1
00:00:00,000 --> 00:00:00,820
2
00:00:00,820 --> 00:00:01,640
Hi.
3
00:00:01,640 --> 00:00:04,500
Today we're going to do another
fun problem that
4
00:00:04,500 --> 00:00:07,750
involves rolling two dice.
5
00:00:07,750 --> 00:00:11,210
So if you guys happen to
frequent casinos, this problem
6
00:00:11,210 --> 00:00:13,750
might be really useful
for you.
7
00:00:13,750 --> 00:00:14,680
I'm just kidding.
8
00:00:14,680 --> 00:00:18,310
But in all seriousness, this
problem is a good problem,
9
00:00:18,310 --> 00:00:22,760
because it's going to remind
us how and when to use the
10
00:00:22,760 --> 00:00:24,060
discrete uniform law.
11
00:00:24,060 --> 00:00:26,650
Don't worry, I'll review
what that says.
12
00:00:26,650 --> 00:00:29,140
And it's also going to exercise
your understanding of
13
00:00:29,140 --> 00:00:31,030
conditional probability.
14
00:00:31,030 --> 00:00:32,110
So quick recap.
15
00:00:32,110 --> 00:00:35,140
The discrete uniform law says
that when your sample space is
16
00:00:35,140 --> 00:00:38,670
discrete, and when the outcomes
in your sample space
17
00:00:38,670 --> 00:00:42,750
are equally likely, then to
compute the probability of any
18
00:00:42,750 --> 00:00:47,855
event A, you can simply count
the number of outcomes in A
19
00:00:47,855 --> 00:00:52,730
and divide it by the total
number of possible outcomes.
20
00:00:52,730 --> 00:00:55,130
OK, so coming back
to our problem.
21
00:00:55,130 --> 00:00:59,060
The problem statement tells
us that we roll two fair
22
00:00:59,060 --> 00:01:01,150
six-sided die.
23
00:01:01,150 --> 00:01:05,069
And it also tells us that each
one of the 36 possible
24
00:01:05,069 --> 00:01:08,940
outcomes is assumed to
be equally likely.
25
00:01:08,940 --> 00:01:12,220
So you know alarm bell should
be going off in your head.
26
00:01:12,220 --> 00:01:14,810
Our sample space is
clearly discrete.
27
00:01:14,810 --> 00:01:16,930
And it says explicitly that all
28
00:01:16,930 --> 00:01:18,380
outcomes are equally likely.
29
00:01:18,380 --> 00:01:22,020
So clearly, we can use the
discrete uniform law.
30
00:01:22,020 --> 00:01:25,850
And again, this is helpful
because it reduces a problem
31
00:01:25,850 --> 00:01:30,090
of computing probabilities
to a problem of counting.
32
00:01:30,090 --> 00:01:32,950
OK, and before we go any
further, I just want to review
33
00:01:32,950 --> 00:01:34,850
what this graph is plotting.
34
00:01:34,850 --> 00:01:38,990
You've seen it a few times, but
just to clarify, on one
35
00:01:38,990 --> 00:01:42,430
axis, we're plotting the outcome
of the first die roll,
36
00:01:42,430 --> 00:01:44,960
and on the second axis, we're
plotting the outcome of the
37
00:01:44,960 --> 00:01:46,810
second die roll.
38
00:01:46,810 --> 00:01:50,750
So if you got a 4 on your first
die, and you get a 1 on
39
00:01:50,750 --> 00:01:54,590
your second die, that
corresponds to this point over
40
00:01:54,590 --> 00:01:56,940
4 and up 1.
41
00:01:56,940 --> 00:02:01,060
OK, so part a asks us to
find the probability
42
00:02:01,060 --> 00:02:04,500
that doubles our rolls.
43
00:02:04,500 --> 00:02:07,480
So let's use some shorthand.
44
00:02:07,480 --> 00:02:10,960
We're going to let D be the
event that doubles are rolled.
45
00:02:10,960 --> 00:02:14,720
And we want to compute
the probability of D.
