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PROFESSOR: So now we start
on a unit of about a half

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a dozen lectures on probability
theory which most students have

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been exposed to, to some
degree, in high school.

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We'll be taking a more
thorough and theoretical look

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at the subject in our six
lectures but, before we begin,

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let's give a little pitch
for the significance of it.

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There's been extensive
debate among the faculty

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that probability theory belongs
right up there with physics

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and chemistry and
math as something

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that should be a fundamental
requirement for all students

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to know.

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It plays an absolutely
fundamental role

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in the hard sciences,
and the social sciences,

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and in engineering that
pervades all those subjects.

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And it's hard to imagine
somebody legitimately being

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called fully-educated if they
don't understand the basics

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of probability theory.

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Historically, probability
theory starts off

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in a somewhat disreputable
way in the 17th and early 18th

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centuries with the
analysis of gambling,

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but then it goes on to be
the basis for the insurance

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industry and underwriting,
predicting life expectancies,

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so that you could understand
what kind of premiums

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to charge.

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And then it goes on to allow
the interpretation of noisy data

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with errors in it and
the degree to which it

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confirms scientific and
social science hypotheses.

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But true to the
historical basis,

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let's begin with an
example from gambling

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that illustrates the
first idea of probability

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and then we're going
to be working up

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to a methodology for inventing
probability models, called

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the tree model.

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So let's begin with
an example from poker

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and I'd like to ask a question.

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If I deal a hand of
five cards in poker,

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what's the probability of
getting exactly two jacks?

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So there are 13 ranks
and there are four kinds

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of jacks-- space,
hearts, diamonds,

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clubs-- what's the probability
that, among my five cards,

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I'm going to get two of them?

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Well, that's really
a counting problem

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because I'm going to
ask, first of all, how

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many possible five-card
hands are there?

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We can think of these as the
outcomes of a random experiment

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of just picking five cards.

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And there are 52 choose
5 five-card hands

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in a 52-card deck.

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Then, there are 4
choose 2 ways of picking

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the suits for the two
jacks that we have

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and so the total number of hands
that have two jacks is simply 4

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choose 2 times 52 minus
4, the remaining 48 cards,

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choose the remaining 3
cards in the five-card hand.

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And then what we would say
is that the probability

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of two jacks is
basically the number

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of hands with two jacks divided
by the total number of hands.

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Turns out to be about 0.04
and, under this interpretation,

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basically, what we're
thinking of probability

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as telling us is, what
fraction of the time

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do I get what I want?

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What's the fraction of the
time that I quote, "win" ,

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if winning consists of getting
a pair of jacks and, by symmetry

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and the fact that we think of
one hand is as likely to come

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up as another, this fraction
of hands that equal two jacks,

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it makes sense to think of that
as that's the probability that

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we'll get that hand.

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If we think of all the hands
as being equally likely,

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we yank 1 out of the
deck, the fraction of time

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that we would expect to get
two jacks is this number.

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About 0.04.

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So, the general
setup of probability,

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the first idea based
on this illustration

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with a pair of jacks,
is that, abstractly, we

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have some random
experiment that's

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capable of producing outcomes.

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These are mathematical
black boxes called outcomes.

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Now, a certain set
of the outcomes,

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we will think of as an event
that we're interested in

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whether or not it happens.

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We could think of
it as the event

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of getting two jacks or the
event of winning some game.

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Then we define the
probability of an event

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as simply the fraction
of the outcomes

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in the event divided by the
total number of outcomes.

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Among all the outcomes,
what fraction of outcomes

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are in the event?

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And we define that to be the
probability of the event.

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That's the first naive idea
about probability theory

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and it applies to a lot
of cases, but not always.

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So now, let's begin
with an example which

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illustrates why this first
idea needs to be refined

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and it doesn't really
give us the kind of theory

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of probability that we'd like.

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So let's turn to a game that
was really famous in the 1970s.

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An enormously popular
TV game hosted

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by a man named Monty Hall.

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The actual name of the TV show
was called Let's Make a Deal,

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but we'll refer to it
as the Monty Hall game,

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and the way that this
Let's Make A Deal show

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worked was, roughly, that
there were three doors.

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This is an actual
picture of the stage set.

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Door 1, door 2, door 3.

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And by the way, this game
show still has a fan base.

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There's a website for
it that you can look at.

