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PROFESSOR: OK.
10
00:00:22,490 --> 00:00:26,380
So today's lecture will be on
the subject of counting.
11
00:00:26,380 --> 00:00:29,620
So counting, I guess, is
a pretty simple affair
12
00:00:29,620 --> 00:00:33,060
conceptually, but it's a
topic that can also get
13
00:00:33,060 --> 00:00:34,410
to be pretty tricky.
14
00:00:34,410 --> 00:00:37,840
The reason we're going to talk
about counting is that there's
15
00:00:37,840 --> 00:00:41,000
a lot of probability problems
whose solution actually
16
00:00:41,000 --> 00:00:45,050
reduces to successfully counting
the cardinalities of
17
00:00:45,050 --> 00:00:46,110
various sets.
18
00:00:46,110 --> 00:00:49,270
So we're going to see the basic,
simplest methods that
19
00:00:49,270 --> 00:00:52,350
one can use to count
systematically in various
20
00:00:52,350 --> 00:00:53,660
situations.
21
00:00:53,660 --> 00:00:56,600
So in contrast to previous
lectures, we're not going to
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00:00:56,600 --> 00:00:59,700
introduce any significant
new concepts of a
23
00:00:59,700 --> 00:01:01,460
probabilistic nature.
24
00:01:01,460 --> 00:01:04,340
We're just going to use the
probability tools that we
25
00:01:04,340 --> 00:01:05,680
already know.
26
00:01:05,680 --> 00:01:08,460
And we're going to apply them
in situations where there's
27
00:01:08,460 --> 00:01:10,840
also some counting involved.
28
00:01:10,840 --> 00:01:13,000
Now, today we're going
to just touch the
29
00:01:13,000 --> 00:01:14,490
surface of this subject.
30
00:01:14,490 --> 00:01:16,650
There's a whole field of
mathematics called
31
00:01:16,650 --> 00:01:19,920
combinatorics who are people who
actually spend their whole
32
00:01:19,920 --> 00:01:24,220
lives counting more and
more complicated sets.
33
00:01:24,220 --> 00:01:27,580
We were not going to get
anywhere close to the full
34
00:01:27,580 --> 00:01:31,360
complexity of the field, but
we'll get just enough tools
35
00:01:31,360 --> 00:01:36,260
that allow us to address
problems of the type that one
36
00:01:36,260 --> 00:01:39,730
encounters in most common
situations.
37
00:01:39,730 --> 00:01:43,250
So the basic idea, the basic
principle is something that
38
00:01:43,250 --> 00:01:45,400
we've already discussed.
39
00:01:45,400 --> 00:01:49,820
So counting methods apply in
situations where we have
40
00:01:49,820 --> 00:01:53,660
probabilistic experiments with
a finite number of outcomes
41
00:01:53,660 --> 00:01:56,070
and where every outcome--
42
00:01:56,070 --> 00:01:57,770
every possible outcome--
43
00:01:57,770 --> 00:02:00,260
has the same probability
of occurring.
44
00:02:00,260 --> 00:02:04,440
So we have our sample space,
omega, and it's got a bunch of
45
00:02:04,440 --> 00:02:06,520
discrete points in there.
46
00:02:06,520 --> 00:02:10,539
And the cardinality of the set
omega is some capital N. So,
47
00:02:10,539 --> 00:02:14,960
in particular, we assume that
the sample points are equally
48
00:02:14,960 --> 00:02:18,490
likely, which means that every
element of the sample space
49
00:02:18,490 --> 00:02:22,420
has the same probability
equal to 1 over N.
50
00:02:22,420 --> 00:02:26,650
And then we are interested in a
subset of the sample space,
51
00:02:26,650 --> 00:02:29,860
call it A. And that
subset consists
52
00:02:29,860 --> 00:02:31,350
of a number of elements.
53
00:02:31,350 --> 00:02:36,080
Let the cardinality of that
subset be equal to little n.
54
00:02:36,080 --> 00:02:39,370
And then to find the probability
of that set, all
55
00:02:39,370 --> 00:02:42,060
we need to do is to add the
probabilities of the
56
00:02:42,060 --> 00:02:43,610
individual elements.
57
00:02:43,610 --> 00:02:47,030
There's little n elements, and
each one has probability one
58
00:02:47,030 --> 00:02:50,340
over capital N. And
that's the answer.
59
00:02:50,340 --> 00:02:53,250
So this means that to solve
problems in this context, all
60
00:02:53,250 --> 00:02:56,580
that we need to be able to do
is to figure out the number
61
00:02:56,580 --> 00:03:00,800
capital N and to figure out
the number little n.
62
00:03:00,800 --> 00:03:04,330
Now, if somebody gives you a set
by just giving you a list
63
00:03:04,330 --> 00:03:07,660
and gives you another set,
again, giving you a list, it's
64
00:03:07,660 --> 00:03:09,120
easy to count there element.
65
00:03:09,120 --> 00:03:11,510
You just count how much
there is on the list.
66
00:03:11,510 --> 00:03:15,030
But sometimes the sets are
described in some more
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00:03:15,030 --> 00:03:20,590
implicit way, and we may have to
do a little bit more work.
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00:03:20,590 --> 00:03:22,360
There's various tricks that are
69
00:03:22,360 --> 00:03:24,410
involved in counting properly.
70
00:03:24,410 --> 00:03:27,080
And the most common
one is to--
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00:03:27,080 --> 00:03:31,350
when you consider a set of
possible outcomes, to describe
72
00:03:31,350 --> 00:03:33,800
the construction of those
possible outcomes through a
73
00:03:33,800 --> 00:03:35,440
sequential process.
74
00:03:35,440 --> 00:03:38,130
So think of a probabilistic
experiment that involves a
75
00:03:38,130 --> 00:03:42,560
number of stages, and in each
one of the stages there's a
76
00:03:42,560 --> 00:03:45,890
number of possible choices
that there may be.
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00:03:45,890 --> 00:03:48,630
The overall experiment consists
of carrying out all
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00:03:48,630 --> 00:03:49,930
the stages to the end.
79
00:03:49,930 --> 00:03:52,855
80
00:03:52,855 --> 00:03:55,830
And the number of points in the
sample space is how many
81
00:03:55,830 --> 00:03:59,920
final outcomes there can be in
this multi-stage experiment.
82
00:03:59,920 --> 00:04:02,910
So in this picture we have an
experiment in which of the
83
00:04:02,910 --> 00:04:07,230
first stage we have
four choices.
84
00:04:07,230 --> 00:04:11,640
In the second stage, no matter
what happened in the first
85
00:04:11,640 --> 00:04:16,010
stage, the way this is drawn
we have three choices.
86
00:04:16,010 --> 00:04:19,720
No matter whether we ended up
here, there, or there, we have
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00:04:19,720 --> 00:04:22,860
three choices in the
second stage.
88
00:04:22,860 --> 00:04:27,830
And then there's a third stage
and at least in this picture,
89
00:04:27,830 --> 00:04:31,520
no matter what happened in the
first two stages, in the third
90
00:04:31,520 --> 00:04:35,930
stage we're going to have
two possible choices.
91
00:04:35,930 --> 00:04:41,070
So how many leaves are there
at the end of this tree?
92
00:04:41,070 --> 00:04:42,170
That's simple.
93
00:04:42,170 --> 00:04:45,460
It's just the product of
these three numbers.
94
00:04:45,460 --> 00:04:48,520
The number of possible leaves
that we have out there is 4
95
00:04:48,520 --> 00:04:50,430
times 3 times 2.
96
00:04:50,430 --> 00:04:54,030
Number of choices at each stage
gets multiplied, and
97
00:04:54,030 --> 00:04:57,680
that gives us the number
of overall choices.
98
00:04:57,680 --> 00:05:01,230
So this is the general rule, the
general trick that we are
99
00:05:01,230 --> 00:05:03,520
going to use over and over.
100
00:05:03,520 --> 00:05:07,660
So let's apply it to some very
simple problems as a warm up.
101
00:05:07,660 --> 00:05:10,530
How many license plates can you
make if you're allowed to
102
00:05:10,530 --> 00:05:17,140
use three letters and then
followed by four digits?
103
00:05:17,140 --> 00:05:20,020
At least if you're dealing with
the English alphabet, you
104
00:05:20,020 --> 00:05:23,460
have 26 choices for
the first letter.
105
00:05:23,460 --> 00:05:27,100
Then you have 26 choices
for the second letter.
106
00:05:27,100 --> 00:05:30,300
And then 26 choices for
the third letter.
107
00:05:30,300 --> 00:05:31,750
And then we start the digits.
108
00:05:31,750 --> 00:05:34,970
We have 10 choices for the first
digit, 10 choices for
109
00:05:34,970 --> 00:05:37,780
the second digit, 10 choices for
the third, 10 choices for
110
00:05:37,780 --> 00:05:40,030
the last one.
111
00:05:40,030 --> 00:05:43,010
Let's make it a little more
complicated, suppose that
112
00:05:43,010 --> 00:05:47,100
we're interested in license
plates where no letter can be
113
00:05:47,100 --> 00:05:50,270
repeated and no digit
can be repeated.
114
00:05:50,270 --> 00:05:53,040
So you have to use different
letters, different digits.
115
00:05:53,040 --> 00:05:55,110
How many license plates
can you make?
116
00:05:55,110 --> 00:05:56,780
OK, let's choose the
first letter,
117
00:05:56,780 --> 00:05:59,130
and we have 26 choices.
118
00:05:59,130 --> 00:06:02,350
Now, I'm ready to choose my
second letter, how many
119
00:06:02,350 --> 00:06:04,610
choices do I have?
