Hi.
In the session, we'll be solving
the Monty Hall problem.
And this problem is based on an old game show that was
called "Let's Make a Deal." And the host of this game
show, his name was Monty Hall, which is why this problem is
now known as the Monty Hall problem.
And this problem is actually pretty well-known, because
there was some disagreement at the time over what the right
answer to this problem should be.
Even some really smart people didn't agree on what the right
answer should be.
And part of what might explain that disagreement is that they
probably were considering slightly different variations
of the problem, because as in all probability problems, the
assumptions that you're working with are very
important, because otherwise you may be solving an actually
different problem.
And so what we'll do first is really layout concretely what
all the assumptions are, what the rules of the game are.
And then we'll go through the methodology to solve for the
actual answer.
So the game is actually relatively simple.
So you're on a game show and you're
presented with three doors.
These doors are closed.
And behind one of these doors is a prize, let's say, a car.
And behind the other two doors, there's nothing.
You don't know which one it is.
And the rules of the game are that, first, you get to choose
any one of these three.
So you pick one of the doors that you want.
They don't show you what's behind that door, but your
friend, who actually knows which door has the prize
behind it, will look at the remaining doors.
So let's, just for example, let's say you chose door one.
Your friend will look at the other two doors
and open one of them.
And you will make sure that the one
that he opens is empty.
That is the prize not behind that one.
And at this point, one of the doors is open and its empty,
you have your original door plus another unopened door.
And you're given an option-- you could either stay with
your initial choice or you can switch to the
other unopened door.
And whichever one is your final choice, they
will open that door.
And if there's a price behind it, you win, and if there not,
then you don't win.
So the question that we're trying to answer is what is
the better strategy here?
Is the better strategy to stay with your initial choice or is
it better to switch to the other unopened door?
OK, so it turns out that the specific rules here actually
are very important.
Specifically, the rule about how your friend chooses to
open doors.
And the fact that he will always open one of the two
other door that you haven't picked and he will make sure
that that door doesn't have a prize behind it.
And let's see how that actually
plays out in this problem.
So the simplest way, I think, of thinking about this problem
is just to think about under what circumstances does
staying with your initial choice win?
So if you think about it, the only way that you can win by
staying with your initial choice is if your initial
choice happened to be the door that has a prize behind it.
And because you're sticking with the initial choice, you
can actually kind of forget about the rest of the game,
about opening of the other door and about switching.
It's as if you're playing a simpler game, which is just
you have three doors, one of them has a prize behind it,
and you choose one of them.
And if you guessed right, then you win.
If you didn't, then you don't win.
And because the another important assumption is that
the prize has an equal probability of being behind
any one of three doors so one third, one third, one third.
Because of that, then if you stay with your first choice,
you win only if your first choice happened
to the right one.
And that is the case with probably one third.
So with that simple argument you can convince yourself that
the probability of winning, given the strategy of staying
with your first choice, is one third.
Now, let's think about the other
strategy, which is to switch.
So under what circumstances does switching win for you?
Well, if your first choice happened to be the right door,
then switching away from that door will always lose.
But let's say, that happens with probably one third.
But the rest of the time with probably 2/3, your first
choice would be wrong.
So let's give an example here.
Let's say, the prize, which I'll denote by happy face, is
behind door two.
And your first choice was door one.
So your first choice was wrong.
Now, your friend can open door two, because door two has the
prize behind it.
He also doesn't open the door that you initially picked.
So he has to open door three.
So door three is open, and now you have an option of sticking
with your first choice-- door one--
or switching to door two.
So in this case, it's obvious to see that
switching wins for you.
And now, if instead, you picked door one first, and the
prize was behind door three, again, you are wrong.
And again, your friend is forced to open door two.
And switching, again, wins for you.
And so if you think about it, switching will win for you, as
long as your initial pick was wrong.
If your initial pick was wrong, then the prize is
behind one of the doors.
Your friend has to open one of the doors, but he can't open
the door that has the prize behind it.
So he has to open the other bad door, leaving the good
door with the prize behind it, as the one that
you can switch to.
And so by switching you will win in this scenario.
And what is the probability of that happening?
Well, that happens if your initial pick was wrong, which
happens with probably 2/3.
So the final answer then, it's pretty simple, the probability
of winning if you stay is one third, and the probability of
winning if you switch is 2/3.
And so maybe counterintuitively the result
is that it's actually better for you, twice as good for
you, to switch rather than stay.
And so that was the argument, the kind of simple argument.
We can also be more methodical about this and actually list
out all of the possible outcomes.
Because it's relatively small problem-- there's only three
doors-- we can actually just list out all
the possible outcomes.
So for example, if you chose door one first, and the prize
was behind door one, your friend has a choice.
He can open door two or door three, because
they're both empty.
And then in that case, if you stay, you win, you picked the
door correctly.
And if you switch to two or three, then you lose.
But if you chose door one, the prize is behind door two, then
your friend has to open door three, he is forced to do
that, then staying with lose but switching would win.
And so on for the other cases.
And so again, this is just an exhaustive list of all the
possible outcomes, from which you can see that, in fact,
staying wins, only if your first choice was correct.
And switching wins in all the other cases.
And so one third of the time, staying would win, 2/3 of the
time switching would win.
OK, so now, we have the answer.