46
00:02:14,720 --> 00:02:17,570
I argue before we can use the
discrete uniform law.
47
00:02:17,570 --> 00:02:22,470
So if we apply that, we just
get the number of outcomes
48
00:02:22,470 --> 00:02:26,050
that comprise the event "doubles
rolled" divided by
49
00:02:26,050 --> 00:02:30,360
36, because there are 36
possible outcomes, which you
50
00:02:30,360 --> 00:02:34,380
can see just by counting
the dots in this graph.
51
00:02:34,380 --> 00:02:37,710
Six possible outcomes for the
first die, six possible
52
00:02:37,710 --> 00:02:38,970
outcomes for the second die.
53
00:02:38,970 --> 00:02:39,555
That's how you--
54
00:02:39,555 --> 00:02:42,800
6 times 6 is 36.
55
00:02:42,800 --> 00:02:45,440
So I've been assuming this
entire time that you know what
56
00:02:45,440 --> 00:02:46,250
doubles are.
57
00:02:46,250 --> 00:02:50,470
For those of you who don't know,
doubles is essentially
58
00:02:50,470 --> 00:02:54,190
when that number on the first
die matches the number on the
59
00:02:54,190 --> 00:02:56,120
second die.
60
00:02:56,120 --> 00:03:00,480
So this outcome here 1-1 is
part of the event "doubles
61
00:03:00,480 --> 00:03:08,076
rolled." Similarly, 2-2, 3-3,
4-4, 5-5, and 6-6--
62
00:03:08,076 --> 00:03:11,510
these six points comprise the
event "doubles rolled."
63
00:03:11,510 --> 00:03:16,000
So we can go ahead and
put 6 over 36,
64
00:03:16,000 --> 00:03:18,340
which is equal to 1/6.
65
00:03:18,340 --> 00:03:20,730
So we're done with part a.
66
00:03:20,730 --> 00:03:22,530
We haven't seen any
conditioning yet.
67
00:03:22,530 --> 00:03:24,840
The conditioning comes
in part b.
68
00:03:24,840 --> 00:03:28,060
So in part b we're still
interested in the event D, in
69
00:03:28,060 --> 00:03:31,020
the event that "doubles are
rolled." But now we want to
70
00:03:31,020 --> 00:03:35,710
compute this probability
conditioned on the event that
71
00:03:35,710 --> 00:03:39,930
the sum of the results is
less than or equal to 4.
72
00:03:39,930 --> 00:03:45,150
So I'm going to use this
shorthand sum less than or
73
00:03:45,150 --> 00:03:50,180
equal to 4 to denote the event
that the role results in the
74
00:03:50,180 --> 00:03:53,010
sum of 4 or smaller.
75
00:03:53,010 --> 00:03:55,570
So there's two ways we're
going to go about
76
00:03:55,570 --> 00:03:57,660
solving part b.
77
00:03:57,660 --> 00:03:59,900
Let's just jump right
into the first way.
78
00:03:59,900 --> 00:04:03,330
The first way is applying the
definition of conditional
79
00:04:03,330 --> 00:04:04,430
probability.
80
00:04:04,430 --> 00:04:10,440
So hopefully you remember that
this is just probability of D
81
00:04:10,440 --> 00:04:14,480
intersect sum less than or
equal to 4, divided by
82
00:04:14,480 --> 00:04:19,870
probability of sum less
than or equal to 4.
83
00:04:19,870 --> 00:04:23,980
Now, sum less than or equal to
4 and D intersect sum less
84
00:04:23,980 --> 00:04:26,950
than or equal to 4 are
just two events.
85
00:04:26,950 --> 00:04:29,740
And so we can apply the discrete
uniform law to
86
00:04:29,740 --> 00:04:33,170
calculate both the numerator
and the denominator.
87
00:04:33,170 --> 00:04:36,050
So let's start with the
denominator first because it
88
00:04:36,050 --> 00:04:38,070
seems a little bit easier.