120
00:06:04,610 --> 00:06:08,090
I have 25, because I already
used one letter.
121
00:06:08,090 --> 00:06:11,900
I have the 25 remaining letters
to choose from.
122
00:06:11,900 --> 00:06:14,330
For the next letter,
how many choices?
123
00:06:14,330 --> 00:06:17,030
Well, I used up two of
my letters, so I
124
00:06:17,030 --> 00:06:19,840
only have 24 available.
125
00:06:19,840 --> 00:06:22,950
And then we start with the
digits, 10 choices for the
126
00:06:22,950 --> 00:06:26,910
first digit, 9 choices for the
second, 8 for the third, 7 for
127
00:06:26,910 --> 00:06:28,160
the last one.
128
00:06:28,160 --> 00:06:30,480
129
00:06:30,480 --> 00:06:31,730
All right.
130
00:06:31,730 --> 00:06:38,710
So, now, let's bring some
symbols in a related problem.
131
00:06:38,710 --> 00:06:44,190
You are given a set that
consists of n elements and
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00:06:44,190 --> 00:06:47,040
you're supposed to take
those n elements and
133
00:06:47,040 --> 00:06:50,050
put them in a sequence.
134
00:06:50,050 --> 00:06:52,290
That is to order them.
135
00:06:52,290 --> 00:06:56,240
Any possible ordering of those
elements is called a
136
00:06:56,240 --> 00:06:57,560
permutation.
137
00:06:57,560 --> 00:07:02,630
So for example, if we have the
set 1, 2, 3, 4, a possible
138
00:07:02,630 --> 00:07:05,105
permutation is the
list 2, 3, 4, 1.
139
00:07:05,105 --> 00:07:08,560
140
00:07:08,560 --> 00:07:10,680
That's one possible
permutation.
141
00:07:10,680 --> 00:07:13,180
And there's lots of possible
permutations, of course, the
142
00:07:13,180 --> 00:07:15,560
question is how many
are there.
143
00:07:15,560 --> 00:07:20,390
OK, let's think about building
this permutation by choosing
144
00:07:20,390 --> 00:07:21,460
one at a time.
145
00:07:21,460 --> 00:07:26,330
Which of these elements goes
into each one of these slots?
146
00:07:26,330 --> 00:07:28,740
How many choices for the number
that goes into the
147
00:07:28,740 --> 00:07:31,160
first slot or the elements?
148
00:07:31,160 --> 00:07:34,430
Well, we can choose any one of
the available elements, so we
149
00:07:34,430 --> 00:07:35,680
have n choices.
150
00:07:35,680 --> 00:07:38,650
151
00:07:38,650 --> 00:07:42,220
Let's say this element goes
here, having used up that
152
00:07:42,220 --> 00:07:45,730
element, we're left with n minus
1 elements and we can
153
00:07:45,730 --> 00:07:49,530
pick any one of these and bring
it into the second slot.
154
00:07:49,530 --> 00:07:52,340
So here we have n choices, here
we're going to have n
155
00:07:52,340 --> 00:07:55,940
minus 1 choices, then how
many we put there will
156
00:07:55,940 --> 00:07:58,060
have n minus 2 choices.
157
00:07:58,060 --> 00:08:00,220
And you go down until the end.
158
00:08:00,220 --> 00:08:02,160
What happens at this point
when you are to
159
00:08:02,160 --> 00:08:03,880
pick the last element?
160
00:08:03,880 --> 00:08:06,650
Well, you've used n minus of
them, there's only one
161
00:08:06,650 --> 00:08:08,070
left in your bag.
162
00:08:08,070 --> 00:08:09,520
You're forced to use that one.
163
00:08:09,520 --> 00:08:13,920
So the last stage, you're going
to have only one choice.
164
00:08:13,920 --> 00:08:18,050
So, basically, the number of
possible permutations is the
165
00:08:18,050 --> 00:08:21,860
product of all integers
from n down to one, or
166
00:08:21,860 --> 00:08:24,020
from one up to n.
167
00:08:24,020 --> 00:08:26,550
And there's a symbol that we
use for this number, it's
168
00:08:26,550 --> 00:08:29,210
called n factorial.
169
00:08:29,210 --> 00:08:32,990
So n factorial is the number of
permutations of n objects.
170
00:08:32,990 --> 00:08:37,320
The number of ways that you can
order n objects that are
171
00:08:37,320 --> 00:08:39,100
given to you.
172
00:08:39,100 --> 00:08:42,100
Now, a different equation.
173
00:08:42,100 --> 00:08:44,310
We have n elements.
174
00:08:44,310 --> 00:08:48,680
Let's say the elements
are 1, 1,2, up to n.
175
00:08:48,680 --> 00:08:51,310
And it's a set.
176
00:08:51,310 --> 00:08:54,490
And we want to create
a subset.
177
00:08:54,490 --> 00:08:58,460
How many possible subsets
are there?
178
00:08:58,460 --> 00:09:02,950
So speaking of subsets means
looking at each one of the
179
00:09:02,950 --> 00:09:06,880
elements and deciding whether
you're going to put it in to
180
00:09:06,880 --> 00:09:08,440
subsets or not.
181
00:09:08,440 --> 00:09:13,240
For example, I could choose
to put 1 in, but 2 I'm not
182
00:09:13,240 --> 00:09:17,100
putting it in, 3 I'm not putting
it in, 4 I'm putting
183
00:09:17,100 --> 00:09:18,630
it, and so on.
184
00:09:18,630 --> 00:09:21,200
So that's how you
create a subset.
185
00:09:21,200 --> 00:09:23,660
You look at each one of the
elements and you say, OK, I'm
186
00:09:23,660 --> 00:09:27,310
going to put it in the subset,
or I'm not going to put it.
187
00:09:27,310 --> 00:09:30,900
So think of these as consisting
of stages.
188
00:09:30,900 --> 00:09:33,240
At each stage you look at
one element, and you
189
00:09:33,240 --> 00:09:35,090
make a binary decision.
190
00:09:35,090 --> 00:09:38,410
Do I put it in the
subset, or not?
191
00:09:38,410 --> 00:09:41,940
So therefore, how many
subsets are there?
192
00:09:41,940 --> 00:09:45,060
Well, I have two choices
for the first element.
193
00:09:45,060 --> 00:09:47,740
Am I going to put in
the subset, or not?
194
00:09:47,740 --> 00:09:50,630
I have two choices for the
next element, and so on.
195
00:09:50,630 --> 00:09:53,450
196
00:09:53,450 --> 00:09:57,390
For each one of the elements,
we have two choices.
197
00:09:57,390 --> 00:10:02,090
So the overall number of choices
is 2 to the power n.
198
00:10:02,090 --> 00:10:03,710
So, conclusion--
199
00:10:03,710 --> 00:10:10,150
the number of subsets, often n
element set, is 2 to the n.
200
00:10:10,150 --> 00:10:15,050
201
00:10:15,050 --> 00:10:20,430
So in particular, if we take n
equal to 1, let's check that
202
00:10:20,430 --> 00:10:22,190
our answer makes sense.
203
00:10:22,190 --> 00:10:26,420
If we have n equal to one, how
many subsets does it have?
204
00:10:26,420 --> 00:10:29,675
So we're dealing with
a set of just one.
205
00:10:29,675 --> 00:10:30,925
What are the subsets?
206
00:10:30,925 --> 00:10:33,830
207
00:10:33,830 --> 00:10:37,290
One subset is this one.
208
00:10:37,290 --> 00:10:41,530
Do we have other subsets
of the one element set?
209
00:10:41,530 --> 00:10:43,920
Yes, we have the empty set.
210
00:10:43,920 --> 00:10:44,870
That's the second one.
211
00:10:44,870 --> 00:10:48,860
These are the two possible
subsets of
212
00:10:48,860 --> 00:10:50,850
this particular set.
213
00:10:50,850 --> 00:10:56,790
So 2 subsets when n is equal to
1, that checks the answer.
214
00:10:56,790 --> 00:10:58,040
All right.
215
00:10:58,040 --> 00:11:00,290
216
00:11:00,290 --> 00:11:07,590
OK, so having gone so far, we
can do our first example now.
217
00:11:07,590 --> 00:11:12,990
So we are given a die
and we're going
218
00:11:12,990 --> 00:11:16,620
to roll it 6 times.
219
00:11:16,620 --> 00:11:20,030
OK, let's make some assumptions
about the rolls.
220
00:11:20,030 --> 00:11:29,560
Let's assume that the rolls are
independent, and that the
221
00:11:29,560 --> 00:11:30,985
die is also fair.
222
00:11:30,985 --> 00:11:34,180
223
00:11:34,180 --> 00:11:38,110
So this means that the
probability of any particular
224
00:11:38,110 --> 00:11:40,350
outcome of the die rolls--
225
00:11:40,350 --> 00:11:43,960
for example, so we have 6 rolls,
one particular outcome
226
00:11:43,960 --> 00:11:48,760
could be 3,3,1,6,5.
227
00:11:48,760 --> 00:11:51,220
So that's one possible
outcome.
228
00:11:51,220 --> 00:11:54,060
What's the probability
of this outcome?
229
00:11:54,060 --> 00:11:57,470
There's probability 1/6 that
this happens, 1/6 that this
230
00:11:57,470 --> 00:12:00,530
happens, 1/6 that this
happens, and so on.
231
00:12:00,530 --> 00:12:04,250
So the probability that
the outcome is this
232
00:12:04,250 --> 00:12:07,540
is 1/6 to the sixth.
233
00:12:07,540 --> 00:12:10,970
234
00:12:10,970 --> 00:12:13,690
What did I use to come
up with this answer?