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Even 40 years later,
people are still caught up

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in the dynamics of the game.

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So there are these
three doors and the idea

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is that behind
the doors, they're

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going to have a prize
behind one of them

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and some kind of booby
prize, often a goat held

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by a beautiful woman
holding a goat on a leash

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just to keep things
visually interesting,

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and that's what you
got if you lost.

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And contestants were going
to somehow or other pick

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a door and hope that
the prize was behind it.

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There's a picture of the staff.

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There's Monty Hall
and the woman who was

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his assistant, Carol Merrill.

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Her job was to pick doors
to open and show them

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to contestants to see
what was behind them.

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OK.

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So here are the rules
for the Monty Hall game.

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The actual quiz show
had more flexible rules

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but, for simplicity, we're going
to define a simple, precise,

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and fixed set of rules.

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The rules are that, behind the
three doors, two of the doors

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are going to have goats
and one of the doors

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is going to have
a prize behind it.

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Often the prize is something
like an automobile.

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Something really desirable.

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So we can assume that the
staff, on purpose, will

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place the price at random
behind the three doors

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because they don't
want anybody to have

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a guess that some doors
are more likely than others

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to have the prize and
they're not allowed to cheat.

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That is, once
they've decided which

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door is going to
have the price, it's

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supposed to stay there
throughout the game.

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They can't move it in
response to which door

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that the contestants pick.

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That would be cheating.

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OK.

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Next, the contestant
is given an opportunity

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to pick one of the doors.

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They're all closed and
it's hard to understand

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how the contestant
would make a choice,

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but if the contestant was
worried about the staff trying

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to outguess him on
where to put the goat

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and where to put the
prize, the contestant

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should just pick all the
doors with equally likelihood.

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Then he can't be beaten by
their trying to outguess him.

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He can only be beaten
by if they cheated him

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by moving the goat
after he picked

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or moving the prize
after he picked.

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At this point,
once the contestant

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has picked a door-- let's
say he picks door 2--

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then Monty instructs
Carol to open a door

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with a goat behind it.

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So he's going to choose
an unpicked door.

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If the contestant
has picked door 2,

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that means that door 1 and
door 3 are unpicked doors,

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and Monty tells Carol,
open either door 1

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or door 3, whichever
one-- or perhaps both--

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have a goat behind them.

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And so Carol is going to
open one of those doors

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and show a goat
and everybody knows

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that they're going to
see a goat because that's

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the way the game works.

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And then at this point,
when the contestant

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has seen that there's a door
that has a goat behind it

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and they're sitting
on a picked door

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and there's another unopened
door that hasn't been picked,

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the contestant's
job is to decide

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whether to stick with the door
that they originally picked

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or switch to the
other unopened door.

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So if they picked door 2
and Carol opened door 3,

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they could stick
with door 2 or they

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could switch to
the closed door 1

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and hope that maybe 1
has the price behind it.

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Those are the rules of the game.

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Now, the game got
a lot of prominence

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in a magazine column written
by a woman named Marilyn Vos

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Savant.

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The name of the magazine
column was called Ask Marilyn

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and she advertises herself as
having the highest recorded

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IQ of all time, some
IQ of 200, and so she

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runs a popular science
and math column

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with various kinds of puzzles.

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And she took up the analysis
of the Monty Hall statistics

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and came to a conclusion
and the conclusion

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caused a firestorm of response.

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Letters from all
sorts of readers,

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even quite sophisticated
PhD Mathematicians who

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were arguing with her conclusion
about the way the game worked

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and the probability
of winning according

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to how the contested behaved.

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The debate basically came
down to these two positions.

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Position 1 said that
sticking and switching

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were equally good.

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It really didn't matter
what the contestant did,

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whether they stuck with the
door that they originally

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picked or switched
to the unpicked door

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after the third
door had been opened

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and that their likelihood of
finding the prize was the same.

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And the other argument,
very emphatically,

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said switching is much better.

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You should really
switch no matter what.

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And how can we
resolve this question?

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Well, the general
method that we're

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proposing for
dealing with problems

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like this where we're
really trying to figure out,

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what is the probability model?

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Is to draw a tree that shows,
step-by-step, the progress

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of the process or
experiment that's

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going to yield a random output
and try to assign probabilities

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to each of the
branches of the tree

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as you go and use
that as a guide

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for how to assign
probabilities to outcomes.

