation.

243
00:14:26,990 --> 00:14:29,500
Consider the unit cube.

244
00:14:29,500 --> 00:14:31,810
And for simplicity,
I'm only taking

245
00:14:31,810 --> 00:14:33,710
a cube in two dimensions.

246
00:14:33,710 --> 00:14:35,710
But in general, think
of the unit cube

247
00:14:35,710 --> 00:14:40,390
in n dimensions, which is an
object that has unit volume.

248
00:14:40,390 --> 00:14:46,590
Inside that unit cube, there
is a complicated subset

249
00:14:46,590 --> 00:14:51,000
which is described maybe by
some very complicated formulas.

250
00:14:51,000 --> 00:14:52,920
And you want to
calculate the volume

251
00:14:52,920 --> 00:14:56,070
of this complicated subset.

252
00:14:56,070 --> 00:14:58,630
The description of the
subset is so complicated

253
00:14:58,630 --> 00:15:03,280
that using integration,
multiple integrals, and calculus

254
00:15:03,280 --> 00:15:05,290
is practically impossible.

255
00:15:05,290 --> 00:15:06,700
What can you do?

256
00:15:06,700 --> 00:15:11,080
What you can do is to start
throwing at random points

257
00:15:11,080 --> 00:15:13,330
inside that unit cube.

258
00:15:13,330 --> 00:15:14,470
So you throw points.

259
00:15:14,470 --> 00:15:15,640
Some fault inside.

260
00:15:15,640 --> 00:15:18,070
Some fall outside.

261
00:15:18,070 --> 00:15:21,070
You count the frequency
with which the points

262
00:15:21,070 --> 00:15:24,520
happen to be inside your set.

263
00:15:24,520 --> 00:15:29,620
And as long as you're
throwing the points uniformly

264
00:15:29,620 --> 00:15:34,150
over the cube, then
the probability

265
00:15:34,150 --> 00:15:39,970
of your complicated set is going
to be the volume of that set.

266
00:15:39,970 --> 00:15:42,190
You estimate the
probability by counting

267
00:15:42,190 --> 00:15:46,930
the frequency with which
you get points in that set.

268
00:15:46,930 --> 00:15:50,830
And so, by using these
observed frequencies,

269
00:15:50,830 --> 00:15:55,210
you can estimate the
volume of a set--

270
00:15:55,210 --> 00:15:57,670
something that might
be very difficult to do

271
00:15:57,670 --> 00:16:00,100
through other numerical methods.

272
00:16:00,100 --> 00:16:04,840
It turns out that these days,
physicists and many engineers

273
00:16:04,840 --> 00:16:07,885
use methods of this
kind quite often

274
00:16:07,885 --> 00:16:11,610
and in many important
applications.