235
00:12:13,690 --> 00:12:17,550
I used independence, so I
multiplied the probability of
236
00:12:17,550 --> 00:12:20,360
the first roll gives me a 2,
times the probability that the
237
00:12:20,360 --> 00:12:22,990
second roll gives me
a 3, and so on.
238
00:12:22,990 --> 00:12:26,790
And then I used the assumption
that the die is fair, so that
239
00:12:26,790 --> 00:12:30,300
the probability of 2 is
1/6, the probably of 3
240
00:12:30,300 --> 00:12:32,240
is 1/6, and so on.
241
00:12:32,240 --> 00:12:34,810
So if I were to spell it out,
it's the probability that we
242
00:12:34,810 --> 00:12:37,740
get the 2 in the first roll,
times the probability of 3 in
243
00:12:37,740 --> 00:12:40,850
the second roll, times the
probability of the
244
00:12:40,850 --> 00:12:42,800
5 in the last roll.
245
00:12:42,800 --> 00:12:46,455
So by independence, I can
multiply probabilities.
246
00:12:46,455 --> 00:12:49,530
And because the die is fair,
each one of these numbers is
247
00:12:49,530 --> 00:12:53,910
1/6 to the sixth.
248
00:12:53,910 --> 00:12:58,170
And so the same calculation
would apply no matter what
249
00:12:58,170 --> 00:13:00,610
numbers I would put in here.
250
00:13:00,610 --> 00:13:03,920
So all possible outcomes
are equally likely.
251
00:13:03,920 --> 00:13:06,630
252
00:13:06,630 --> 00:13:08,400
Let's start with this.
253
00:13:08,400 --> 00:13:12,650
So since all possible outcomes
are equally likely to find an
254
00:13:12,650 --> 00:13:15,870
answer to a probability
question, if we're dealing
255
00:13:15,870 --> 00:13:21,430
with some particular event, so
the event is that all rolls
256
00:13:21,430 --> 00:13:22,900
give different numbers.
257
00:13:22,900 --> 00:13:31,160
That's our event A. And our
sample space is some set
258
00:13:31,160 --> 00:13:32,790
capital omega.
259
00:13:32,790 --> 00:13:35,890
We know that the answer is going
to be the cardinality of
260
00:13:35,890 --> 00:13:40,200
the set A, divided by the
cardinality of the set omega.
261
00:13:40,200 --> 00:13:42,830
So let's deal with the
easy one first.
262
00:13:42,830 --> 00:13:45,960
How many elements are there
in the sample space?
263
00:13:45,960 --> 00:13:48,880
How many possible outcomes
are there when you
264
00:13:48,880 --> 00:13:51,570
roll a dice 6 times?
265
00:13:51,570 --> 00:13:54,600
You have 6 choices for
the first roll.
266
00:13:54,600 --> 00:13:57,840
You have 6 choices for the
second roll and so on.
267
00:13:57,840 --> 00:14:00,330
So the overall number
of outcomes is going
268
00:14:00,330 --> 00:14:03,950
to be 6 to the sixth.
269
00:14:03,950 --> 00:14:08,200
So number of elements in
the sample space is 6
270
00:14:08,200 --> 00:14:10,470
to the sixth power.
271
00:14:10,470 --> 00:14:14,820
And I guess this checks
with this.
272
00:14:14,820 --> 00:14:18,480
We have 6 to the sixth outcomes,
each one has this
273
00:14:18,480 --> 00:14:20,570
much probability,
so the overall
274
00:14:20,570 --> 00:14:23,230
probability is equal to one.
275
00:14:23,230 --> 00:14:24,460
Right?
276
00:14:24,460 --> 00:14:28,690
So the probability of an
individual outcome is one over
277
00:14:28,690 --> 00:14:32,400
how many possible outcomes
we have, which is this.
278
00:14:32,400 --> 00:14:33,810
All right.
279
00:14:33,810 --> 00:14:36,620
So how about the numerator?
280
00:14:36,620 --> 00:14:42,430
We are interested in outcomes
in which the numbers that we
281
00:14:42,430 --> 00:14:44,585
get are all different.
282
00:14:44,585 --> 00:14:48,770
283
00:14:48,770 --> 00:14:54,080
So what is an outcome in which
the numbers are all different?
284
00:14:54,080 --> 00:14:56,310
So the die has 6 faces.
285
00:14:56,310 --> 00:14:58,200
We roll it 6 times.
286
00:14:58,200 --> 00:15:00,340
We're going to get 6
different numbers.
287
00:15:00,340 --> 00:15:03,780
This means that we're going to
exhaust all the possible
288
00:15:03,780 --> 00:15:07,640
numbers, but they can appear
in any possible sequence.
289
00:15:07,640 --> 00:15:13,190
So an outcome that makes this
event happen is a list of the
290
00:15:13,190 --> 00:15:16,250
numbers from 1 to 6,
but arranged in
291
00:15:16,250 --> 00:15:18,160
some arbitrary order.
292
00:15:18,160 --> 00:15:23,200
So the possible outcomes that
make event A happen are just
293
00:15:23,200 --> 00:15:25,990
the permutations of the
numbers from 1 to 6.
294
00:15:25,990 --> 00:15:31,070
295
00:15:31,070 --> 00:15:33,900
One possible outcome that makes
our events to happen--
296
00:15:33,900 --> 00:15:35,440
it would be this.
297
00:15:35,440 --> 00:15:39,000
298
00:15:39,000 --> 00:15:42,050
Here we have 6 possible numbers,
but any other list of
299
00:15:42,050 --> 00:15:44,070
this kind in which none
of the numbers is
300
00:15:44,070 --> 00:15:46,650
repeated would also do.
301
00:15:46,650 --> 00:15:51,660
So number of outcomes that make
the event happen is the
302
00:15:51,660 --> 00:15:53,980
number of permutations
of 6 elements.
303
00:15:53,980 --> 00:15:56,060
So it's 6 factorial.
304
00:15:56,060 --> 00:15:59,340
And so the final answer is
going to be 6 factorial
305
00:15:59,340 --> 00:16:02,830
divided by 6 to the sixth.
306
00:16:02,830 --> 00:16:06,580
All right, so that's a typical
way that's one solves problems
307
00:16:06,580 --> 00:16:07,800
of this kind.
308
00:16:07,800 --> 00:16:10,660
We know how to count
certain things.
309
00:16:10,660 --> 00:16:14,260
For example, here we knew how to
count permutations, and we
310
00:16:14,260 --> 00:16:16,830
used our knowledge to count the
elements of the set that
311
00:16:16,830 --> 00:16:18,080
we need to deal with.
312
00:16:18,080 --> 00:16:24,380
313
00:16:24,380 --> 00:16:30,970
So now let's get to a slightly
more difficult problem.
314
00:16:30,970 --> 00:16:37,390
We're given once more a
set with n elements.
315
00:16:37,390 --> 00:16:40,620
316
00:16:40,620 --> 00:16:46,000
We already know how many subsets
that set has, but now
317
00:16:46,000 --> 00:16:50,940
we would be interested in
subsets that have exactly k
318
00:16:50,940 --> 00:16:54,480
elements in them.
319
00:16:54,480 --> 00:17:05,819
So we start with our big set
that has n elements, and we
320
00:17:05,819 --> 00:17:11,890
want to construct a subset
that has k elements.
321
00:17:11,890 --> 00:17:14,200
Out of those n I'm
going to choose k
322
00:17:14,200 --> 00:17:16,079
and put them in there.
323
00:17:16,079 --> 00:17:18,180
In how many ways
can I do this?
324
00:17:18,180 --> 00:17:20,710
More concrete way of thinking
about this problem--
325
00:17:20,710 --> 00:17:24,960
you have n people in some group
and you want to form a
326
00:17:24,960 --> 00:17:28,270
committee by picking people from
that group, and you want
327
00:17:28,270 --> 00:17:31,020
to form a committee
with k people.
328
00:17:31,020 --> 00:17:32,460
Where k is a given number.
329
00:17:32,460 --> 00:17:34,670
For example, a 5 person
committee.
330
00:17:34,670 --> 00:17:37,510
How many 5 person committees
are possible if you're
331
00:17:37,510 --> 00:17:39,450
starting with 100 people?
332
00:17:39,450 --> 00:17:40,960
So that's what we
want to count.
333
00:17:40,960 --> 00:17:44,210
How many k element subsets
are there?
334
00:17:44,210 --> 00:17:48,030
We don't yet know the answer,
but let's give a name to it.
335
00:17:48,030 --> 00:17:52,210
And the name is going to be this
particular symbol, which
336
00:17:52,210 --> 00:17:55,220
we read as n choose k.
337
00:17:55,220 --> 00:18:00,130
Out of n elements, we want
to choose k of them.
338
00:18:00,130 --> 00:18:02,000
OK.
339
00:18:02,000 --> 00:18:04,810
That may be a little tricky.
340
00:18:04,810 --> 00:18:10,170
So what we're going to do is
to instead figure out a
341
00:18:10,170 --> 00:18:15,590
somewhat easier problem,
which is going to be--
342
00:18:15,590 --> 00:18:20,840
in how many ways can I pick k
out of these people and puts
343
00:18:20,840 --> 00:18:25,450
them in a particular order?
344
00:18:25,450 --> 00:18:30,430
So how many possible ordered
lists can I make that consist
345
00:18:30,430 --> 00:18:31,840
of k people?
346
00:18:31,840 --> 00:18:35,240
By ordered, I mean that we take
those k people and we say
347
00:18:35,240 --> 00:18:38,010
this is the first person
in the community.
