PROFESSOR: So now we start on a unit of about a half
a dozen lectures on probability theory which most students have
been exposed to, to some degree, in high school.
We'll be taking a more thorough and theoretical look
at the subject in our six lectures but, before we begin,
let's give a little pitch for the significance of it.
There's been extensive debate among the faculty
that probability theory belongs right up there with physics
and chemistry and math as something
that should be a fundamental requirement for all students
to know.
It plays an absolutely fundamental role
in the hard sciences, and the social sciences,
and in engineering that pervades all those subjects.
And it's hard to imagine somebody legitimately being
called fully-educated if they don't understand the basics
of probability theory.
Historically, probability theory starts off
in a somewhat disreputable way in the 17th and early 18th
centuries with the analysis of gambling,
but then it goes on to be the basis for the insurance
industry and underwriting, predicting life expectancies,
so that you could understand what kind of premiums
to charge.
And then it goes on to allow the interpretation of noisy data
with errors in it and the degree to which it
confirms scientific and social science hypotheses.
But true to the historical basis,
let's begin with an example from gambling
that illustrates the first idea of probability
and then we're going to be working up
to a methodology for inventing probability models, called
the tree model.
So let's begin with an example from poker
and I'd like to ask a question.
If I deal a hand of five cards in poker,
what's the probability of getting exactly two jacks?
So there are 13 ranks and there are four kinds
of jacks-- space, hearts, diamonds,
clubs-- what's the probability that, among my five cards,
I'm going to get two of them?
Well, that's really a counting problem
because I'm going to ask, first of all, how
many possible five-card hands are there?
We can think of these as the outcomes of a random experiment
of just picking five cards.
And there are 52 choose 5 five-card hands
in a 52-card deck.
Then, there are 4 choose 2 ways of picking
the suits for the two jacks that we have
and so the total number of hands that have two jacks is simply 4
choose 2 times 52 minus 4, the remaining 48 cards,
choose the remaining 3 cards in the five-card hand.
And then what we would say is that the probability
of two jacks is basically the number
of hands with two jacks divided by the total number of hands.
Turns out to be about 0.04 and, under this interpretation,
basically, what we're thinking of probability
as telling us is, what fraction of the time
do I get what I want?
What's the fraction of the time that I quote, "win" ,
if winning consists of getting a pair of jacks and, by symmetry
and the fact that we think of one hand is as likely to come
up as another, this fraction of hands that equal two jacks,
it makes sense to think of that as that's the probability that
we'll get that hand.
If we think of all the hands as being equally likely,
we yank 1 out of the deck, the fraction of time
that we would expect to get two jacks is this number.
About 0.04.
So, the general setup of probability,
the first idea based on this illustration
with a pair of jacks, is that, abstractly, we
have some random experiment that's
capable of producing outcomes.
These are mathematical black boxes called outcomes.
Now, a certain set of the outcomes,
we will think of as an event that we're interested in
whether or not it happens.
We could think of it as the event
of getting two jacks or the event of winning some game.
Then we define the probability of an event
as simply the fraction of the outcomes
in the event divided by the total number of outcomes.
Among all the outcomes, what fraction of outcomes
are in the event?
And we define that to be the probability of the event.
That's the first naive idea about probability theory
and it applies to a lot of cases, but not always.
So now, let's begin with an example which
illustrates why this first idea needs to be refined
and it doesn't really give us the kind of theory
of probability that we'd like.
So let's turn to a game that was really famous in the 1970s.
An enormously popular TV game hosted
by a man named Monty Hall.
The actual name of the TV show was called Let's Make a Deal,
but we'll refer to it as the Monty Hall game,
and the way that this Let's Make A Deal show
worked was, roughly, that there were three doors.
This is an actual picture of the stage set.
Door 1, door 2, door 3.
And by the way, this game show still has a fan base.
There's a website for it that you can look at.
Even 40 years later, people are still caught up
in the dynamics of the game.
So there are these three doors and the idea
is that behind the doors, they're
going to have a prize behind one of them
and some kind of booby prize, often a goat held
by a beautiful woman holding a goat on a leash
just to keep things visually interesting,
and that's what you got if you lost.
And contestants were going to somehow or other pick
a door and hope that the prize was behind it.
There's a picture of the staff.
There's Monty Hall and the woman who was
his assistant, Carol Merrill.
Her job was to pick doors to open and show them
to contestants to see what was behind them.
OK.
So here are the rules for the Monty Hall game.
The actual quiz show had more flexible rules
but, for simplicity, we're going to define a simple, precise,
and fixed set of rules.
