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Now, conditional
probability will

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let us explain a lot of
the confused arguments

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that people brought
up about Monty Hall.

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And we'll see that it is
a little bit confusing

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and where there is some
correct sounding arguments that

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give you the wrong answer.

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So let's go back and
look at our Monty Hall

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tree that allowed us to
derive the sample space

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00:00:22,785 --> 00:00:27,400
and probability space for the
whole process of the prize

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00:00:27,400 --> 00:00:29,600
being placed and the
contest picking a door

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and Carol opening a door.

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Now, this tree was
way more complicated

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than we needed if all
we were trying to do

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was figure out the probability
of winning if you switch.

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But having the
tree will allow us

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to discuss a whole
bunch of other events

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in their probabilities
that will get

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us a grip on some of
the arguments that

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gave the wrong answer.

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So let's look at the
event, first of all

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that the goat is at 2.

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Now, this is the branch
where the prize is at 2.

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And so in all the other
branches the goat is at 2,

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l which means that we have
these eight of the 12 outcomes

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in the event-- goat is at 2.

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Now, let's also look at the
event that the prize is at 1.

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That's just this
branch of the tree, OK?

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So one of the arguments is
that when the contestant is

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at the point where they've seen
that the open door and they're

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trying to decide whether
to stick or switch,

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they know that the
goat is at the door 2.

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Say without loss of
generality that that

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was the door that they
got to look at behind,

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that Carol opened.

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And so we want to ask
the probability, given

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that he picked 1, what's the
probability that the prize is

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at 1 given that
the goat is at 2?

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That means that if
you're at door 1

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then you should stick if
that probability is high

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and otherwise you
shouldn't stick.

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So we can look at this event,
the prize at 1 given the goat

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at 2, and what we can see
is that it's taking up

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exactly half of the
outcomes for goat at 2

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and the same kind of outcomes--
red ones and green ones.

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The red ones are worth an
1/18 and the green ones

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are worth a 1/9 in
probability, and that

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implies that the probability
that the prize is at 1 given

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that the goat is at 2 is 1/2.

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It really is.

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And that's the argument
that people were saying.

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They said, look, when
the contestant sees

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that the goat is at
door 2, and they're

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trying to decide whether
the goat-- the prize is

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at the door-- is it door
1 or at the other door,

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and it's equally likely.

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And so it doesn't matter
whether they stick or switch.

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That's a correct
argument but it's not

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calculating the probability
of the stick strategy winning.

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Why?

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Well, because there's
more information that's

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available than goat is at 2.

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The contestant not only
knows that the goat

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is at 2 and trying to
figure out the probability

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that the prize is at
1, but the contestant

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knows what door he picked.

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So let's suppose that the
contestant did pick door 1

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and learned that the goat
was at door 2, that's

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a different event.

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If the blue one is marked
off at the places where

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the contestant
picks one, this is

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where the door is picked-- is
1 and here's 1 and here's 1.

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This 1 splits into one event,
this 1 splits into one event,

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but this choice of 1
splits into two outcomes.

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And so when we look at the
event that both the goat is at 2

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and the contestant
picked 1, which

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is what the contest
really knows when they get

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to see that there's
a goat at door 2,

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we wind up with the overlap
of just three outcomes.

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Two outcomes that have
probability 1/8 and one outcome

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that has probability a 1/9.

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It's just those three.

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And the result is that the
probability that the prize is

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at 1 given that you
picked 1 and the goat

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is at 2-- so this is
the event-- goat at 2

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and picked 1, these
three outcomes.

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The prize is at 1 is
these two outcomes, which

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are each worth an
1/18 and this is

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00:04:28,490 --> 00:04:29,800
an outcome that's worth a 1/9.

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So the prize at
1 outcomes amount

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to 1/2 of the total probability
of this event, goat at 2

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picked at 1.

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So, again, the
probability that the prize

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is at 1 given that the
contestant picked 1 and saw

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00:04:43,630 --> 00:04:47,430
the goat at 2 is a 1/2 also.

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That's confusing.

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00:04:49,460 --> 00:04:52,420
So it seems as though the
contestant may as well stick

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because at the point that he
has to decide whether to stick

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or switch, and he knows where--
he sees where the goat is

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00:04:59,060 --> 00:05:01,210
and he knows what
door he's picked,

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00:05:01,210 --> 00:05:03,610
it's 50-50 whether he
should stick or switch.

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00:05:03,610 --> 00:05:07,520
The probability that the prize
is at door 1 that he picked

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is a 1/2, so it really doesn't
matter if he stays there

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00:05:10,740 --> 00:05:15,240
or if he decides to switch
to the unopened door.

