Now, conditional probability will
let us explain a lot of the confused arguments
that people brought up about Monty Hall.
And we'll see that it is a little bit confusing
and where there is some correct sounding arguments that
give you the wrong answer.
So let's go back and look at our Monty Hall
tree that allowed us to derive the sample space
and probability space for the whole process of the prize
being placed and the contest picking a door
and Carol opening a door.
Now, this tree was way more complicated
than we needed if all we were trying to do
was figure out the probability of winning if you switch.
But having the tree will allow us
to discuss a whole bunch of other events
in their probabilities that will get
us a grip on some of the arguments that
gave the wrong answer.
So let's look at the event, first of all
that the goat is at 2.
Now, this is the branch where the prize is at 2.
And so in all the other branches the goat is at 2,
l which means that we have these eight of the 12 outcomes
in the event-- goat is at 2.
Now, let's also look at the event that the prize is at 1.
That's just this branch of the tree, OK?
So one of the arguments is that when the contestant is
at the point where they've seen that the open door and they're
trying to decide whether to stick or switch,
they know that the goat is at the door 2.
Say without loss of generality that that
was the door that they got to look at behind,
that Carol opened.
And so we want to ask the probability, given
that he picked 1, what's the probability that the prize is
at 1 given that the goat is at 2?
That means that if you're at door 1
then you should stick if that probability is high
and otherwise you shouldn't stick.
So we can look at this event, the prize at 1 given the goat
at 2, and what we can see is that it's taking up
exactly half of the outcomes for goat at 2
and the same kind of outcomes-- red ones and green ones.
The red ones are worth an 1/18 and the green ones
are worth a 1/9 in probability, and that
implies that the probability that the prize is at 1 given
that the goat is at 2 is 1/2.
It really is.
And that's the argument that people were saying.
They said, look, when the contestant sees
that the goat is at door 2, and they're
trying to decide whether the goat-- the prize is
at the door-- is it door 1 or at the other door,
and it's equally likely.
And so it doesn't matter whether they stick or switch.
That's a correct argument but it's not
calculating the probability of the stick strategy winning.
Why?
Well, because there's more information that's
available than goat is at 2.
The contestant not only knows that the goat
is at 2 and trying to figure out the probability
that the prize is at 1, but the contestant
knows what door he picked.
So let's suppose that the contestant did pick door 1
and learned that the goat was at door 2, that's
a different event.
If the blue one is marked off at the places where
the contestant picks one, this is
where the door is picked-- is 1 and here's 1 and here's 1.
This 1 splits into one event, this 1 splits into one event,
but this choice of 1 splits into two outcomes.
And so when we look at the event that both the goat is at 2
and the contestant picked 1, which
is what the contest really knows when they get
to see that there's a goat at door 2,
we wind up with the overlap of just three outcomes.
Two outcomes that have probability 1/8 and one outcome
that has probability a 1/9.
It's just those three.
And the result is that the probability that the prize is
at 1 given that you picked 1 and the goat
is at 2-- so this is the event-- goat at 2
and picked 1, these three outcomes.
The prize is at 1 is these two outcomes, which
are each worth an 1/18 and this is
an outcome that's worth a 1/9.
So the prize at 1 outcomes amount
to 1/2 of the total probability of this event, goat at 2
picked at 1.
So, again, the probability that the prize
is at 1 given that the contestant picked 1 and saw
the goat at 2 is a 1/2 also.
That's confusing.
So it seems as though the contestant may as well stick
because at the point that he has to decide whether to stick
or switch, and he knows where-- he sees where the goat is
and he knows what door he's picked,
it's 50-50 whether he should stick or switch.
The probability that the prize is at door 1 that he picked
is a 1/2, so it really doesn't matter if he stays there
or if he decides to switch to the unopened door.
But wait a minute, that's not right
because the contestant not only knows what door he picked,
not only knows that there's a goat behind a given door
that Carol has opened, but he knows that Carol
has opened that door.
That's how he got to know that the goat was there.
So let's go back and look at the tree.
What basically the previous two arguments
are conditioning on the wrong events.
It's a typical mistake and one that you really
have to watch out for.
So if you use the correct event, what we're looking at
is the contestant knows that they've picked door 1.
That's the outcomes of picked door 1 are marked here.
In addition, the contestant will get
to know, for example, in a play of the game
that Carol has opened door 2.
Carol opening door 2 is quite a different event
from the goat being at 2.
This is a picture of the outcomes in Carol
opening door 2, and we're interested in the intersection
of them.
That is, just this guy that's in both
and this guy that's in both.
There they are.
And so what we can do is identify
that the event that you picked 1 and that Carol opened
door 2 consists simply of two outcomes-- one worth an 1/18
and one worth a 1/9.
Now, of these two outcomes, which one has the prize at 1?
Well, only that one.
Remember the first component here is where the prize is.
And so the prize at 1 event among the given
picked 1 and opened 2 is just this red outcome.
Now, the red outcome has probability
1/18 and the green outcome has probability
that's twice as much.
So that means that relative to this event, the probability
that the prize is at 1 given that you picked 1 and opened 2
is actually 1/18 over 1/18 plus 1/9, or 1/3.
So given that you picked 1 and you get to see what Carrol did,
the probability the prize is at the door you picked
is only 1/3, which means that if you stick you only
have a 1/3 chance of winning.
You should switch.
And if you do, you'll have a 2/3 probability of winning.
So when we finally condition on everything that we know,
which is the contestant knows what door he picked
and what door Carol opened, then we
discover that it correctly-- as we deduced previously--
that the probability of switching wins is 2/3.
So we're not trying to rederive the fact
that the probability of switching wins is 2/3.
We're trying to illustrate a very basic blunder that you
have to watch out for, which is when you're trying to reason
about some situation and you condition on some event
that you think summarizes what's going on,
if you don't get the conditioning event right,
you're going to get the wrong answer.
So it's easy to see how many people got confused,
and, in fact, finding the right event can be tricky.
When in doubt, the 4 step method with constructing
the tree where you're not even thinking
about conditional probabilities but you're just examining
the individual outcomes is a good fall back
to avoid these kinds of confusing situations.