348
00:18:38,010 --> 00:18:39,600
That's the second person
in the committee.
349
00:18:39,600 --> 00:18:42,070
That's the third person in
the committee and so on.
350
00:18:42,070 --> 00:18:46,100
So in how many ways
can we do this?
351
00:18:46,100 --> 00:18:50,840
Out of these n, we want to
choose just k of them and put
352
00:18:50,840 --> 00:18:52,010
them in slots.
353
00:18:52,010 --> 00:18:53,970
One after the other.
354
00:18:53,970 --> 00:18:57,480
So this is pretty much like the
license plate problem we
355
00:18:57,480 --> 00:19:00,680
solved just a little earlier.
356
00:19:00,680 --> 00:19:06,390
So we have n choices for who
we put as the top person in
357
00:19:06,390 --> 00:19:07,350
the community.
358
00:19:07,350 --> 00:19:11,490
We can pick anyone and have
them be the first person.
359
00:19:11,490 --> 00:19:13,000
Then I'm going to choose
the second
360
00:19:13,000 --> 00:19:14,640
person in the committee.
361
00:19:14,640 --> 00:19:16,700
I've used up 1 person.
362
00:19:16,700 --> 00:19:21,530
So I'm going to have n
minus 1 choices here.
363
00:19:21,530 --> 00:19:25,840
And now, at this stage I've used
up 2 people, so I have n
364
00:19:25,840 --> 00:19:28,640
minus 2 choices here.
365
00:19:28,640 --> 00:19:31,240
And this keeps going on.
366
00:19:31,240 --> 00:19:34,110
Well, what is going to
be the last number?
367
00:19:34,110 --> 00:19:36,310
Is it's n minus k?
368
00:19:36,310 --> 00:19:39,980
Well, not really.
369
00:19:39,980 --> 00:19:44,090
I'm starting subtracting numbers
after the second one,
370
00:19:44,090 --> 00:19:48,250
so by the end I will have
subtracted k minus 1.
371
00:19:48,250 --> 00:19:54,800
So that's how many choices I
will have for the last person.
372
00:19:54,800 --> 00:19:58,270
So this is the number
of ways--
373
00:19:58,270 --> 00:20:02,390
the product of these numbers
there gives me the number of
374
00:20:02,390 --> 00:20:08,420
ways that I can create ordered
lists consisting of k people
375
00:20:08,420 --> 00:20:11,700
out of the n that
we started with.
376
00:20:11,700 --> 00:20:15,120
Now, you can do a little bit of
algebra and check that this
377
00:20:15,120 --> 00:20:17,910
expression here is the same
as that expression.
378
00:20:17,910 --> 00:20:19,200
Why is this?
379
00:20:19,200 --> 00:20:22,520
This factorial has all the
products from 1 up to n.
380
00:20:22,520 --> 00:20:25,140
This factorial has all
the products from 1
381
00:20:25,140 --> 00:20:26,710
up to n minus k.
382
00:20:26,710 --> 00:20:28,240
So you get cancellations.
383
00:20:28,240 --> 00:20:31,860
And what's left is all the
products starting from the
384
00:20:31,860 --> 00:20:37,610
next number after here, which
is this particular number.
385
00:20:37,610 --> 00:20:42,350
So the number of possible ways
of creating such ordered lists
386
00:20:42,350 --> 00:20:46,330
is n factorial divided by
n minus k factorial.
387
00:20:46,330 --> 00:20:49,480
388
00:20:49,480 --> 00:20:53,180
Now, a different way that I
could make an ordered list--
389
00:20:53,180 --> 00:20:57,950
instead of picking the people
one at a time, I could first
390
00:20:57,950 --> 00:21:01,700
choose my k people who are going
to be in the committee,
391
00:21:01,700 --> 00:21:04,080
and then put them in order.
392
00:21:04,080 --> 00:21:07,680
And tell them out of these k,
you are the first, you are the
393
00:21:07,680 --> 00:21:10,010
second, you are the third.
394
00:21:10,010 --> 00:21:12,590
Starting with this k
people, in how many
395
00:21:12,590 --> 00:21:15,580
ways can I order them?
396
00:21:15,580 --> 00:21:18,150
That's the number
of permutations.
397
00:21:18,150 --> 00:21:20,820
398
00:21:20,820 --> 00:21:25,180
Starting with a set with k
objects, in how many ways can
399
00:21:25,180 --> 00:21:28,340
I put them in a specific
order?
400
00:21:28,340 --> 00:21:31,140
How many specific orders
are there?
401
00:21:31,140 --> 00:21:32,390
That's basically the question.
402
00:21:32,390 --> 00:21:34,600
In how many ways can
I permute these k
403
00:21:34,600 --> 00:21:36,290
people and arrange them.
404
00:21:36,290 --> 00:21:38,450
So the number of ways
that you can do
405
00:21:38,450 --> 00:21:42,660
this step is k factorial.
406
00:21:42,660 --> 00:21:48,330
So in how many ways can I
start with a set with n
407
00:21:48,330 --> 00:21:52,020
elements, go through this
process, and end up with a
408
00:21:52,020 --> 00:21:55,560
sorted list with k elements?
409
00:21:55,560 --> 00:21:57,620
By the rule that--
410
00:21:57,620 --> 00:22:02,160
when we have stages, the total
number of stages is how many
411
00:22:02,160 --> 00:22:05,370
choices we had in the first
stage, times how many choices
412
00:22:05,370 --> 00:22:08,510
we had in the second stage.
413
00:22:08,510 --> 00:22:12,670
The number of ways that this
process can happen is this
414
00:22:12,670 --> 00:22:14,890
times that.
415
00:22:14,890 --> 00:22:18,340
This is a different way that
that process could happen.
416
00:22:18,340 --> 00:22:22,640
And the number of possible
of ways is this number.
417
00:22:22,640 --> 00:22:27,600
No matter which way we carry out
that process, in the end
418
00:22:27,600 --> 00:22:34,610
we have the possible ways of
arranging k people out of the
419
00:22:34,610 --> 00:22:36,770
n that we started with.
420
00:22:36,770 --> 00:22:40,730
So the final answer that we get
when we count should be
421
00:22:40,730 --> 00:22:44,220
either this, or this
times that.
422
00:22:44,220 --> 00:22:47,620
Both are equally valid ways of
counting, so both should give
423
00:22:47,620 --> 00:22:49,050
us the same answer.
424
00:22:49,050 --> 00:22:52,950
So we get this equality here.
425
00:22:52,950 --> 00:22:56,370
So these two expressions
corresponds to two different
426
00:22:56,370 --> 00:23:01,660
ways of constructing ordered
lists of k people starting
427
00:23:01,660 --> 00:23:05,580
with n people initially.
428
00:23:05,580 --> 00:23:09,120
And now that we have this
relation, we can send the k
429
00:23:09,120 --> 00:23:11,540
factorial to the denominator.
430
00:23:11,540 --> 00:23:13,940
And that tells us what
that number, n choose
431
00:23:13,940 --> 00:23:16,250
k, is going to be.
432
00:23:16,250 --> 00:23:20,060
So this formula-- it's written
here in red, because you're
433
00:23:20,060 --> 00:23:22,150
going to see it a zillion
times until
434
00:23:22,150 --> 00:23:23,740
the end of the semester--
435
00:23:23,740 --> 00:23:25,950
they are called the binomial
coefficients.
436
00:23:25,950 --> 00:23:31,170
437
00:23:31,170 --> 00:23:34,600
And they tell us the number of
possible ways that we can
438
00:23:34,600 --> 00:23:38,380
create a k element subset,
starting with a
439
00:23:38,380 --> 00:23:41,270
set that has n elements.
440
00:23:41,270 --> 00:23:44,430
It's always good to do a sanity
check to formulas by
441
00:23:44,430 --> 00:23:46,710
considering extreme cases.
442
00:23:46,710 --> 00:23:52,810
So let's take the case where
k is equal to n.
443
00:23:52,810 --> 00:23:56,820
444
00:23:56,820 --> 00:23:59,420
What's the right answer
in this case?
445
00:23:59,420 --> 00:24:02,905
How many n elements subsets
are there out
446
00:24:02,905 --> 00:24:04,950
of an element set?
447
00:24:04,950 --> 00:24:07,580
Well, your subset needs
to include every one.
448
00:24:07,580 --> 00:24:09,400
You don't have any choices.
449
00:24:09,400 --> 00:24:10,750
There's only one choice.
450
00:24:10,750 --> 00:24:12,600
It's the set itself.
451
00:24:12,600 --> 00:24:15,700
So the answer should
be equal to 1.
452
00:24:15,700 --> 00:24:19,980
That's the number of n element
subsets, starting with a set
453
00:24:19,980 --> 00:24:21,340
with n elements.
454
00:24:21,340 --> 00:24:25,250
Let's see if the formula gives
us the right answer.
455
00:24:25,250 --> 00:24:31,750
We have n factorial divided by
k, which is n in our case--
456
00:24:31,750 --> 00:24:32,630
n factorial.
457
00:24:32,630 --> 00:24:36,700
And then n minus k
is 0 factorial.
458
00:24:36,700 --> 00:24:42,070
So if our formula is correct, we
should have this equality.
459
00:24:42,070 --> 00:24:45,620
And what's the way to
make that correct?
460
00:24:45,620 --> 00:24:47,880
Well, it depends what kind
of meaning do we
461
00:24:47,880 --> 00:24:49,420
give to this symbol?
462
00:24:49,420 --> 00:24:53,510
How do we define
zero factorial?
463
00:24:53,510 --> 00:24:55,750
I guess in some ways
it's arbitrary.