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00:05:15,240 --> 00:05:18,020
But wait a minute,
that's not right

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because the contestant not
only knows what door he picked,

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not only knows that there's
a goat behind a given door

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that Carol has opened,
but he knows that Carol

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00:05:30,330 --> 00:05:32,020
has opened that door.

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That's how he got to know
that the goat was there.

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So let's go back and
look at the tree.

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What basically the
previous two arguments

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are conditioning on
the wrong events.

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It's a typical mistake
and one that you really

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00:05:43,530 --> 00:05:45,360
have to watch out for.

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So if you use the correct
event, what we're looking at

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is the contestant knows
that they've picked door 1.

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That's the outcomes of picked
door 1 are marked here.

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In addition, the
contestant will get

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00:05:59,710 --> 00:06:01,515
to know, for example,
in a play of the game

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00:06:01,515 --> 00:06:04,260
that Carol has opened door 2.

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00:06:04,260 --> 00:06:07,390
Carol opening door 2 is
quite a different event

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from the goat being at 2.

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This is a picture of
the outcomes in Carol

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opening door 2, and we're
interested in the intersection

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of them.

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That is, just this
guy that's in both

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00:06:18,740 --> 00:06:21,070
and this guy that's in both.

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There they are.

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00:06:22,640 --> 00:06:25,255
And so what we
can do is identify

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that the event that you
picked 1 and that Carol opened

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00:06:29,030 --> 00:06:34,250
door 2 consists simply of two
outcomes-- one worth an 1/18

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00:06:34,250 --> 00:06:36,540
and one worth a 1/9.

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Now, of these two outcomes,
which one has the prize at 1?

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Well, only that one.

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00:06:41,070 --> 00:06:43,750
Remember the first component
here is where the prize is.

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00:06:43,750 --> 00:06:49,920
And so the prize at 1
event among the given

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00:06:49,920 --> 00:06:54,040
picked 1 and opened 2 is
just this red outcome.

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00:06:54,040 --> 00:06:57,730
Now, the red outcome
has probability

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00:06:57,730 --> 00:07:01,050
1/18 and the green
outcome has probability

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00:07:01,050 --> 00:07:03,400
that's twice as much.

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00:07:03,400 --> 00:07:08,210
So that means that relative
to this event, the probability

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00:07:08,210 --> 00:07:12,680
that the prize is at 1 given
that you picked 1 and opened 2

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00:07:12,680 --> 00:07:19,200
is actually 1/18 over
1/18 plus 1/9, or 1/3.

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00:07:19,200 --> 00:07:23,890
So given that you picked 1 and
you get to see what Carrol did,

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00:07:23,890 --> 00:07:26,780
the probability the prize
is at the door you picked

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00:07:26,780 --> 00:07:31,250
is only 1/3, which means
that if you stick you only

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00:07:31,250 --> 00:07:33,450
have a 1/3 chance of winning.

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00:07:33,450 --> 00:07:36,200
You should switch.

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00:07:36,200 --> 00:07:40,310
And if you do, you'll have a
2/3 probability of winning.

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00:07:40,310 --> 00:07:43,740
So when we finally condition
on everything that we know,

151
00:07:43,740 --> 00:07:46,490
which is the contestant
knows what door he picked

152
00:07:46,490 --> 00:07:48,770
and what door Carol
opened, then we

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00:07:48,770 --> 00:07:52,580
discover that it correctly--
as we deduced previously--

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00:07:52,580 --> 00:07:55,530
that the probability of
switching wins is 2/3.

155
00:07:55,530 --> 00:07:57,890
So we're not trying
to rederive the fact

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00:07:57,890 --> 00:08:00,370
that the probability of
switching wins is 2/3.

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00:08:00,370 --> 00:08:03,160
We're trying to illustrate a
very basic blunder that you

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00:08:03,160 --> 00:08:05,970
have to watch out for, which
is when you're trying to reason

159
00:08:05,970 --> 00:08:09,150
about some situation and
you condition on some event

160
00:08:09,150 --> 00:08:12,520
that you think summarizes
what's going on,

161
00:08:12,520 --> 00:08:14,800
if you don't get the
conditioning event right,

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00:08:14,800 --> 00:08:17,540
you're going to get
the wrong answer.

163
00:08:17,540 --> 00:08:20,680
So it's easy to see how
many people got confused,

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00:08:20,680 --> 00:08:23,550
and, in fact, finding the
right event can be tricky.

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00:08:23,550 --> 00:08:27,560
When in doubt, the 4 step
method with constructing

166
00:08:27,560 --> 00:08:29,460
the tree where you're
not even thinking

167
00:08:29,460 --> 00:08:33,130
about conditional probabilities
but you're just examining

168
00:08:33,130 --> 00:08:36,330
the individual outcomes
is a good fall back

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00:08:36,330 --> 00:08:39,750
to avoid these kinds of
confusing situations.