Let's try to figure out and convince ourselves that it is
actually right, because you might think before going
through this process that maybe it doesn't matter
whether you stay or you switch, they both have the
same probably of winning, or maybe even staying is better.
So why is staying worse and switching better?
Well, the first argument really is something that we've
already talked about.
By staying, you're essentially banking on your first choice
being correct, which is a relatively poor bet, because
you have only one in three chance of being right.
But by switching, you're actually banking on your first
choice being wrong, which is a relatively better bet, because
you're more likely to be wrong than right in your first
choice, because you're just picking blindly.
OK, so that is one intuitive explanation for why
switching is better.
Another slightly different way to think about it is that
instead of picking single doors, you're actually picking
groups of doors.
So let's say that your first pick was door one.
Then you're actually really deciding between door one or
doors two and three combined.
So why is that?
It's because by staying with door one, you're
staying with door one.
But by switching, you're actually getting two doors for
the price of one, because you know that your friend will
reveal one of these to be empty, and the other one will
stay closed.
But switching really kind of buys you both of these.
And so because it buys you two opportunities to win, you get
2/3 chance of winning, versus a one third chance.
Another way of thinking about this is to increase the scale
of the problem, and maybe that will help visualize the
counterintuitive answer.
So instead of having three doors, imagine that you have
1,000 doors that are closed.
And again, one prize is behind one of the doors.
And the rules are similar-- you pick one door first, and
then your friend will open 998 other doors.
And these doors are guaranteed to be empty.
And now you're left with your initial door plus one other
door that is unopened.
So now the question is should you stay with your first
choice or switch to your other choice?
And it should be more intuitively obvious now that
the better decision would be to switch, because you're
overwhelmingly more likely to have picked incorrectly for
your first pick.
You have only 1 in 1,000 chance of getting it right.
So that is kind of just taking this to a bigger extreme and
really driving home the intuition.
OK, so what we've really discovered is that the fact
that the rules of the game are that your friend has to open
one of the other two doors and cannot reveal the prize plays
a big role in this problem.
And that is an important assumption.
OK, so now let's think about a slightly
different variation now.
So a different strategy.
Instead of just always staying or always switching, we have a
specific other strategy, which is that you will choose door
one first and then, depending on what your friend does, you
will act accordingly.
So if your friend opens door two, you will not switch.
And if your friend opens door three, you will switch.
So let's draw out exactly what happens here.
So you have door one that you've chosen.
And the prize can be behind doors one, two, or three.
And again, it's equally likely.
So the probabilities of these branches are one third, one
third, and one third.
And now given that, your friend in this scenario has a
choice between opening doors two or three.
And so because of doors, you chose one, the prize actually
is behind one, and so two and three are both empty, so he
can choose whichever one he wants to open.
And the problem actually hasn't specified how your
friend actually decides between this.
So we'll leave it in general.
So we'll say that the probability p, your friend
will open two, door two, in this case.
And with the remaining probability 1 minus p, he will
open door three.
What about in this case?
Well, you chose door one.
The prize is actually behind door two.
So following the rules of the game, your friend is forced to
open door three.
So this happens with probability 1.
And similarly, if the prize is behind door three, your friend
is forced to open door two, which, again, happens with
probably 1.
So now let's see how this strategy works.
When do you win?
You win when, according to the strategy, your final choice is
the right door.
So according to the strategy, in this case, your friend
opened door two.
And according to your strategy, if door two is open,
you don't switch.
So you stay with your first choice of one.
And that happens to the right one, so you win in this case.
But what about here?
Your friend opened door three, and by your strategy, you do
switch, which is the wrong choice here, so you lose.
Here, you switch, because you open door three, and you
switch to the right door, so that wins.
And this one, you don't switch, and you lose.
All right, so what is the final probability of winning?
And the final probably of winning is the probability of
getting to these two outcomes, which happens with probability
one third times p plus one third times 1.
So one third.
So the final answer is one third p plus one third.
And notice now that the answer isn't just a number.
Like in this case, the answer was one third and 2/3.
And it didn't actually matter how your friend chose between
these two doors when he had a choice.
But in this case, it actually doesn't matter, because p
stays in the answer.
But one thing that we can do is we can compare this with
these strategies.
So what we see is that, well p is a probability, so it has to
be between 0 and 1.
So this probability winning for this strategy is somewhere
between one third times 0 plus one third, which is one third.
And one third times 1 plus one third, which is 2/3.
So the strategy is somewhere between 2/3 and one third.
So what we see is that no matter what, this strategy is
at least as good as staying all the time, because that was
only one third.
And no matter what it can't be any better than switching,
which was 2/3.
So you can also come up with lots of other different
strategies and see what the probabilities of winning are
in that case.
OK, so what have we learned in this problem?
What are the key takeaways?
One important takeaway is that it's important to really
understand a problem and arrive at a concrete and
precise set of assumptions.
So really have a precise problem that you're solving.
And another important takeaway is to think about your final
answer, make sure that that actually makes sense to you,
make sure that you can justify it somehow intuitively.
In that case, you can actually convince yourself that your
answer is actually correct, because sometimes go through a
lot of formulas, and sometimes your formula may have an error
in there somewhere.
But you could take the final answer and ask yourself does
this actually makes sense intuitively?
That's often a very good check and sometimes you can catch
errors in your calculations that way.
OK so we'll see next time.