89
00:04:38,070 --> 00:04:42,930
So sum less than or equal to
4, let's figure this out.
90
00:04:42,930 --> 00:04:45,970
Well, 1-1 gives us a sum
of 2, that's less
91
00:04:45,970 --> 00:04:48,000
than or equal to 4.
92
00:04:48,000 --> 00:04:49,535
2-1 gives us 3.
93
00:04:49,535 --> 00:04:51,346
3-1 gives us 4.
94
00:04:51,346 --> 00:04:55,960
4-1 gives us 5, so we don't want
to include this or this,
95
00:04:55,960 --> 00:04:57,160
or this point.
96
00:04:57,160 --> 00:04:59,380
And you can sort of convince
yourself that the next point
97
00:04:59,380 --> 00:05:01,420
we want to include
is this one.
98
00:05:01,420 --> 00:05:05,040
That corresponds to 2-2, which
is 4, so it makes sense that
99
00:05:05,040 --> 00:05:09,410
these guys should form the
boundary, because all dots
100
00:05:09,410 --> 00:05:12,540
sort of up and to the right
will have a bigger sum.
101
00:05:12,540 --> 00:05:14,350
3-1 gives us 4.
102
00:05:14,350 --> 00:05:18,170
And 1-2 gives us 3.
103
00:05:18,170 --> 00:05:20,172
So these six points--
104
00:05:20,172 --> 00:05:22,850
1, 2, 3, 4, 5, 6--
105
00:05:22,850 --> 00:05:26,970
are the outcomes that comprise
the event sum less than or
106
00:05:26,970 --> 00:05:28,910
equal to 4.
107
00:05:28,910 --> 00:05:34,480
So we can go ahead and write in
the denominator, 6 over 36,
108
00:05:34,480 --> 00:05:38,310
because we just counted the
outcomes in sum less than or
109
00:05:38,310 --> 00:05:40,620
equal to, 4 and divided
it by the number
110
00:05:40,620 --> 00:05:43,170
of outcomes in omega.
111
00:05:43,170 --> 00:05:45,510
Now, let's compute
the numerator.
112
00:05:45,510 --> 00:05:49,410
D intersect sum less
than or equal to 4.
113
00:05:49,410 --> 00:05:51,920
So we already found the
blue check marks.
114
00:05:51,920 --> 00:05:55,380
Those correspond to sum less
than or equal to 4.
115
00:05:55,380 --> 00:05:58,350
Out of the points that have blue
check marks, which one
116
00:05:58,350 --> 00:05:59,740
correspond to doubles?
117
00:05:59,740 --> 00:06:02,590
Well, they're actually
already circled.
118
00:06:02,590 --> 00:06:04,370
It's just these two points.
119
00:06:04,370 --> 00:06:08,920
So we don't even need to circle
those, so we get 2 over
120
00:06:08,920 --> 00:06:11,830
36, using the discrete
uniform law.
121
00:06:11,830 --> 00:06:15,080
And you see that these two
36s cancel each other.
122
00:06:15,080 --> 00:06:19,980
So you just get 2/6 or 1/3.
123
00:06:19,980 --> 00:06:23,780
So that is one way of solving
part b, but I want to take
124
00:06:23,780 --> 00:06:25,930
you, guys, through a different
way, which I think is
125
00:06:25,930 --> 00:06:30,240
important, and that make sure
you really understand what
126
00:06:30,240 --> 00:06:32,630
conditioning means.
127
00:06:32,630 --> 00:06:39,000
So another way that you can
solve part b is to say, OK, we
128
00:06:39,000 --> 00:06:41,710
are now in the universe, we
are in the conditional
129
00:06:41,710 --> 00:06:44,730
universe, where we know
the sum of our
130
00:06:44,730 --> 00:06:49,090
results is 4 or smaller.
131
00:06:49,090 --> 00:06:55,600
And so that means our new sample
space is really just
132
00:06:55,600 --> 00:06:58,125
this set of six points.