464
00:24:55,750 --> 00:24:58,110
We're going to define it
in a way that makes
465
00:24:58,110 --> 00:24:59,640
this formula right.
466
00:24:59,640 --> 00:25:03,870
So the definition that we will
be using is that whenever you
467
00:25:03,870 --> 00:25:08,700
have 0 factorial, it's going
to stand for the number 1.
468
00:25:08,700 --> 00:25:12,030
So let's check that this is
also correct, at the other
469
00:25:12,030 --> 00:25:13,380
extreme case.
470
00:25:13,380 --> 00:25:17,670
If we let k equal to 0, what
does the formula give us?
471
00:25:17,670 --> 00:25:20,710
It gives us, again, n factorial
divided by 0
472
00:25:20,710 --> 00:25:23,090
factorial times n factorial.
473
00:25:23,090 --> 00:25:27,560
According to our convention,
this again is equal to 1.
474
00:25:27,560 --> 00:25:33,680
So there is one subset of our
set that we started with that
475
00:25:33,680 --> 00:25:35,260
has zero elements.
476
00:25:35,260 --> 00:25:37,450
Which subset is it?
477
00:25:37,450 --> 00:25:39,980
It's the empty set.
478
00:25:39,980 --> 00:25:45,190
So the empty set is the single
subset of the set that we
479
00:25:45,190 --> 00:25:49,360
started with that happens to
have exactly zero elements.
480
00:25:49,360 --> 00:25:52,510
So the formula checks in this
extreme case as well.
481
00:25:52,510 --> 00:25:55,820
So we're comfortable using it.
482
00:25:55,820 --> 00:26:01,180
Now these factorials and these
coefficients are really messy
483
00:26:01,180 --> 00:26:03,050
algebraic objects.
484
00:26:03,050 --> 00:26:07,740
There's lots of beautiful
identities that they satisfy,
485
00:26:07,740 --> 00:26:10,350
which you can prove
algebraically sometimes by
486
00:26:10,350 --> 00:26:13,930
using induction and having
cancellations happen
487
00:26:13,930 --> 00:26:15,310
all over the place.
488
00:26:15,310 --> 00:26:17,780
But it's really messy.
489
00:26:17,780 --> 00:26:22,540
Sometimes you can bypass those
calculations by being clever
490
00:26:22,540 --> 00:26:24,630
and using your understanding
of what these
491
00:26:24,630 --> 00:26:26,620
coefficients stand for.
492
00:26:26,620 --> 00:26:31,490
So here's a typical example.
493
00:26:31,490 --> 00:26:35,450
What is the sum of those
binomial coefficients?
494
00:26:35,450 --> 00:26:40,130
I fix n, and sum over
all possible cases.
495
00:26:40,130 --> 00:26:44,110
So if you're an algebra genius,
you're going to take
496
00:26:44,110 --> 00:26:49,830
this expression here, plug it in
here, and then start doing
497
00:26:49,830 --> 00:26:51,460
algebra furiously.
498
00:26:51,460 --> 00:26:54,970
And half an hour later, you
may get the right answer.
499
00:26:54,970 --> 00:26:56,425
But now let's try
to be clever.
500
00:26:56,425 --> 00:26:59,470
501
00:26:59,470 --> 00:27:01,380
What does this really do?
502
00:27:01,380 --> 00:27:04,200
What does that formula count?
503
00:27:04,200 --> 00:27:07,280
We're considering k
element subsets.
504
00:27:07,280 --> 00:27:09,040
That's this number.
505
00:27:09,040 --> 00:27:12,360
And we're considering the number
of k element subsets
506
00:27:12,360 --> 00:27:14,840
for different choices of k.
507
00:27:14,840 --> 00:27:18,890
The first term in this sum
counts how many 0-element
508
00:27:18,890 --> 00:27:20,450
subsets we have.
509
00:27:20,450 --> 00:27:23,680
The next term in this sum counts
how many 1-element
510
00:27:23,680 --> 00:27:24,660
subsets we have.
511
00:27:24,660 --> 00:27:30,010
The next term counts how many
2-element subsets we have.
512
00:27:30,010 --> 00:27:33,130
So in the end, what
have we counted?
513
00:27:33,130 --> 00:27:36,660
We've counted the total
number of subsets.
514
00:27:36,660 --> 00:27:38,430
We've considered all possible
cardinalities.
515
00:27:38,430 --> 00:27:43,420
516
00:27:43,420 --> 00:27:46,850
We've counted the number
of subsets of size k.
517
00:27:46,850 --> 00:27:49,740
We've considered all
possible sizes k.
518
00:27:49,740 --> 00:27:52,230
The overall count is
going to be the
519
00:27:52,230 --> 00:27:54,356
total number of subsets.
520
00:27:54,356 --> 00:27:57,880
521
00:27:57,880 --> 00:28:00,800
And we know what this is.
522
00:28:00,800 --> 00:28:03,740
A couple of slides ago, we
discussed that this number is
523
00:28:03,740 --> 00:28:05,480
equal to 2 to the n.
524
00:28:05,480 --> 00:28:11,550
So, nice, clean and simple
answer, which is easy to guess
525
00:28:11,550 --> 00:28:15,110
once you give an interpretation
to the
526
00:28:15,110 --> 00:28:17,580
algebraic expression that you
have in front of you.
527
00:28:17,580 --> 00:28:21,610
528
00:28:21,610 --> 00:28:22,280
All right.
529
00:28:22,280 --> 00:28:27,410
So let's move again to sort of
an example in which those
530
00:28:27,410 --> 00:28:31,960
binomial coefficients are
going to show up.
531
00:28:31,960 --> 00:28:34,700
So here's the setting--
532
00:28:34,700 --> 00:28:40,900
n independent coin tosses,
and each coin toss has a
533
00:28:40,900 --> 00:28:45,770
probability, P, of resulting
in heads.
534
00:28:45,770 --> 00:28:48,320
So this is our probabilistic
experiment.
535
00:28:48,320 --> 00:28:51,200
Suppose we do 6 tosses.
536
00:28:51,200 --> 00:28:53,980
What's the probability that we
get this particular sequence
537
00:28:53,980 --> 00:28:56,320
of outcomes?
538
00:28:56,320 --> 00:28:59,800
Because of independence, we
can multiply probability.
539
00:28:59,800 --> 00:29:02,400
So it's going to be the
probability that the first
540
00:29:02,400 --> 00:29:05,570
toss results in heads, times
the probability that the
541
00:29:05,570 --> 00:29:08,770
second toss results in tails,
times the probability that the
542
00:29:08,770 --> 00:29:12,050
third one results in tails,
times probability of heads,
543
00:29:12,050 --> 00:29:14,610
times probability of heads,
times probability of heads,
544
00:29:14,610 --> 00:29:20,930
which is just P to the fourth
times (1 minus P) squared.
545
00:29:20,930 --> 00:29:24,360
So that's the probability of
this particular sequence.
546
00:29:24,360 --> 00:29:26,980
How about a different
sequence?
547
00:29:26,980 --> 00:29:32,830
If I had 4 tails and 2 heads,
but in a different order--
548
00:29:32,830 --> 00:29:39,130
549
00:29:39,130 --> 00:29:42,870
let's say if we considered
this particular outcome--
550
00:29:42,870 --> 00:29:45,480
would the answer be different?
551
00:29:45,480 --> 00:29:49,020
We would still have P, times P,
times P, times P, times (1
552
00:29:49,020 --> 00:29:51,070
minus P), times (1 minus P).
553
00:29:51,070 --> 00:29:54,670
We would get again,
the same answer.
554
00:29:54,670 --> 00:29:59,510
So what you observe from just
this example is that, more
555
00:29:59,510 --> 00:30:03,240
generally, the probability
of obtaining a particular
556
00:30:03,240 --> 00:30:08,930
sequence of heads and tails is
P to a power, equal to the
557
00:30:08,930 --> 00:30:10,300
number of heads.
558
00:30:10,300 --> 00:30:12,240
So here we had 4 heads.
559
00:30:12,240 --> 00:30:15,100
So there's P to the
fourth showing up.
560
00:30:15,100 --> 00:30:21,116
And then (1 minus P) to the
power number of tails.
561
00:30:21,116 --> 00:30:26,970
So every k head sequence--
562
00:30:26,970 --> 00:30:32,310
every outcome in which we have
exactly k heads, has the same
563
00:30:32,310 --> 00:30:37,180
probability, which is going to
be P to the k, (1 minus p), to
564
00:30:37,180 --> 00:30:38,930
the (n minus k).
565
00:30:38,930 --> 00:30:43,310
This is the probability of any
particular sequence that has
566
00:30:43,310 --> 00:30:44,980
exactly k heads.
567
00:30:44,980 --> 00:30:46,980
So that's the probability
of a particular
568
00:30:46,980 --> 00:30:48,920
sequence with k heads.
569
00:30:48,920 --> 00:30:53,160
So now let's ask the question,
what is the probability that
570
00:30:53,160 --> 00:30:57,980
my experiment results in exactly
k heads, but in some
571
00:30:57,980 --> 00:30:59,930
arbitrary order?
572
00:30:59,930 --> 00:31:02,500
So the heads could
show up anywhere.
573
00:31:02,500 --> 00:31:04,080
So there's a number
of different ways
574
00:31:04,080 --> 00:31:05,370
that this can happen.
575
00:31:05,370 --> 00:31:11,220
What's the overall probability
that this event takes place?
576
00:31:11,220 --> 00:31:15,560
So the probability of an event
taking place is the sum of the
577
00:31:15,560 --> 00:31:19,390
probabilities of all the
individual ways that
578
00:31:19,390 --> 00:31:22,020
the event can occur.