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So let's begin, first
of all, by finding out

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what the outcomes
are, and we're going

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to be analyzing the
switch strategy.

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So, just for definiteness,
let's suppose

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that the contestant adopts the
strategy that they pick a door,

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Carol opens a door
that shows a goat,

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and they're going to switch
to the non-goat closed door

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that they did not
originally pick.

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They're going to switch
to the other door

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that they can
switch to and we're

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going to ask about,
what are the outcomes

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and consequences of
winning or losing

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if you adopt that strategy?

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Well, the tree of
possibilities goes like this.

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The first step in this
process that we've described

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is that the staff picks
a prize location, a door

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to put the prize behind, and so
there are three possibilities.

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They could put the prize behind
door 1, door 2, and door 3.

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OK Well, let's examine
the possibility that they

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put the prize behind door 1.

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So the next stage
is they pick a door

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and if the prize is behind
one and they pick a door,

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again, there are
three possible doors

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that the contestant might pick.

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The contestant has no
idea where the price is

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and so the contestant can choose
either door 1 or door 2 or door

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3.

246
00:11:43,770 --> 00:11:48,190
At that point, the third
event in this random process,

247
00:11:48,190 --> 00:11:53,110
or experiment, is that
Carol opens a door that

248
00:11:53,110 --> 00:11:55,160
has a goat behind it.

249
00:11:55,160 --> 00:11:58,710
So let's examine
those possibilities.

250
00:11:58,710 --> 00:12:01,390
So, one possibility
is that the prize

251
00:12:01,390 --> 00:12:06,680
is behind one and the contestant
picks door one, initially.

252
00:12:06,680 --> 00:12:11,970
Well that means that Carol can
open either door 2 or door 3

253
00:12:11,970 --> 00:12:15,120
in that circumstance
because both of them

254
00:12:15,120 --> 00:12:16,410
have goats behind them.

255
00:12:16,410 --> 00:12:18,480
On the other hand,
if the prize is at 1

256
00:12:18,480 --> 00:12:24,080
and the contestant picks
door 2, the two closed doors

257
00:12:24,080 --> 00:12:25,860
have-- one has the
prize, 1, and the other

258
00:12:25,860 --> 00:12:27,330
doesn't have the prize, 3.

259
00:12:27,330 --> 00:12:29,390
Carol has to open door three.

260
00:12:29,390 --> 00:12:31,490
Likewise, if the
contestant picks

261
00:12:31,490 --> 00:12:33,640
door 3 when the prize
is behind door 1,

262
00:12:33,640 --> 00:12:38,050
Carol has to open door 2.

263
00:12:38,050 --> 00:12:40,100
Here she's got a two-way branch.

264
00:12:40,100 --> 00:12:44,270
She can choose to open either
of the two goat doors, 2 or 3.

265
00:12:44,270 --> 00:12:46,730
Here there's only one
unopened door with a goat,

266
00:12:46,730 --> 00:12:48,830
she's got to open 3 there, too.

267
00:12:48,830 --> 00:12:49,370
OK.

268
00:12:49,370 --> 00:12:51,880
And that describes the
outcomes of the experiment.

269
00:12:51,880 --> 00:12:53,580
That's the process
of the experiment

270
00:12:53,580 --> 00:12:56,920
and these nodes at the end,
these leaves of the tree,

271
00:12:56,920 --> 00:13:00,930
describe the final
outcomes on this branch.

272
00:13:00,930 --> 00:13:03,400
Now, if you look at
the classification

273
00:13:03,400 --> 00:13:05,900
of these outcomes according
to winning and losing,

274
00:13:05,900 --> 00:13:08,440
well, we're looking at
the switch strategy.

275
00:13:08,440 --> 00:13:12,800
So if the price was behind 1
and the contestant picked door 1

276
00:13:12,800 --> 00:13:16,480
initially, then their
strategy is to switch

277
00:13:16,480 --> 00:13:20,490
and they're going to switch
away from the prize door.

278
00:13:20,490 --> 00:13:25,390
So whichever door Carol opened
to reveal the goat, 2 or 3,

279
00:13:25,390 --> 00:13:27,540
the contestant is going
to switch to the other one

280
00:13:27,540 --> 00:13:28,770
and they're going to lose.