The rules are that, behind the three doors, two of the doors
are going to have goats and one of the doors
is going to have a prize behind it.
Often the prize is something like an automobile.
Something really desirable.
So we can assume that the staff, on purpose, will
place the price at random behind the three doors
because they don't want anybody to have
a guess that some doors are more likely than others
to have the prize and they're not allowed to cheat.
That is, once they've decided which
door is going to have the price, it's
supposed to stay there throughout the game.
They can't move it in response to which door
that the contestants pick.
That would be cheating.
OK.
Next, the contestant is given an opportunity
to pick one of the doors.
They're all closed and it's hard to understand
how the contestant would make a choice,
but if the contestant was worried about the staff trying
to outguess him on where to put the goat
and where to put the prize, the contestant
should just pick all the doors with equally likelihood.
Then he can't be beaten by their trying to outguess him.
He can only be beaten by if they cheated him
by moving the goat after he picked
or moving the prize after he picked.
At this point, once the contestant
has picked a door-- let's say he picks door 2--
then Monty instructs Carol to open a door
with a goat behind it.
So he's going to choose an unpicked door.
If the contestant has picked door 2,
that means that door 1 and door 3 are unpicked doors,
and Monty tells Carol, open either door 1
or door 3, whichever one-- or perhaps both--
have a goat behind them.
And so Carol is going to open one of those doors
and show a goat and everybody knows
that they're going to see a goat because that's
the way the game works.
And then at this point, when the contestant
has seen that there's a door that has a goat behind it
and they're sitting on a picked door
and there's another unopened door that hasn't been picked,
the contestant's job is to decide
whether to stick with the door that they originally picked
or switch to the other unopened door.
So if they picked door 2 and Carol opened door 3,
they could stick with door 2 or they
could switch to the closed door 1
and hope that maybe 1 has the price behind it.
Those are the rules of the game.
Now, the game got a lot of prominence
in a magazine column written by a woman named Marilyn Vos
Savant.
The name of the magazine column was called Ask Marilyn
and she advertises herself as having the highest recorded
IQ of all time, some IQ of 200, and so she
runs a popular science and math column
with various kinds of puzzles.
And she took up the analysis of the Monty Hall statistics
and came to a conclusion and the conclusion
caused a firestorm of response.
Letters from all sorts of readers,
even quite sophisticated PhD Mathematicians who
were arguing with her conclusion about the way the game worked
and the probability of winning according
to how the contested behaved.
The debate basically came down to these two positions.
Position 1 said that sticking and switching
were equally good.
It really didn't matter what the contestant did,
whether they stuck with the door that they originally
picked or switched to the unpicked door
after the third door had been opened
and that their likelihood of finding the prize was the same.
And the other argument, very emphatically,
said switching is much better.
You should really switch no matter what.
And how can we resolve this question?
Well, the general method that we're
proposing for dealing with problems
like this where we're really trying to figure out,
what is the probability model?
Is to draw a tree that shows, step-by-step, the progress
of the process or experiment that's
going to yield a random output and try to assign probabilities
to each of the branches of the tree
as you go and use that as a guide
for how to assign probabilities to outcomes.
So let's begin, first of all, by finding out
what the outcomes are, and we're going
to be analyzing the switch strategy.
So, just for definiteness, let's suppose
that the contestant adopts the strategy that they pick a door,
Carol opens a door that shows a goat,
and they're going to switch to the non-goat closed door
that they did not originally pick.
They're going to switch to the other door
that they can switch to and we're
going to ask about, what are the outcomes
and consequences of winning or losing
if you adopt that strategy?
Well, the tree of possibilities goes like this.
The first step in this process that we've described
is that the staff picks a prize location, a door
to put the prize behind, and so there are three possibilities.
They could put the prize behind door 1, door 2, and door 3.
OK Well, let's examine the possibility that they
put the prize behind door 1.
So the next stage is they pick a door
and if the prize is behind one and they pick a door,
again, there are three possible doors
that the contestant might pick.
The contestant has no idea where the price is
and so the contestant can choose either door 1 or door 2 or door
3.
At that point, the third event in this random process,
or experiment, is that Carol opens a door that
has a goat behind it.
So let's examine those possibilities.
So, one possibility is that the prize
is behind one and the contestant picks door one, initially.
Well that means that Carol can open either door 2 or door 3
in that circumstance because both of them
have goats behind them.
On the other hand, if the prize is at 1
and the contestant picks door 2, the two closed doors
have-- one has the prize, 1, and the other
doesn't have the prize, 3.
Carol has to open door three.
Likewise, if the contestant picks
door 3 when the prize is behind door 1,
Carol has to open door 2.
