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PROFESSOR: OK.

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Let's take another look
at the Monty Hall tree

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that we worked
with to figure out

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the probability of the
switch strategy winning.

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Now this tree was just
absolutely literal

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and absurdly complicated
and large for what

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we're trying to analyze.

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That is, we literally
thought of each

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of the three outcomes of whether
the prize was behind door one,

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door two, door three,
and then exactly which

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door the contestant picked
next-- either door one, door

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two, door three, and so on.

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And it's clear that this tree
has a symmetric structure.

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That is, these three
sub-trees-- whether the prize

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is behind one, the
prize is behind two,

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or the prize is behind three--
all have the same structure.

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And we could have gotten
away with analyzing one,

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and deduced that that's what
happened with the other two

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branches.

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In fact, that really
was what we did.

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But why not make that
explicit in our analysis

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from the first place?

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Instead of it having
this triple tree,

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let's just look at the
tree with one branch.

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So there's the central
branch that I'm keeping.

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Namely, we're assuming
that the first move is

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to have the prize at door
two, and then the door

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picked has three choices.

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And then the door opened has
two choices or one, depending

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on where the prize was placed
relative to the contestant's

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pick here.

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If the prize was at two, and
the contestant picked door two,

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then Carol has a choice of two
doors-- one or three-- to open.

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On the other hand, if
the prize was at two,

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and the contestant
picked door one,

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Carol has no choice but to open
door three with probability 1.

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OK.

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Now when we're
looking at this tree,

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the first branch is kind
of fixed and forced.

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So we don't really need it
as part of the analysis.

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What we're really
looking at is analyzing

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what happens in the experiment
starting at the stage

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where the contestant
picks a door.

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So let's just enlarge this tree
to get a better look at it.

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There's the same tree
where we're starting off,

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where we're assuming that
the prize is at door two,

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and then the door picked
can be either door one, door

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two, or door three,
and the door opened

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then can be door three, door
one, door three, or door one

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according to the
constraints on Carol.

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Now a better way to understand
this tree [? as ?] instead

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of saying the prize
is at door two,

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and this is where the
contestant chooses door two,

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and then have to worry about
all the other branches that

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are symmetrically like this,
we could have reformulated

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the tree model in the
first place by saying,

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let's just consider the cases
that wherever the prize is,

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there are three possibilities.

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The contestant picks the
door where their prize is,

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or they pick the
next door-- let's

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say in some counter-clockwise
direction from where

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the prize is.

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Or going around among
the doors in a circle,

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they pick a door
that's two doors away

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from where the prize is.

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And if we reformulate
it that way,

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then these are the cases--
either the contestant picks

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the prize door, or they
pick the first door that

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doesn't have the prize, or
they pick the second door

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that doesn't have the prize.

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And each of those occur
with probability of 1/3.

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And likewise, once they've
picked door number one

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with no prize, then
that means that Carol

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has the choice of only one
door that she can open,

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because the other
unpicked door has a prize.

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So she's got to open the
second no-prize door,

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because the contestant has
picked the first no-prize door.

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Likewise, if the contestant
picks the prize door,

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Carol can pick either
of the non-prize door--

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non-prize one or non-prize two.

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Both are losses.

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And likewise here, where
Carol's move is forced,

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and the contestant will win.

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So now we're in great shape,
because I've really gotten rid

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of the rest of the tree.

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It's not as though I'm
analyzing 1/3 of it,

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and the 1/3 analysis is going
to apply to the other parts

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by symmetry.

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But I've captured
the whole story here

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by simply relativizing the
first move instead of it

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being literally door one,
door two, door three.

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I don't care what actual
door the contestant picks.

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All I care about is whether
the contestant picks the prize

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door, or the first door
that's not the prize door,

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or the second door that's
not the prize door.

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This would have been
a much better tree

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to start off with the
first place, at least

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for the purpose of analyzing
the probability of winning.

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Now we're going to
get some mileage out

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of the more complicated tree
in a later video segment

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when we start talking about
conditional probabilities

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of what are the probabilities
of things happening

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at various stages
in the experiment.

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And so we will want to have some
of these other vertices that

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represent stages
of the experiment.

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But if we'd really
been thinking solely

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about how to analyze the
probability of winning

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with the switch
strategy, this would've

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been a much better
tree to start off with.

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But wait.

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Let's look at this tree.

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First of all, there
really isn't any need

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to model this branch
of the experiment,

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because at this point, once
we're talking about switching,

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if the contestant has
picked a non-prize door,

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they win-- period.

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Carol's move is forced,
and it's going to be a win.

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We might have well just
collapsed the win down

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to say that as soon as they pick
a non-prize door they've won,

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and who cares what
happens after that?

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Same thing down here.

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So the tree really
could've simplified to one

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where you pick a no-prize
door and you win,

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you pick the other no-prize
door and you win, or you

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pick a prize door,
and then Carol

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has a choice of opening either
of the other two no-prize

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[INAUDIBLE]

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OK.

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Because after all,
what's the point

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in distinguishing between
whether you pick prize door

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one, or you pick prize door two?

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You win in both cases.

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And really, all
we care about, we

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could have condensed the
entire tree down to one

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where either you pick the
prize door with probability

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of 1/3, in which case
you're guaranteed

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to lose no matter what happens.

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Or you pick the
non-prize door, which

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you do with probability
2/3, in which case

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you win no matter what happens.

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And that is a really
simple tree, OK?

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There it is.

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And what we can
read off immediately

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is that with the switch strategy
the probability of winning

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is 2/3.

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So the switch wins if and
only if the prize door is not

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picked.

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And that means that the
probability that switch wins

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is 2/3, which is what we
already figured out using

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the more complicated tree.

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But this way of getting
at it is a lot clearer.

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So the message here is that
the tree that you come up

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with to model the
experimental outcomes

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is really a modeling process.

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And there may be
many models that work

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to capture a given scenario.

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And it will often pay off to
try to find a simpler tree

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to make the analysis simpler.