281
00:13:28,770 --> 00:13:33,180
So both of these outcomes count
as losses for the contestant.

282
00:13:33,180 --> 00:13:37,400
On the other hand, if the
prize was behind door 1

283
00:13:37,400 --> 00:13:39,760
and the contestant
picked door 2,

284
00:13:39,760 --> 00:13:43,460
then Carol opens the
non-prize door, 3,

285
00:13:43,460 --> 00:13:45,860
and the contestant
switches from 2.

286
00:13:45,860 --> 00:13:49,120
The only choice they have is
to switch to 1, the prize door.

287
00:13:49,120 --> 00:13:50,390
They win.

288
00:13:50,390 --> 00:13:52,610
And this other
case is symmetric.

289
00:13:52,610 --> 00:13:54,910
And that summarizes
the wins and losses

290
00:13:54,910 --> 00:13:56,160
in this branch of the tree.

291
00:13:56,160 --> 00:13:57,660
Now, of course, the
rest of the tree

292
00:13:57,660 --> 00:14:00,710
is symmetric so we don't need
to talk it through again.

293
00:14:00,710 --> 00:14:03,430
This is just simply the case
where the prize is behind 2.

294
00:14:03,430 --> 00:14:04,980
The contestant has
the same choices

295
00:14:04,980 --> 00:14:07,070
and [? Marilyn ?]
has the same choices

296
00:14:07,070 --> 00:14:10,530
of which unopened door
to choose and likewise

297
00:14:10,530 --> 00:14:12,440
if the prize is behind 3.

298
00:14:12,440 --> 00:14:15,020
So if we look at
this tree, the tree

299
00:14:15,020 --> 00:14:17,360
is telling us that this
is an experiment which

300
00:14:17,360 --> 00:14:21,990
we think of as having twelve
outcomes, four in each

301
00:14:21,990 --> 00:14:23,340
of these major branches.

302
00:14:23,340 --> 00:14:27,090
So there are twelve outcomes
of this random experiment,

303
00:14:27,090 --> 00:14:34,590
of which, six are losses and
six are wins for the contestant

304
00:14:34,590 --> 00:14:38,450
and so we discover that there's
six wins and six losses.

305
00:14:38,450 --> 00:14:40,720
Now, the way that
this game works,

306
00:14:40,720 --> 00:14:46,670
if you think about it-- if
the switching strategy wins,

307
00:14:46,670 --> 00:14:49,460
that means that the
sticking strategy

308
00:14:49,460 --> 00:14:52,790
would have lost because
if switching wins,

309
00:14:52,790 --> 00:14:56,590
it meant that you switched to
the door that had the prize

310
00:14:56,590 --> 00:14:59,310
and so if you
hadn't switched, you

311
00:14:59,310 --> 00:15:01,580
must have been at a door
that didn't have the prize

312
00:15:01,580 --> 00:15:02,740
and likewise.

313
00:15:02,740 --> 00:15:07,780
If switching loses, then
you must have switched

314
00:15:07,780 --> 00:15:10,390
from the door with the prize
to a door that didn't have

315
00:15:10,390 --> 00:15:12,910
the prize-- switching-- and
that means if you'd stuck,

316
00:15:12,910 --> 00:15:14,150
you would have won.

317
00:15:14,150 --> 00:15:16,550
So what we can say is that
really analyzing the switch

318
00:15:16,550 --> 00:15:18,990
strategy enables us
to analyze the stick

319
00:15:18,990 --> 00:15:21,890
strategy simultaneously
because you win by sticking if

320
00:15:21,890 --> 00:15:23,810
and only if you
lose by switching.

321
00:15:23,810 --> 00:15:25,430
Now this simplification
doesn't hold

322
00:15:25,430 --> 00:15:27,220
when there's more
than three doors,

323
00:15:27,220 --> 00:15:30,090
and that's another
exercise, but for now, it's

324
00:15:30,090 --> 00:15:33,140
telling us that if we
analyze the switch strategy,

325
00:15:33,140 --> 00:15:35,610
we also understand
the stick strategy.

326
00:15:35,610 --> 00:15:40,630
And of course, that means that
if you use the stick strategy

327
00:15:40,630 --> 00:15:45,550
then the six wins become losses
and the six losses become wins

328
00:15:45,550 --> 00:15:49,900
and, again, there are six ways
to lose and six ways to win.