133
00:06:58,125 --> 00:07:01,430
134
00:07:01,430 --> 00:07:05,720
And one thing that it's worth
noting is that conditioning
135
00:07:05,720 --> 00:07:09,780
never changes the relative
frequencies or relative
136
00:07:09,780 --> 00:07:12,510
likelihoods of the different
outcomes.
137
00:07:12,510 --> 00:07:16,530
So because all outcomes were
equally likely in our original
138
00:07:16,530 --> 00:07:19,770
sample space omega, in the
conditional worlds the
139
00:07:19,770 --> 00:07:22,280
outcomes are also
equally likely.
140
00:07:22,280 --> 00:07:25,400
So using that argument, we could
say that in our sort of
141
00:07:25,400 --> 00:07:27,330
blue conditional universe
all of the
142
00:07:27,330 --> 00:07:29,570
outcomes are equally likely.
143
00:07:29,570 --> 00:07:32,910
And therefore, we can apply a
conditional version of the
144
00:07:32,910 --> 00:07:34,540
discrete uniform law.
145
00:07:34,540 --> 00:07:37,800
So namely, to compute the
probability of some event in
146
00:07:37,800 --> 00:07:39,160
that conditional world.
147
00:07:39,160 --> 00:07:42,760
So the conditional probability
that "doubles are rolled", we
148
00:07:42,760 --> 00:07:46,440
need only count the number of
outcomes in that event and
149
00:07:46,440 --> 00:07:49,430
divide it by the total
number of outcomes.
150
00:07:49,430 --> 00:07:55,000
So in the conditional world,
there's only two outcomes that
151
00:07:55,000 --> 00:07:58,670
comprise the event "doubles
rolled." These are the only
152
00:07:58,670 --> 00:08:00,770
two circles in the blue
region, right?
153
00:08:00,770 --> 00:08:04,290
So applying the conditional
version number
154
00:08:04,290 --> 00:08:05,770
law, we have two.
155
00:08:05,770 --> 00:08:09,240
And then we need to divide
by the size of omega.
156
00:08:09,240 --> 00:08:12,920
So our conditional universe,
we've already said, has six
157
00:08:12,920 --> 00:08:14,320
possible dots.
158
00:08:14,320 --> 00:08:18,020
So we just divide by 6, and you
see that we get the same
159
00:08:18,020 --> 00:08:20,420
answer of 1/3.
160
00:08:20,420 --> 00:08:23,010
And so again, we used two
different strategies.
161
00:08:23,010 --> 00:08:26,040
I happen to prefer the second
one, because it's slightly
162
00:08:26,040 --> 00:08:30,260
faster and it makes you think
about what does conditioning
163
00:08:30,260 --> 00:08:31,480
really mean.
164
00:08:31,480 --> 00:08:34,840
Conditioning means you're now
restricting your attention to
165
00:08:34,840 --> 00:08:36,179
a conditional universe.
166
00:08:36,179 --> 00:08:39,700
And given that you're in this
conditional universe where the
167
00:08:39,700 --> 00:08:42,730
sum was less than or equal
to 4, what is then the
168
00:08:42,730 --> 00:08:46,192
probability that doubles
also happened?
169
00:08:46,192 --> 00:08:48,790
OK, hopefully you, guys,
are following.
170
00:08:48,790 --> 00:08:51,610
Let's move on to part c.
171
00:08:51,610 --> 00:08:55,680
So part c asks for the
probability that at least one
172
00:08:55,680 --> 00:08:58,510
die roll is a 6.
173
00:08:58,510 --> 00:09:03,190
So I'm going to use the letter
S to denote this, the
174
00:09:03,190 --> 00:09:06,120
probability that at least
one die roll is a 6.
175
00:09:06,120 --> 00:09:08,250
So let's go back to
our picture and
176
00:09:08,250 --> 00:09:11,310
we'll use a green marker.
177
00:09:11,310 --> 00:09:17,120
So hopefully you agree that
anything in this column
178
00:09:17,120 --> 00:09:20,520
corresponds to at least one 6.