579
00:31:22,020 --> 00:31:24,410
So it's the sum of the
probabilities of all the
580
00:31:24,410 --> 00:31:27,650
outcomes that make
the event happen.
581
00:31:27,650 --> 00:31:31,420
The different ways that we can
obtain k heads are the number
582
00:31:31,420 --> 00:31:37,940
of different sequences that
contain exactly k heads.
583
00:31:37,940 --> 00:31:44,430
We just figured out that any
sequence with exactly k heads
584
00:31:44,430 --> 00:31:47,110
has this probability.
585
00:31:47,110 --> 00:31:51,110
So to do this summation, we just
need to take the common
586
00:31:51,110 --> 00:31:56,030
probability of each individual
k head sequence, times how
587
00:31:56,030 --> 00:31:58,140
many terms we have
in this sum.
588
00:31:58,140 --> 00:32:01,320
589
00:32:01,320 --> 00:32:07,020
So what we're left to do now
is to figure out how many k
590
00:32:07,020 --> 00:32:09,990
head sequences are there.
591
00:32:09,990 --> 00:32:15,448
How many outcomes are there in
which we have exactly k heads.
592
00:32:15,448 --> 00:32:18,940
593
00:32:18,940 --> 00:32:21,270
OK.
594
00:32:21,270 --> 00:32:27,590
So what are the ways that I can
describe to you a sequence
595
00:32:27,590 --> 00:32:30,600
with k heads?
596
00:32:30,600 --> 00:32:34,970
I can take my n slots
that corresponds to
597
00:32:34,970 --> 00:32:36,220
the different tosses.
598
00:32:36,220 --> 00:32:42,920
599
00:32:42,920 --> 00:32:45,420
I'm interested in particular
sequences that
600
00:32:45,420 --> 00:32:47,750
have exactly k heads.
601
00:32:47,750 --> 00:32:53,590
So what I need to do is to
choose k slots and assign
602
00:32:53,590 --> 00:32:54,850
heads to them.
603
00:32:54,850 --> 00:33:05,530
604
00:33:05,530 --> 00:33:11,580
So to specify a sequence that
has exactly k heads is the
605
00:33:11,580 --> 00:33:17,380
same thing as drawing this
picture and telling you which
606
00:33:17,380 --> 00:33:23,640
are the k slots that happened
to have heads.
607
00:33:23,640 --> 00:33:30,110
So I need to choose out of those
n slots, k of them, and
608
00:33:30,110 --> 00:33:31,635
assign them heads.
609
00:33:31,635 --> 00:33:35,290
In how many ways can I
choose this k slots?
610
00:33:35,290 --> 00:33:41,640
Well, it's the question of
starting with a set of n slots
611
00:33:41,640 --> 00:33:47,080
and choosing k slots out
of the n available.
612
00:33:47,080 --> 00:33:55,540
So the number of k head
sequences is the same as the
613
00:33:55,540 --> 00:34:04,300
number of k element subsets of
the set of slots that we
614
00:34:04,300 --> 00:34:10,520
started with, which are
the n slots 1 up to n.
615
00:34:10,520 --> 00:34:12,800
We know what that number is.
616
00:34:12,800 --> 00:34:18,770
We counted, before, the number
of k element subsets, starting
617
00:34:18,770 --> 00:34:20,290
with a set with n elements.
618
00:34:20,290 --> 00:34:23,030
And we gave a symbol to that
number, which is that
619
00:34:23,030 --> 00:34:24,850
thing, n choose k.
620
00:34:24,850 --> 00:34:28,110
So this is the final answer
that we obtain.
621
00:34:28,110 --> 00:34:32,449
So these are the so-called
binomial probabilities.
622
00:34:32,449 --> 00:34:35,190
And they gave us the
probabilities for different
623
00:34:35,190 --> 00:34:39,580
numbers of heads starting with
a fair coin that's being
624
00:34:39,580 --> 00:34:42,050
tossed a number of times.
625
00:34:42,050 --> 00:34:46,170
This formula is correct, of
course, for reasonable values
626
00:34:46,170 --> 00:34:52,370
of k, meaning its correct for
k equals 0, 1, up to n.
627
00:34:52,370 --> 00:34:57,650
If k is bigger than n, what's
the probability of k heads?
628
00:34:57,650 --> 00:35:01,340
If k is bigger than n, there's
no way to obtain k heads, so
629
00:35:01,340 --> 00:35:03,480
that probability is,
of course, zero.
630
00:35:03,480 --> 00:35:07,610
So these probabilities only
makes sense for the numbers k
631
00:35:07,610 --> 00:35:10,405
that are possible, given
that we have n tosses.
632
00:35:10,405 --> 00:35:13,200
633
00:35:13,200 --> 00:35:16,850
And now a question similar
to the one we had in
634
00:35:16,850 --> 00:35:18,480
the previous slide.
635
00:35:18,480 --> 00:35:22,910
If I write down this
summation--
636
00:35:22,910 --> 00:35:28,240
even worse algebra than the one
in the previous slide--
637
00:35:28,240 --> 00:35:35,840
what do you think this number
will turn out to be?
638
00:35:35,840 --> 00:35:39,930
It should be 1 because this
is the probability
639
00:35:39,930 --> 00:35:42,930
of obtaining k heads.
640
00:35:42,930 --> 00:35:45,470
When we do the summation, what
we're doing is we're
641
00:35:45,470 --> 00:35:48,550
considering the probability of
0 heads, plus the probability
642
00:35:48,550 --> 00:35:50,780
of 1 head, plus the probability
of 2 heads, plus
643
00:35:50,780 --> 00:35:52,420
the probability of n heads.
644
00:35:52,420 --> 00:35:54,780
We've exhausted all the
possibilities in our
645
00:35:54,780 --> 00:35:55,720
experiment.
646
00:35:55,720 --> 00:35:58,730
So the overall probability,
when you exhaust all
647
00:35:58,730 --> 00:36:01,160
possibilities, must
be equal to 1.
648
00:36:01,160 --> 00:36:04,180
So that's yet another beautiful
formula that
649
00:36:04,180 --> 00:36:06,960
evaluates into something
really simple.
650
00:36:06,960 --> 00:36:11,460
And if you tried to prove this
identity algebraically, of
651
00:36:11,460 --> 00:36:16,030
course, you would have to
suffer quite a bit.
652
00:36:16,030 --> 00:36:20,130
So now armed with the binomial
probabilities, we can do the
653
00:36:20,130 --> 00:36:21,380
harder problems.
654
00:36:21,380 --> 00:36:23,480
655
00:36:23,480 --> 00:36:27,340
So let's take the same
experiment again.
656
00:36:27,340 --> 00:36:32,610
We flip a coin independently
10 times.
657
00:36:32,610 --> 00:36:37,450
So these 10 tosses
are independent.
658
00:36:37,450 --> 00:36:40,000
We flip it 10 times.
659
00:36:40,000 --> 00:36:43,985
We don't see the result, but
somebody comes and tells us,
660
00:36:43,985 --> 00:36:47,890
you know, there were exactly
3 heads in the 10
661
00:36:47,890 --> 00:36:49,688
tosses that you had.
662
00:36:49,688 --> 00:36:50,930
OK?
663
00:36:50,930 --> 00:36:53,280
So a certain event happened.
664
00:36:53,280 --> 00:36:57,450
And now you're asked to find
the probability of another
665
00:36:57,450 --> 00:37:01,400
event, which is that the first
2 tosses were heads.
666
00:37:01,400 --> 00:37:08,990
Let's call that event A. OK.
667
00:37:08,990 --> 00:37:14,320
So are we in the setting
of discrete
668
00:37:14,320 --> 00:37:16,850
uniform probability laws?
669
00:37:16,850 --> 00:37:21,760
When we toss a coin multiple
times, is it the case that all
670
00:37:21,760 --> 00:37:24,130
outcomes are equally likely?
671
00:37:24,130 --> 00:37:27,850
All sequences are
equally likely?
672
00:37:27,850 --> 00:37:30,515
That's the case if you
have a fair coin--
673
00:37:30,515 --> 00:37:32,630
that all sequences are
equally likely.
674
00:37:32,630 --> 00:37:37,170
But if your coin is not fair,
of course, heads/heads is
675
00:37:37,170 --> 00:37:39,630
going to have a different
probability than tails/tails.
676
00:37:39,630 --> 00:37:43,720
If your coin is biased towards
heads, then heads/heads is
677
00:37:43,720 --> 00:37:45,330
going to be more likely.
678
00:37:45,330 --> 00:37:49,440
So we're not quite in
the uniform setting.
679
00:37:49,440 --> 00:37:53,450
Our overall sample space, omega,
does not have equally
680
00:37:53,450 --> 00:37:55,680
likely elements.
681
00:37:55,680 --> 00:37:57,880
Do we care about that?
682
00:37:57,880 --> 00:37:59,700
Not necessarily.
683
00:37:59,700 --> 00:38:04,570
All the action now happens
inside the event B that we are
684
00:38:04,570 --> 00:38:06,510
told has occurred.
685
00:38:06,510 --> 00:38:10,000
So we have our big sample
space, omega.
686
00:38:10,000 --> 00:38:13,860
Elements of that sample space
are not equally likely.
687
00:38:13,860 --> 00:38:17,390
We are told that a certain
event B occurred.
688
00:38:17,390 --> 00:38:21,830
And inside that event B, we're
asked to find the conditional
689
00:38:21,830 --> 00:38:26,100
probability that A has
also occurred.
690
00:38:26,100 --> 00:38:30,850
Now here's the lucky thing,
inside the event B, all
691
00:38:30,850 --> 00:38:33,270
outcomes are equally likely.