329
00:15:49,900 --> 00:15:52,960
So the first false
conclusion from this

330
00:15:52,960 --> 00:15:55,450
is by reasoning about it as
though they were poker hands,

331
00:15:55,450 --> 00:15:57,360
and the false
conclusion says, look,

332
00:15:57,360 --> 00:16:01,730
sticking and switching win with
the same number of outcomes

333
00:16:01,730 --> 00:16:04,085
and lose with the same
number of outcomes.

334
00:16:04,085 --> 00:16:05,960
So it really doesn't
matter whether you stick

335
00:16:05,960 --> 00:16:08,450
or switch because the
probability of winning,

336
00:16:08,450 --> 00:16:11,120
in both cases, is
half the outcome.

337
00:16:11,120 --> 00:16:12,530
6 out of 12.

338
00:16:12,530 --> 00:16:14,570
The probability doesn't matter.

339
00:16:14,570 --> 00:16:16,740
It makes no difference
whether you stick or switch.

340
00:16:16,740 --> 00:16:26,210
And that's wrong, and
we will see why soon.

341
00:16:26,210 --> 00:16:30,200
The other false
argument is that we

342
00:16:30,200 --> 00:16:35,440
think about what happens
after Carol has opened a door.

343
00:16:35,440 --> 00:16:36,230
So, where are we?

344
00:16:36,230 --> 00:16:39,280
The contestant has picked
a door, has no idea

345
00:16:39,280 --> 00:16:41,680
where the goat or the prize is.

346
00:16:41,680 --> 00:16:45,430
Carol opens the door and
shows the contestant a goat.

347
00:16:45,430 --> 00:16:46,640
What's left?

348
00:16:46,640 --> 00:16:48,940
Well, there's two
closed doors left.

349
00:16:48,940 --> 00:16:51,830
One is the door with
the prize and the other

350
00:16:51,830 --> 00:16:54,610
is the door without the price
that has a goat behind it

351
00:16:54,610 --> 00:16:58,960
and, by symmetry of the
doors, the contestant

352
00:16:58,960 --> 00:17:02,040
has no idea what's behind
the door that he picked

353
00:17:02,040 --> 00:17:04,020
or the remaining unopened door.

354
00:17:04,020 --> 00:17:06,650
They're equally likely
to contain the prize

355
00:17:06,650 --> 00:17:10,680
and so the argument is,
again, that whether you stick

356
00:17:10,680 --> 00:17:14,339
or switch between those
two doors that haven't yet

357
00:17:14,339 --> 00:17:16,300
been opened, it doesn't
really matter and so,

358
00:17:16,300 --> 00:17:19,970
again, the stick strategy
and the switch strategy each

359
00:17:19,970 --> 00:17:22,849
win with the same
50-50 probability.

360
00:17:22,849 --> 00:17:25,859
And that's wrong, too.

361
00:17:25,859 --> 00:17:26,670
What's wrong?

362
00:17:26,670 --> 00:17:29,970
Well, let's go back and look
at this tree a little bit more

363
00:17:29,970 --> 00:17:32,820
carefully to understand
what's going on.

364
00:17:32,820 --> 00:17:34,960
And the first thing to
notice about the tree

365
00:17:34,960 --> 00:17:40,730
is that the structure of the
tree leading to the leaves

366
00:17:40,730 --> 00:17:41,720
is not the same.

367
00:17:45,350 --> 00:17:48,500
Here's a leaf that
has degree [? 2. ?]

368
00:17:48,500 --> 00:17:50,510
One way to get in
and only one way out

369
00:17:50,510 --> 00:17:52,570
and here's a leaf
that has degree 3.

370
00:17:52,570 --> 00:17:56,030
One way in and two ways out, if
we think of going from the root

371
00:17:56,030 --> 00:17:57,090
to the leaf.

372
00:17:57,090 --> 00:18:00,510
And so it's not clear that
these branches, these leaves,

373
00:18:00,510 --> 00:18:02,200
should be treated the same way.

374
00:18:02,200 --> 00:18:04,830
Well let's think about it more
carefully, about-- how are we

375
00:18:04,830 --> 00:18:08,410
going to assign probabilities
to the various steps

376
00:18:08,410 --> 00:18:09,660
of the experiment?