179
00:09:20,520 --> 00:09:24,850
So this point, this point, this
point, this point, this
180
00:09:24,850 --> 00:09:30,630
point, and this point your first
die landed on a 6, so at
181
00:09:30,630 --> 00:09:32,720
least one 6 is satisfied.
182
00:09:32,720 --> 00:09:38,440
Similarly, if your second die
has a 6, then we're also OK.
183
00:09:38,440 --> 00:09:41,960
So I claim we want to look
at these 11 points.
184
00:09:41,960 --> 00:09:43,885
Let me just check that,
yeah, 6 plus 5--
185
00:09:43,885 --> 00:09:44,850
11.
186
00:09:44,850 --> 00:09:50,110
So using the discrete uniform
law again, we get
187
00:09:50,110 --> 00:09:52,150
11 divided by 36.
188
00:09:52,150 --> 00:09:54,660
189
00:09:54,660 --> 00:09:58,210
OK, last problem, we're
almost done.
190
00:09:58,210 --> 00:10:02,930
So again, we're interested in
the event S again, so the
191
00:10:02,930 --> 00:10:05,810
event that at least one
die roll is a 6.
192
00:10:05,810 --> 00:10:09,340
But now we want to compute the
probability of that event in
193
00:10:09,340 --> 00:10:12,380
the conditional world where
the two dice land
194
00:10:12,380 --> 00:10:13,630
on different numbers.
195
00:10:13,630 --> 00:10:16,020
196
00:10:16,020 --> 00:10:19,590
So I'm going to call this
probability of S. Let's see,
197
00:10:19,590 --> 00:10:21,910
I'm running out of letters.
198
00:10:21,910 --> 00:10:26,350
Let's for lack of a better
letter, my name is Katie, so
199
00:10:26,350 --> 00:10:29,640
we'll just use a K. We want to
compute the probability of S
200
00:10:29,640 --> 00:10:33,970
given K. And instead of using
the definition of conditional
201
00:10:33,970 --> 00:10:37,660
probability, like we did back in
part b, we're going to use
202
00:10:37,660 --> 00:10:39,520
the faster route.
203
00:10:39,520 --> 00:10:48,820
So essentially, we're going to
find the number of outcomes in
204
00:10:48,820 --> 00:10:51,715
the conditional world.
205
00:10:51,715 --> 00:10:54,720
206
00:10:54,720 --> 00:10:57,770
And then we're also going to
compute the number of outcomes
207
00:10:57,770 --> 00:11:00,710
that comprise S in the
conditional world.
208
00:11:00,710 --> 00:11:04,610
So let's take a look at this.
209
00:11:04,610 --> 00:11:10,160
We are conditioning on the event
that the two dice land
210
00:11:10,160 --> 00:11:11,590
on different numbers.
211
00:11:11,590 --> 00:11:15,860
So hopefully you agree with me
that every single dot that is
212
00:11:15,860 --> 00:11:18,760
not on the diagonal, so every
single dot that doesn't
213
00:11:18,760 --> 00:11:23,160
correspond to doubles, is a
dot that we care about.
214
00:11:23,160 --> 00:11:27,970
So our conditional universe of
that the two dice land on
215
00:11:27,970 --> 00:11:34,066
"different numbers", that
corresponds to these dots.
216
00:11:34,066 --> 00:11:37,550
217
00:11:37,550 --> 00:11:42,900
And it corresponds
to these dots.
218
00:11:42,900 --> 00:11:46,740
I don't want to get this one.
219
00:11:46,740 --> 00:11:49,650
OK, that's good.
220
00:11:49,650 --> 00:11:54,100
So let's see, how many outcomes
do we have in our
221
00:11:54,100 --> 00:11:56,660
conditional world?
222
00:11:56,660 --> 00:11:59,315
And I'm sorry I don't know why
I didn't include this.
223
00:11:59,315 --> 00:12:00,940
This is absolutely included.