692
00:38:33,270 --> 00:38:35,920
693
00:38:35,920 --> 00:38:40,710
The outcomes inside B are the
sequences of 10 tosses that
694
00:38:40,710 --> 00:38:42,970
have exactly 3 heads.
695
00:38:42,970 --> 00:38:47,370
Every 3-head sequence has
this probability.
696
00:38:47,370 --> 00:38:50,790
So the elements of
B are equally
697
00:38:50,790 --> 00:38:52,760
likely with each other.
698
00:38:52,760 --> 00:38:55,800
699
00:38:55,800 --> 00:39:01,030
Once we condition on the event
B having occurred, what
700
00:39:01,030 --> 00:39:03,740
happens to the probabilities
of the different outcomes
701
00:39:03,740 --> 00:39:05,430
inside here?
702
00:39:05,430 --> 00:39:09,790
Well, conditional probability
laws keep the same proportions
703
00:39:09,790 --> 00:39:11,710
as the unconditional ones.
704
00:39:11,710 --> 00:39:15,930
The elements of B were equally
likely when we started, so
705
00:39:15,930 --> 00:39:21,590
they're equally likely once we
are told that B has occurred.
706
00:39:21,590 --> 00:39:26,440
So to do with this problem, we
need to just transport us to
707
00:39:26,440 --> 00:39:30,680
this smaller universe and think
about what's happening
708
00:39:30,680 --> 00:39:32,920
in that little universe.
709
00:39:32,920 --> 00:39:36,150
In that little universe,
all elements of
710
00:39:36,150 --> 00:39:39,930
B are equally likely.
711
00:39:39,930 --> 00:39:43,860
So to find the probability of
some subset of that set, we
712
00:39:43,860 --> 00:39:47,250
only need to count the
cardinality of B, and count
713
00:39:47,250 --> 00:39:51,090
the cardinality of A.
So let's do that.
714
00:39:51,090 --> 00:39:53,780
Number of outcomes in B--
715
00:39:53,780 --> 00:40:00,290
in how many ways can we get
3 heads out of 10 tosses?
716
00:40:00,290 --> 00:40:03,190
That's the number we considered
before, and
717
00:40:03,190 --> 00:40:06,250
it's 10 choose 3.
718
00:40:06,250 --> 00:40:11,020
This is the number of
3-head sequences
719
00:40:11,020 --> 00:40:13,840
when you have 10 tosses.
720
00:40:13,840 --> 00:40:20,580
Now let's look at the event A.
The event A is that the first
721
00:40:20,580 --> 00:40:26,150
2 tosses where heads, but we're
living now inside this
722
00:40:26,150 --> 00:40:31,220
universe B. Given that B
occurred, how many elements
723
00:40:31,220 --> 00:40:34,760
does A have in there?
724
00:40:34,760 --> 00:40:41,536
In how many ways can A happen
inside the B universe.
725
00:40:41,536 --> 00:40:46,860
If you're told that the
first 2 were heads--
726
00:40:46,860 --> 00:40:49,470
sorry.
727
00:40:49,470 --> 00:40:54,540
So out of the outcomes in B that
have 3 heads, how many
728
00:40:54,540 --> 00:40:56,630
start with heads/heads?
729
00:40:56,630 --> 00:41:00,370
Well, if it starts with
heads/heads, then the only
730
00:41:00,370 --> 00:41:04,830
uncertainty is the location
of the third head.
731
00:41:04,830 --> 00:41:07,940
So we started with heads/heads,
we're going to
732
00:41:07,940 --> 00:41:13,020
have three heads, the question
is, where is that third head
733
00:41:13,020 --> 00:41:14,090
going to be.
734
00:41:14,090 --> 00:41:16,540
It has eight possibilities.
735
00:41:16,540 --> 00:41:20,940
So slot 1 is heads, slot 2 is
heads, the third heads can be
736
00:41:20,940 --> 00:41:22,140
anywhere else.
737
00:41:22,140 --> 00:41:25,020
So there's 8 possibilities
for where the third
738
00:41:25,020 --> 00:41:26,270
head is going to be.
739
00:41:26,270 --> 00:41:29,630
740
00:41:29,630 --> 00:41:31,660
OK.
741
00:41:31,660 --> 00:41:36,720
So what we have counted here is
really the cardinality of A
742
00:41:36,720 --> 00:41:43,450
intersection B, which is out of
the elements in B, how many
743
00:41:43,450 --> 00:41:49,410
of them make A happen, divided
by the cardinality of B. And
744
00:41:49,410 --> 00:41:53,860
that gives us the answer, which
is going to be 10 choose
745
00:41:53,860 --> 00:41:57,530
3, divided by 8.
746
00:41:57,530 --> 00:42:01,330
And I should probably redraw a
little bit of the picture that
747
00:42:01,330 --> 00:42:02,510
they have here.
748
00:42:02,510 --> 00:42:06,690
The set A is not necessarily
contained in B. It could also
749
00:42:06,690 --> 00:42:14,750
have stuff outside B. So the
event that the first 2 tosses
750
00:42:14,750 --> 00:42:18,690
are heads can happen with a
total of 3 heads, but it can
751
00:42:18,690 --> 00:42:22,650
also happen with a different
total number of heads.
752
00:42:22,650 --> 00:42:27,340
But once we are transported
inside the set B, what we need
753
00:42:27,340 --> 00:42:32,460
to count is just this part of
A. It's A intersection B and
754
00:42:32,460 --> 00:42:35,180
compare it with the total number
of elements in the set
755
00:42:35,180 --> 00:42:40,310
B. Did I write it the
opposite way?
756
00:42:40,310 --> 00:42:41,700
Yes.
757
00:42:41,700 --> 00:42:46,330
So this is 8 over 10 choose 3.
758
00:42:46,330 --> 00:42:49,260
759
00:42:49,260 --> 00:42:49,640
OK.
760
00:42:49,640 --> 00:42:52,965
So we're going to close with a
more difficult problem now.
761
00:42:52,965 --> 00:42:57,920
762
00:42:57,920 --> 00:43:00,580
OK.
763
00:43:00,580 --> 00:43:05,650
This business of n choose k has
to do with starting with a
764
00:43:05,650 --> 00:43:11,350
set and picking a subset
of k elements.
765
00:43:11,350 --> 00:43:15,080
Another way of thinking of that
is that we start with a
766
00:43:15,080 --> 00:43:20,770
set with n elements and you
choose a subset that has k,
767
00:43:20,770 --> 00:43:24,350
which means that there's n
minus k that are left.
768
00:43:24,350 --> 00:43:29,980
Picking a subset is the same as
partitioning our set into
769
00:43:29,980 --> 00:43:32,510
two pieces.
770
00:43:32,510 --> 00:43:36,010
Now let's generalize this
question and start counting
771
00:43:36,010 --> 00:43:38,150
partitions in general.
772
00:43:38,150 --> 00:43:42,570
Somebody gives you a set
that has n elements.
773
00:43:42,570 --> 00:43:44,670
Somebody gives you also
certain numbers--
774
00:43:44,670 --> 00:43:49,400
n1, n2, n3, let's say,
n4, where these
775
00:43:49,400 --> 00:43:53,740
numbers add up to n.
776
00:43:53,740 --> 00:43:58,740
And you're asked to partition
this set into four subsets
777
00:43:58,740 --> 00:44:01,450
where each one of the subsets
has this particular
778
00:44:01,450 --> 00:44:02,580
cardinality.
779
00:44:02,580 --> 00:44:08,250
So you're asking to cut it into
four pieces, each one
780
00:44:08,250 --> 00:44:11,100
having the prescribed
cardinality.
781
00:44:11,100 --> 00:44:15,090
In how many ways can we
do this partitioning?
782
00:44:15,090 --> 00:44:19,370
n choose k was the answer when
we partitioned in two pieces,
783
00:44:19,370 --> 00:44:21,910
what's the answer
more generally?
784
00:44:21,910 --> 00:44:26,230
For a concrete example of a
partition, you have your 52
785
00:44:26,230 --> 00:44:32,120
card deck and you deal, as in
bridge, by giving 13 cards to
786
00:44:32,120 --> 00:44:34,000
each one of the players.
787
00:44:34,000 --> 00:44:38,080
Assuming that the dealing is
done fairly and with a well
788
00:44:38,080 --> 00:44:43,790
shuffled deck of cards, every
particular partition of the 52
789
00:44:43,790 --> 00:44:50,590
cards into four hands, that is
four subsets of 13 each,
790
00:44:50,590 --> 00:44:52,380
should be equally likely.
791
00:44:52,380 --> 00:44:56,140
So we take the 52 cards and we
partition them into subsets of
792
00:44:56,140 --> 00:44:58,550
13, 13, 13, and 13.
793
00:44:58,550 --> 00:45:01,020
And we assume that all possible
partitions, all
794
00:45:01,020 --> 00:45:04,240
possible ways of dealing the
cards are equally likely.
795
00:45:04,240 --> 00:45:07,560
So we are again in a setting
where we can use counting,
796
00:45:07,560 --> 00:45:10,410
because all the possible
outcomes are equally likely.
797
00:45:10,410 --> 00:45:14,050
So an outcome of the experiment
is the hands that
798
00:45:14,050 --> 00:45:17,070
each player ends up getting.
799
00:45:17,070 --> 00:45:20,170
And when you get the cards in
your hands, it doesn't matter
800
00:45:20,170 --> 00:45:22,000
in which order that
you got them.
801
00:45:22,000 --> 00:45:25,460
It only matters what cards
you have on you.
802
00:45:25,460 --> 00:45:31,160
So it only matters which subset
of the cards you got.