377
00:18:09,660 --> 00:18:12,480
Well, what we're going to
assume, for simplicity,

378
00:18:12,480 --> 00:18:18,450
is that the staff chooses a door
at random to place the prize.

379
00:18:18,450 --> 00:18:21,000
So that means that
each of these branches

380
00:18:21,000 --> 00:18:22,790
occurs with probability 1/3.

381
00:18:22,790 --> 00:18:25,840
1/3 of the time, they put
the prize behind door 1, 1/3

382
00:18:25,840 --> 00:18:28,850
behind door 2, and
1/3 behind door 3.

383
00:18:28,850 --> 00:18:29,590
OK.

384
00:18:29,590 --> 00:18:31,620
Let's continue exploring
the branch where they

385
00:18:31,620 --> 00:18:33,850
put the prize behind door 1.

386
00:18:33,850 --> 00:18:36,920
At that point, the contestant
is going to pick a door

387
00:18:36,920 --> 00:18:39,470
and they can pick
either door 1, 2, or 3

388
00:18:39,470 --> 00:18:43,500
and, absent any
knowledge and also

389
00:18:43,500 --> 00:18:45,200
to be sure that they
can't be outguessed

390
00:18:45,200 --> 00:18:49,837
by the staff realizing that
they mostly prefer door 1.

391
00:18:49,837 --> 00:18:51,420
So if they're going
to switch, they'll

392
00:18:51,420 --> 00:18:54,260
put the prize behind door
1 to fool the contestant.

393
00:18:54,260 --> 00:18:59,550
The contestant's protection
is, pick a door at random.

394
00:18:59,550 --> 00:19:03,650
Choose door 1 1/3 of the time,
and door 2 1/3 of the time,

395
00:19:03,650 --> 00:19:07,390
and door 3 1/3 of the time in
a completely unpredictable way.

396
00:19:07,390 --> 00:19:10,010
And so the contestants
is going to choose

397
00:19:10,010 --> 00:19:13,550
each of those possible
doors as their first choice

398
00:19:13,550 --> 00:19:16,110
with probability 1/3.

399
00:19:16,110 --> 00:19:17,890
Now what happens next?

400
00:19:17,890 --> 00:19:19,300
Well, the next
thing that happens

401
00:19:19,300 --> 00:19:20,720
is that Carol opens a door.

402
00:19:20,720 --> 00:19:23,040
Now this is the case
where Carol has a choice.

403
00:19:23,040 --> 00:19:25,450
The prize is behind
one and the contestant

404
00:19:25,450 --> 00:19:26,910
happened to pick door 1.

405
00:19:26,910 --> 00:19:29,530
That means doors 2
and 3 both have goats

406
00:19:29,530 --> 00:19:31,450
and, again, for
simplicity, let's

407
00:19:31,450 --> 00:19:34,750
assume the Carol, when she
has a choice-- she can open

408
00:19:34,750 --> 00:19:37,150
either door 2 or door
3, here-- does them

409
00:19:37,150 --> 00:19:38,490
with equal probability.

410
00:19:38,490 --> 00:19:40,880
So we're going to assign
probability 1/2 to her opening

411
00:19:40,880 --> 00:19:44,700
door 2 when she has the
choice between 2 or 3

412
00:19:44,700 --> 00:19:48,840
and probability 1/2 that she'll
open door 3 and, by the way,

413
00:19:48,840 --> 00:19:52,400
we saw that those were losing
outcomes for the contestant.

414
00:19:52,400 --> 00:19:54,150
But here, things are
a little different.

415
00:19:54,150 --> 00:19:59,120
If the prize is behind
door 1 and the contestant

416
00:19:59,120 --> 00:20:03,230
has chosen door 2,
Carol has no choice

417
00:20:03,230 --> 00:20:06,080
but to open the
only other unchosen

418
00:20:06,080 --> 00:20:09,370
door with the goat
behind, namely, door 3.

419
00:20:09,370 --> 00:20:11,940
So we could say that
this choice, really,

420
00:20:11,940 --> 00:20:20,380
is probability 1 and I got a
little bit ahead of myself here

421
00:20:20,380 --> 00:20:22,440
but, having filled
in the probabilities

422
00:20:22,440 --> 00:20:24,110
on these edges,
what we figured out

423
00:20:24,110 --> 00:20:28,200
is that the probability of
this topmost branch of losing

424
00:20:28,200 --> 00:20:30,550
is we said, well, 1/3
of the time you go here

425
00:20:30,550 --> 00:20:37,030
and 1/3 of that third you
go here and 1/2 of that time

426
00:20:37,030 --> 00:20:39,100
you go to this vertex.