Here she's got a two-way branch.
She can choose to open either of the two goat doors, 2 or 3.
Here there's only one unopened door with a goat,
she's got to open 3 there, too.
OK.
And that describes the outcomes of the experiment.
That's the process of the experiment
and these nodes at the end, these leaves of the tree,
describe the final outcomes on this branch.
Now, if you look at the classification
of these outcomes according to winning and losing,
well, we're looking at the switch strategy.
So if the price was behind 1 and the contestant picked door 1
initially, then their strategy is to switch
and they're going to switch away from the prize door.
So whichever door Carol opened to reveal the goat, 2 or 3,
the contestant is going to switch to the other one
and they're going to lose.
So both of these outcomes count as losses for the contestant.
On the other hand, if the prize was behind door 1
and the contestant picked door 2,
then Carol opens the non-prize door, 3,
and the contestant switches from 2.
The only choice they have is to switch to 1, the prize door.
They win.
And this other case is symmetric.
And that summarizes the wins and losses
in this branch of the tree.
Now, of course, the rest of the tree
is symmetric so we don't need to talk it through again.
This is just simply the case where the prize is behind 2.
The contestant has the same choices
and [? Marilyn ?] has the same choices
of which unopened door to choose and likewise
if the prize is behind 3.
So if we look at this tree, the tree
is telling us that this is an experiment which
we think of as having twelve outcomes, four in each
of these major branches.
So there are twelve outcomes of this random experiment,
of which, six are losses and six are wins for the contestant
and so we discover that there's six wins and six losses.
Now, the way that this game works,
if you think about it-- if the switching strategy wins,
that means that the sticking strategy
would have lost because if switching wins,
it meant that you switched to the door that had the prize
and so if you hadn't switched, you
must have been at a door that didn't have the prize
and likewise.
If switching loses, then you must have switched
from the door with the prize to a door that didn't have
the prize-- switching-- and that means if you'd stuck,
you would have won.
So what we can say is that really analyzing the switch
strategy enables us to analyze the stick
strategy simultaneously because you win by sticking if
and only if you lose by switching.
Now this simplification doesn't hold
when there's more than three doors,
and that's another exercise, but for now, it's
telling us that if we analyze the switch strategy,
we also understand the stick strategy.
And of course, that means that if you use the stick strategy
then the six wins become losses and the six losses become wins
and, again, there are six ways to lose and six ways to win.
So the first false conclusion from this
is by reasoning about it as though they were poker hands,
and the false conclusion says, look,
sticking and switching win with the same number of outcomes
and lose with the same number of outcomes.
So it really doesn't matter whether you stick
or switch because the probability of winning,
in both cases, is half the outcome.
6 out of 12.
The probability doesn't matter.
It makes no difference whether you stick or switch.
And that's wrong, and we will see why soon.
The other false argument is that we
think about what happens after Carol has opened a door.
So, where are we?
The contestant has picked a door, has no idea
where the goat or the prize is.
Carol opens the door and shows the contestant a goat.
What's left?
Well, there's two closed doors left.
One is the door with the prize and the other
is the door without the price that has a goat behind it
and, by symmetry of the doors, the contestant
has no idea what's behind the door that he picked
or the remaining unopened door.
They're equally likely to contain the prize
and so the argument is, again, that whether you stick
or switch between those two doors that haven't yet
been opened, it doesn't really matter and so,
again, the stick strategy and the switch strategy each
win with the same 50-50 probability.
And that's wrong, too.
What's wrong?
Well, let's go back and look at this tree a little bit more
carefully to understand what's going on.
And the first thing to notice about the tree
is that the structure of the tree leading to the leaves
is not the same.
Here's a leaf that has degree [? 2. ?]
One way to get in and only one way out
and here's a leaf that has degree 3.
One way in and two ways out, if we think of going from the root
to the leaf.
And so it's not clear that these branches, these leaves,
should be treated the same way.
Well let's think about it more carefully, about-- how are we
going to assign probabilities to the various steps
of the experiment?
Well, what we're going to assume, for simplicity,
is that the staff chooses a door at random to place the prize.
So that means that each of these branches
occurs with probability 1/3.
1/3 of the time, they put the prize behind door 1, 1/3
behind door 2, and 1/3 behind door 3.
OK.
Let's continue exploring the branch where they
put the prize behind door 1.
At that point, the contestant is going to pick a door
and they can pick either door 1, 2, or 3
and, absent any knowledge and also
to be sure that they can't be outguessed
by the staff realizing that they mostly prefer door 1.
So if they're going to switch, they'll
put the prize behind door 1 to fool the contestant.