224
00:12:00,940 --> 00:12:03,460
I'm just testing to see if you,
225
00:12:03,460 --> 00:12:05,160
guys, are paying attention.
226
00:12:05,160 --> 00:12:08,720
So we counted before that there
are six dots on the
227
00:12:08,720 --> 00:12:12,350
diagonal, and we know that
there are 36 dots total.
228
00:12:12,350 --> 00:12:15,660
So the number of dots, or
outcomes to use the proper
229
00:12:15,660 --> 00:12:20,670
word, in our conditional world
is 36 minus 6, or 30.
230
00:12:20,670 --> 00:12:24,640
So we get a 30 on
the denominator.
231
00:12:24,640 --> 00:12:28,260
And now we're sort of using a
conditional version of our
232
00:12:28,260 --> 00:12:29,910
discrete uniform law, again.
233
00:12:29,910 --> 00:12:33,960
And the reason why we can do
this is, as I argued before,
234
00:12:33,960 --> 00:12:36,870
that conditioning doesn't change
the relative frequency
235
00:12:36,870 --> 00:12:38,130
of the outcomes.
236
00:12:38,130 --> 00:12:41,560
So in this conditional world,
all of the outcomes are still
237
00:12:41,560 --> 00:12:44,740
equally likely, hence we can
apply this law again.
238
00:12:44,740 --> 00:12:49,680
So now we need to count the
number of outcomes that are in
239
00:12:49,680 --> 00:12:55,090
the orange conditional world,
but that also satisfy at least
240
00:12:55,090 --> 00:12:57,760
one die roll is a 6.
241
00:12:57,760 --> 00:12:59,610
So you can see--
242
00:12:59,610 --> 00:13:02,720
1-- we just need to count the
green circles that are also in
243
00:13:02,720 --> 00:13:03,590
the orange.
244
00:13:03,590 --> 00:13:08,910
So that's 1, 2, 3, 4,
5, 6, 7, 8, 9, 10.
245
00:13:08,910 --> 00:13:17,460
So we get a 10, so our answer
is 10 over 30, or 1/3.
246
00:13:17,460 --> 00:13:19,340
So now we're done with
this problem.
247
00:13:19,340 --> 00:13:23,360
As you see, hopefully, it
wasn't too painful.
248
00:13:23,360 --> 00:13:24,920
And what are the important
takeaways
249
00:13:24,920 --> 00:13:26,560
here for this problem?
250
00:13:26,560 --> 00:13:32,290
Well, one is that whenever you
have a discrete sample space,
251
00:13:32,290 --> 00:13:35,540
in which all of outcomes are
equally likely, you should
252
00:13:35,540 --> 00:13:39,570
think about using the discrete
uniform law, because this law
253
00:13:39,570 --> 00:13:43,290
lets you reduce the problem from
computing probabilities
254
00:13:43,290 --> 00:13:46,740
to just counting outcomes
within events.
255
00:13:46,740 --> 00:13:50,860
And the second takeaway is
the way we thought about
256
00:13:50,860 --> 00:13:51,970
conditioning.
257
00:13:51,970 --> 00:13:55,340
So we talked about one thing,
which is that in your
258
00:13:55,340 --> 00:13:59,340
conditional world, when you
condition, the relative
259
00:13:59,340 --> 00:14:02,310
likelihoods of the various
outcomes don't change.
260
00:14:02,310 --> 00:14:05,640
So in our original universe,
all of the outcomes were
261
00:14:05,640 --> 00:14:06,770
equally likely.
262
00:14:06,770 --> 00:14:09,380
So in our conditional universe,
all of the outcomes
263
00:14:09,380 --> 00:14:10,840
are equally likely.
264
00:14:10,840 --> 00:14:14,900
And we saw it was much faster to
apply a conditional version
265
00:14:14,900 --> 00:14:17,820
of the discrete uniform law.
266
00:14:17,820 --> 00:14:19,380
So that's it for today.
267
00:14:19,380 --> 00:14:21,750
And we'll do more problems
next time.
268
00:14:21,750 --> 00:14:23,000