427
00:20:39,100 --> 00:20:44,110
So it's 1/3 of 1/3 and 1/2
of that, or a weight of 1/18

428
00:20:44,110 --> 00:20:46,740
and, by symmetry,
this gets weight 1/18.

429
00:20:46,740 --> 00:20:49,740
But this way, 1/3 of the time,
the prize is behind door 1.

430
00:20:49,740 --> 00:20:54,900
1/3 of the time, the contestant
picks door 2 and after that,

431
00:20:54,900 --> 00:20:57,360
Carol is was forced
to open door 3.

432
00:20:57,360 --> 00:20:59,690
So this branch occurs
with certainty, as

433
00:20:59,690 --> 00:21:02,430
with probability 1, which
means that we wind up

434
00:21:02,430 --> 00:21:07,910
at this leaf 1/3 of 1/3
of the time for sure,

435
00:21:07,910 --> 00:21:11,540
and its weight is 1/9.

436
00:21:11,540 --> 00:21:14,220
And of course, by symmetry,
the similar weights

437
00:21:14,220 --> 00:21:16,950
get assigned to the
winning and the losing.

438
00:21:16,950 --> 00:21:21,280
So what we've concluded is that,
although there are six wins,

439
00:21:21,280 --> 00:21:25,140
the weight of the wins
is 6/9 because they're

440
00:21:25,140 --> 00:21:29,655
each worth 1/9 of the
time and that winning

441
00:21:29,655 --> 00:21:31,400
will occur 2/3 of the time.

442
00:21:31,400 --> 00:21:34,240
Likewise, there are six
losses but they each

443
00:21:34,240 --> 00:21:39,740
only occur 1/18 of the time
and so we lose 1/3 third

444
00:21:39,740 --> 00:21:42,990
of the time by the
switch strategy.

445
00:21:42,990 --> 00:21:46,260
The summary, then, is
that the probability

446
00:21:46,260 --> 00:21:51,150
of winning if you switch
is 2/3 and, by the remark

447
00:21:51,150 --> 00:21:54,270
that you win with
switching if and only

448
00:21:54,270 --> 00:21:57,470
if you lose with sticking,
it follows that you lose

449
00:21:57,470 --> 00:22:00,320
by sticking 2/3 of the time.

450
00:22:00,320 --> 00:22:03,790
And so sticking is really a
bad strategy and switching

451
00:22:03,790 --> 00:22:06,320
is the dominant way to go.

452
00:22:06,320 --> 00:22:10,220
Now, in class, we back up
this theoretical analysis.

453
00:22:10,220 --> 00:22:13,320
It's very logical but the
question is, is it true?

454
00:22:13,320 --> 00:22:16,120
And you can do
statistical experiments

455
00:22:16,120 --> 00:22:19,120
and have students pick
doors and goats and prizes

456
00:22:19,120 --> 00:22:23,140
and, sure enough, it turns out
that roughly 2/3 of the time,

457
00:22:23,140 --> 00:22:26,920
and closer and closer to 2/3 the
more times you play the game,

458
00:22:26,920 --> 00:22:31,890
the switching strategy
wins 2/3 of the time.

459
00:22:31,890 --> 00:22:35,290
So, the second key idea
in probability theory

460
00:22:35,290 --> 00:22:38,550
is that the outcomes may
have different probabilities.

461
00:22:38,550 --> 00:22:40,180
They may have different weights.

462
00:22:40,180 --> 00:22:43,820
Unlike the poker hand case,
when we look more closely

463
00:22:43,820 --> 00:22:46,880
at a random experiment
with different outcomes,

464
00:22:46,880 --> 00:22:50,940
we will agree that, for various
kinds of reasons of symmetry

465
00:22:50,940 --> 00:22:52,960
or logic and so on,
that it make sense

466
00:22:52,960 --> 00:22:55,200
to assign different
probability weights

467
00:22:55,200 --> 00:22:56,530
to the different outcomes.