The contestant's protection is, pick a door at random.
Choose door 1 1/3 of the time, and door 2 1/3 of the time,
and door 3 1/3 of the time in a completely unpredictable way.
And so the contestants is going to choose
each of those possible doors as their first choice
with probability 1/3.
Now what happens next?
Well, the next thing that happens
is that Carol opens a door.
Now this is the case where Carol has a choice.
The prize is behind one and the contestant
happened to pick door 1.
That means doors 2 and 3 both have goats
and, again, for simplicity, let's
assume the Carol, when she has a choice-- she can open
either door 2 or door 3, here-- does them
with equal probability.
So we're going to assign probability 1/2 to her opening
door 2 when she has the choice between 2 or 3
and probability 1/2 that she'll open door 3 and, by the way,
we saw that those were losing outcomes for the contestant.
But here, things are a little different.
If the prize is behind door 1 and the contestant
has chosen door 2, Carol has no choice
but to open the only other unchosen
door with the goat behind, namely, door 3.
So we could say that this choice, really,
is probability 1 and I got a little bit ahead of myself here
but, having filled in the probabilities
on these edges, what we figured out
is that the probability of this topmost branch of losing
is we said, well, 1/3 of the time you go here
and 1/3 of that third you go here and 1/2 of that time
you go to this vertex.
So it's 1/3 of 1/3 and 1/2 of that, or a weight of 1/18
and, by symmetry, this gets weight 1/18.
But this way, 1/3 of the time, the prize is behind door 1.
1/3 of the time, the contestant picks door 2 and after that,
Carol is was forced to open door 3.
So this branch occurs with certainty, as
with probability 1, which means that we wind up
at this leaf 1/3 of 1/3 of the time for sure,
and its weight is 1/9.
And of course, by symmetry, the similar weights
get assigned to the winning and the losing.
So what we've concluded is that, although there are six wins,
the weight of the wins is 6/9 because they're
each worth 1/9 of the time and that winning
will occur 2/3 of the time.
Likewise, there are six losses but they each
only occur 1/18 of the time and so we lose 1/3 third
of the time by the switch strategy.
The summary, then, is that the probability
of winning if you switch is 2/3 and, by the remark
that you win with switching if and only
if you lose with sticking, it follows that you lose
by sticking 2/3 of the time.
And so sticking is really a bad strategy and switching
is the dominant way to go.
Now, in class, we back up this theoretical analysis.
It's very logical but the question is, is it true?
And you can do statistical experiments
and have students pick doors and goats and prizes
and, sure enough, it turns out that roughly 2/3 of the time,
and closer and closer to 2/3 the more times you play the game,
the switching strategy wins 2/3 of the time.
So, the second key idea in probability theory
is that the outcomes may have different probabilities.
They may have different weights.
Unlike the poker hand case, when we look more closely
at a random experiment with different outcomes,
we will agree that, for various kinds of reasons of symmetry
or logic and so on, that it make sense
to assign different probability weights
to the different outcomes.
It's not the case that the outcomes
have uniform probability, that they're all equally likely.
So, to summarize, what happens, especially-- this example
illustrates the confusion about of probability theory
that was engendered to even some serious experts--
but, in general, intuition is very important,
as in any subject, but it's also dangerous in probability
theory.
Particularly, for beginners who aren't experienced about some
of these traps that you can fall into
and so our proposal is that you be very
wary of intuitive arguments.
They're valuable but you need another disciplined way
to check them, and we propose that you
stick with what we call the four-part method when you're
trying to devise a probability model
for some random experiment.
So, the steps are, first, that you
try to identify the outcomes of the random experiment
and this is where the tree structure comes up.
If you try to model, step-by-step
at each stage of the tree, what the possible sub-steps are
in the overall process that yields the random outcome,
that's where the tree comes in as we
illustrated with Monty Hall.
The next thing to do is, among the outcomes,
identify the ones that you consider
to be of the winning events or the winning outcomes
or the outcomes in the event that you are concerned about
whether or not it happens.
Getting two jacks, picking the door with the prize.
So you need to identify the target event whose probability
you're interested in.
We could call it the winning event, the probability
of winning.
The third key step is to try to use the tree and logic of it
to assign probabilities to the outcomes
and the fourth step, then, is, simply,
to compute the probability of the event which
you do in a very straightforward way
by basically adding up the probabilities of each
of the outcomes in the event.
That is the four-step method.
Now, this Monty Hall tree that we came up with
was very literal and wildly, unnecessarily complicated.
So let's take another look at that and a simpler argument
that will lead us to the same conclusion about how
the Monty Hall game works, and we'll
do that in the next video.