803
00:45:31,160 --> 00:45:31,590
All right.
804
00:45:31,590 --> 00:45:35,820
So what's the cardinality of
the sample space in this
805
00:45:35,820 --> 00:45:37,160
experiment?
806
00:45:37,160 --> 00:45:42,010
So let's do it for the concrete
numbers that we have
807
00:45:42,010 --> 00:45:49,390
for the problem of partitioning
52 cards.
808
00:45:49,390 --> 00:45:52,540
So think of dealing as follows--
you shuffle the deck
809
00:45:52,540 --> 00:45:56,250
perfectly, and then you take the
top 13 cards and give them
810
00:45:56,250 --> 00:45:57,740
to one person.
811
00:45:57,740 --> 00:46:03,230
In how many possible hands are
there for that person?
812
00:46:03,230 --> 00:46:08,970
Out of the 52 cards, I choose 13
at random and give them to
813
00:46:08,970 --> 00:46:10,680
the first person.
814
00:46:10,680 --> 00:46:13,260
Having done that, what
happens next?
815
00:46:13,260 --> 00:46:16,260
I'm left with 39 cards.
816
00:46:16,260 --> 00:46:20,600
And out of those 39 cards, I
pick 13 of them and give them
817
00:46:20,600 --> 00:46:22,250
to the second person.
818
00:46:22,250 --> 00:46:25,790
Now I'm left with 26 cards.
819
00:46:25,790 --> 00:46:30,920
Out of those 26, I choose 13,
give them to the third person.
820
00:46:30,920 --> 00:46:34,040
And for the last person there
isn't really any choice.
821
00:46:34,040 --> 00:46:37,890
Out of the 13, I have to give
that person all 13.
822
00:46:37,890 --> 00:46:40,230
And that number is
just equal to 1.
823
00:46:40,230 --> 00:46:43,530
So we don't care about it.
824
00:46:43,530 --> 00:46:43,910
All right.
825
00:46:43,910 --> 00:46:48,270
So next thing you do is to write
down the formulas for
826
00:46:48,270 --> 00:46:49,450
these numbers.
827
00:46:49,450 --> 00:46:52,450
So, for example, here you
would have 52 factorial,
828
00:46:52,450 --> 00:46:55,880
divided by 13 factorial,
times 39
829
00:46:55,880 --> 00:46:59,040
factorial, and you continue.
830
00:46:59,040 --> 00:47:01,310
And then there are nice
cancellations that happen.
831
00:47:01,310 --> 00:47:05,120
This 39 factorial is going to
cancel the 39 factorial that
832
00:47:05,120 --> 00:47:07,020
comes from there, and so on.
833
00:47:07,020 --> 00:47:10,200
After you do the cancellations
and all the algebra, you're
834
00:47:10,200 --> 00:47:13,380
left with this particular
answer, which is the number of
835
00:47:13,380 --> 00:47:18,120
possible partitions of 52 cards
into four players where
836
00:47:18,120 --> 00:47:21,710
each player gets exactly
13 hands.
837
00:47:21,710 --> 00:47:25,140
If you were to generalize this
formula to the setting that we
838
00:47:25,140 --> 00:47:29,200
have here, the more general
formula is--
839
00:47:29,200 --> 00:47:33,840
you have n factorial, where n is
the number of objects that
840
00:47:33,840 --> 00:47:39,770
you are distributing, divided
by the product of the
841
00:47:39,770 --> 00:47:41,840
factorials of the--
842
00:47:41,840 --> 00:47:46,000
OK, here I'm doing it for
the case where we split
843
00:47:46,000 --> 00:47:49,310
it into four sets.
844
00:47:49,310 --> 00:47:53,740
So that would be the answer when
we partition a set into
845
00:47:53,740 --> 00:47:57,780
four subsets of prescribed
cardinalities.
846
00:47:57,780 --> 00:48:00,120
And you can guess how that
formula would generalize if
847
00:48:00,120 --> 00:48:03,590
you want to split it into
five sets or six sets.
848
00:48:03,590 --> 00:48:03,950
OK.
849
00:48:03,950 --> 00:48:10,190
So far we just figured out the
size of the sample space.
850
00:48:10,190 --> 00:48:14,660
Now we need to look at our
event, which is the event that
851
00:48:14,660 --> 00:48:20,640
each player gets an ace, let's
call that event A. In how many
852
00:48:20,640 --> 00:48:22,800
ways can that event happens?
853
00:48:22,800 --> 00:48:26,970
How many possible hands are
there in which every player
854
00:48:26,970 --> 00:48:29,350
has exactly one ace?
855
00:48:29,350 --> 00:48:33,100
So I need to think about the
sequential process by which I
856
00:48:33,100 --> 00:48:36,860
distribute the cards so that
everybody gets exactly one
857
00:48:36,860 --> 00:48:40,440
ace, and then try to think
in how many ways can that
858
00:48:40,440 --> 00:48:42,210
sequential process happen.
859
00:48:42,210 --> 00:48:45,660
So one way of making sure that
everybody gets exactly one ace
860
00:48:45,660 --> 00:48:46,730
is the following--
861
00:48:46,730 --> 00:48:51,210
I take the four aces and I
distribute them randomly to
862
00:48:51,210 --> 00:48:53,970
the four players, but making
sure that each one gets
863
00:48:53,970 --> 00:48:55,580
exactly one ace.
864
00:48:55,580 --> 00:48:57,510
In how many ways can
that happen?
865
00:48:57,510 --> 00:49:02,210
I take the ace of spades and I
send it to a random person out
866
00:49:02,210 --> 00:49:03,210
of the four.
867
00:49:03,210 --> 00:49:07,430
So there's 4 choices for this.
868
00:49:07,430 --> 00:49:10,280
Then I'm left with 3
aces to distribute.
869
00:49:10,280 --> 00:49:14,050
That person already
gotten an ace.
870
00:49:14,050 --> 00:49:17,270
I take the next ace, and
I give it to one of
871
00:49:17,270 --> 00:49:19,590
the 3 people remaining.
872
00:49:19,590 --> 00:49:22,480
So there's 3 choices
for how to do that.
873
00:49:22,480 --> 00:49:26,970
And then for the next ace,
there's 2 people who have not
874
00:49:26,970 --> 00:49:28,640
yet gotten an ace,
and they give it
875
00:49:28,640 --> 00:49:30,770
randomly to one of them.
876
00:49:30,770 --> 00:49:36,760
So these are the possible ways
of distributing for the 4
877
00:49:36,760 --> 00:49:42,040
aces, so that each person
gets exactly one.
878
00:49:42,040 --> 00:49:44,230
It's actually the same
as this problem.
879
00:49:44,230 --> 00:49:48,930
Starting with a set of four
things, in how many ways can I
880
00:49:48,930 --> 00:49:53,430
partition them into four subsets
where the first set
881
00:49:53,430 --> 00:49:56,220
has one element, the second has
one element, the third one
882
00:49:56,220 --> 00:49:58,270
has another element,
and so on.
883
00:49:58,270 --> 00:50:05,710
So it agrees with that formula
by giving us 4 factorial.
884
00:50:05,710 --> 00:50:06,040
OK.
885
00:50:06,040 --> 00:50:09,400
So there are different ways
of distributing the aces.
886
00:50:09,400 --> 00:50:11,760
And then there's different
ways of distributing the
887
00:50:11,760 --> 00:50:13,460
remaining 48 cards.
888
00:50:13,460 --> 00:50:15,110
How many ways are there?
889
00:50:15,110 --> 00:50:18,920
Well, I have 48 cards that I'm
going to distribute to four
890
00:50:18,920 --> 00:50:22,760
players by giving 12
cards to each one.
891
00:50:22,760 --> 00:50:26,430
It's exactly the same question
as the one we had here, except
892
00:50:26,430 --> 00:50:30,230
that now it's 48 cards,
12 to each person.
893
00:50:30,230 --> 00:50:33,400
And that gives us this
particular count.
894
00:50:33,400 --> 00:50:39,216
So putting all that together
gives us the different ways
895
00:50:39,216 --> 00:50:43,350
that we can distribute the cards
to the four players so
896
00:50:43,350 --> 00:50:45,910
that each one gets
exactly one ace.
897
00:50:45,910 --> 00:50:48,600
The number of possible ways
is going to be this four
898
00:50:48,600 --> 00:50:54,610
factorial, coming from here,
times this number--
899
00:50:54,610 --> 00:50:56,890
this gives us the number of
ways that the event of
900
00:50:56,890 --> 00:50:58,760
interest can happen--
901
00:50:58,760 --> 00:51:02,760
and then the denominator is the
cardinality of our sample
902
00:51:02,760 --> 00:51:04,930
space, which is this number.
903
00:51:04,930 --> 00:51:07,560
So this looks like
a horrible mess.
904
00:51:07,560 --> 00:51:10,590
It turns out that this
expression does simplify to
905
00:51:10,590 --> 00:51:13,180
something really,
really simple.
906
00:51:13,180 --> 00:51:16,420
And if you look at the textbook
for this problem, you
907
00:51:16,420 --> 00:51:18,750
will see an alternative
derivation that gives you a
908
00:51:18,750 --> 00:51:22,720
short cut to the same
numerical answer.
909
00:51:22,720 --> 00:51:23,160
All right.
910
00:51:23,160 --> 00:51:25,240
So that basically concludes
chapter one.
911
00:51:25,240 --> 00:51:29,940
From next time we're going to
consider introducing random
912
00:51:29,940 --> 00:51:32,950
variables and make the subject
even more interesting.
913
00:51:32,950 --> 00:51:34,200