468
00:22:56,530 --> 00:22:58,390
It's not the case
that the outcomes

469
00:22:58,390 --> 00:23:01,760
have uniform probability, that
they're all equally likely.

470
00:23:04,310 --> 00:23:09,960
So, to summarize, what happens,
especially-- this example

471
00:23:09,960 --> 00:23:12,550
illustrates the confusion
about of probability theory

472
00:23:12,550 --> 00:23:15,240
that was engendered to
even some serious experts--

473
00:23:15,240 --> 00:23:18,000
but, in general, intuition
is very important,

474
00:23:18,000 --> 00:23:21,781
as in any subject, but it's
also dangerous in probability

475
00:23:21,781 --> 00:23:22,280
theory.

476
00:23:22,280 --> 00:23:26,870
Particularly, for beginners who
aren't experienced about some

477
00:23:26,870 --> 00:23:28,800
of these traps that
you can fall into

478
00:23:28,800 --> 00:23:31,520
and so our proposal
is that you be very

479
00:23:31,520 --> 00:23:33,120
wary of intuitive arguments.

480
00:23:33,120 --> 00:23:35,680
They're valuable but you
need another disciplined way

481
00:23:35,680 --> 00:23:38,100
to check them, and
we propose that you

482
00:23:38,100 --> 00:23:41,600
stick with what we call the
four-part method when you're

483
00:23:41,600 --> 00:23:44,610
trying to devise a
probability model

484
00:23:44,610 --> 00:23:46,400
for some random experiment.

485
00:23:46,400 --> 00:23:50,330
So, the steps are,
first, that you

486
00:23:50,330 --> 00:23:53,830
try to identify the outcomes
of the random experiment

487
00:23:53,830 --> 00:23:56,410
and this is where the
tree structure comes up.

488
00:23:56,410 --> 00:23:58,910
If you try to
model, step-by-step

489
00:23:58,910 --> 00:24:02,420
at each stage of the tree,
what the possible sub-steps are

490
00:24:02,420 --> 00:24:06,140
in the overall process that
yields the random outcome,

491
00:24:06,140 --> 00:24:07,720
that's where the
tree comes in as we

492
00:24:07,720 --> 00:24:10,110
illustrated with Monty Hall.

493
00:24:10,110 --> 00:24:12,790
The next thing to do
is, among the outcomes,

494
00:24:12,790 --> 00:24:15,010
identify the ones
that you consider

495
00:24:15,010 --> 00:24:18,290
to be of the winning events
or the winning outcomes

496
00:24:18,290 --> 00:24:23,290
or the outcomes in the event
that you are concerned about

497
00:24:23,290 --> 00:24:24,650
whether or not it happens.

498
00:24:24,650 --> 00:24:28,650
Getting two jacks, picking
the door with the prize.

499
00:24:28,650 --> 00:24:32,590
So you need to identify the
target event whose probability

500
00:24:32,590 --> 00:24:33,560
you're interested in.

501
00:24:33,560 --> 00:24:35,685
We could call it the winning
event, the probability

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00:24:35,685 --> 00:24:36,570
of winning.

503
00:24:36,570 --> 00:24:40,470
The third key step is to try
to use the tree and logic of it

504
00:24:40,470 --> 00:24:43,790
to assign probabilities
to the outcomes

505
00:24:43,790 --> 00:24:45,430
and the fourth step,
then, is, simply,

506
00:24:45,430 --> 00:24:47,740
to compute the probability
of the event which

507
00:24:47,740 --> 00:24:49,410
you do in a very
straightforward way

508
00:24:49,410 --> 00:24:52,750
by basically adding up
the probabilities of each

509
00:24:52,750 --> 00:24:54,490
of the outcomes in the event.

510
00:24:54,490 --> 00:24:57,590
That is the four-step method.

511
00:24:57,590 --> 00:25:01,420
Now, this Monty Hall
tree that we came up with

512
00:25:01,420 --> 00:25:08,010
was very literal and wildly,
unnecessarily complicated.

513
00:25:08,010 --> 00:25:11,050
So let's take another look at
that and a simpler argument

514
00:25:11,050 --> 00:25:13,920
that will lead us to the
same conclusion about how

515
00:25:13,920 --> 00:25:15,550
the Monty Hall game
works, and we'll

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00:25:15,550 --> 00:25:17,950
do that in the next video.