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Hi.

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In the session, we'll
be solving

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the Monty Hall problem.

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And this problem is based on
an old game show that was

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called "Let's Make a Deal."
And the host of this game

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show, his name was Monty Hall,
which is why this problem is

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now known as the Monty
Hall problem.

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And this problem is actually
pretty well-known, because

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there was some disagreement at
the time over what the right

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answer to this problem
should be.

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Even some really smart people
didn't agree on what the right

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answer should be.

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And part of what might explain
that disagreement is that they

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probably were considering
slightly different variations

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of the problem, because as in
all probability problems, the

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assumptions that you're
working with are very

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important, because otherwise you
may be solving an actually

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different problem.

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And so what we'll do first is
really layout concretely what

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all the assumptions are, what
the rules of the game are.

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And then we'll go through the
methodology to solve for the

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actual answer.

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So the game is actually
relatively simple.

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So you're on a game
show and you're

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presented with three doors.

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These doors are closed.

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And behind one of these doors is
a prize, let's say, a car.

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And behind the other two
doors, there's nothing.

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You don't know which
one it is.

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And the rules of the game are
that, first, you get to choose

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any one of these three.

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So you pick one of the
doors that you want.

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They don't show you what's
behind that door, but your

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friend, who actually knows
which door has the prize

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behind it, will look at
the remaining doors.

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So let's, just for example,
let's say you chose door one.

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Your friend will look at
the other two doors

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and open one of them.

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And you will make sure
that the one

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that he opens is empty.

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That is the prize not
behind that one.

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And at this point, one of the
doors is open and its empty,

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you have your original door plus
another unopened door.

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And you're given an option--
you could either stay with

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your initial choice or
you can switch to the

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other unopened door.

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And whichever one is your
final choice, they

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will open that door.

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And if there's a price behind
it, you win, and if there not,

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then you don't win.

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So the question that we're
trying to answer is what is

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the better strategy here?

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Is the better strategy to stay
with your initial choice or is

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it better to switch to the
other unopened door?

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OK, so it turns out that the
specific rules here actually

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are very important.

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Specifically, the rule about
how your friend chooses to

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open doors.

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And the fact that he will always
open one of the two

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other door that you haven't
picked and he will make sure

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that that door doesn't have
a prize behind it.

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And let's see how that actually

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plays out in this problem.

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So the simplest way, I think, of
thinking about this problem

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is just to think about under
what circumstances does

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staying with your initial
choice win?

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So if you think about it, the
only way that you can win by

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staying with your initial choice
is if your initial

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choice happened to be the door
that has a prize behind it.

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And because you're sticking with
the initial choice, you

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can actually kind of forget
about the rest of the game,

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about opening of the other
door and about switching.

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It's as if you're playing a
simpler game, which is just

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you have three doors, one of
them has a prize behind it,

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and you choose one of them.

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And if you guessed right,
then you win.

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If you didn't, then
you don't win.

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And because the another
important assumption is that

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the prize has an equal
probability of being behind

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any one of three doors so one
third, one third, one third.

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Because of that, then if you
stay with your first choice,

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you win only if your first
choice happened

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to the right one.

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And that is the case with
probably one third.

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So with that simple argument you
can convince yourself that

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the probability of winning,
given the strategy of staying

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with your first choice,
is one third.

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Now, let's think
about the other

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strategy, which is to switch.

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So under what circumstances does
switching win for you?

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Well, if your first choice
happened to be the right door,

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then switching away from that
door will always lose.

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But let's say, that happens
with probably one third.

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But the rest of the time with
probably 2/3, your first

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choice would be wrong.

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So let's give an example here.

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Let's say, the prize, which I'll
denote by happy face, is

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behind door two.

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And your first choice
was door one.

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So your first choice
was wrong.

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Now, your friend can open door
two, because door two has the

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prize behind it.

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He also doesn't open the door
that you initially picked.

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So he has to open door three.

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So door three is open, and now
you have an option of sticking

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with your first choice--
door one--

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or switching to door two.

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So in this case, it's
obvious to see that

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switching wins for you.

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And now, if instead, you picked
door one first, and the

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prize was behind door three,
again, you are wrong.

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And again, your friend is
forced to open door two.

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And switching, again,
wins for you.

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And so if you think about it,
switching will win for you, as

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long as your initial
pick was wrong.

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If your initial pick was wrong,
then the prize is

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behind one of the doors.

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Your friend has to open one of
the doors, but he can't open

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the door that has the
prize behind it.

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So he has to open the other
bad door, leaving the good

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door with the prize behind
it, as the one that

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you can switch to.

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And so by switching you will
win in this scenario.

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And what is the probability
of that happening?

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Well, that happens if your
initial pick was wrong, which

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happens with probably 2/3.

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So the final answer then, it's
pretty simple, the probability

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of winning if you stay is one
third, and the probability of

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winning if you switch is 2/3.

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And so maybe counterintuitively
the result

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is that it's actually better
for you, twice as good for

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you, to switch rather
than stay.

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And so that was the argument,
the kind of simple argument.

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We can also be more methodical
about this and actually list

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out all of the possible
outcomes.

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Because it's relatively small
problem-- there's only three

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doors-- we can actually
just list out all

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the possible outcomes.

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So for example, if you chose
door one first, and the prize

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was behind door one, your
friend has a choice.

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He can open door two or
door three, because

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they're both empty.

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And then in that case, if you
stay, you win, you picked the

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door correctly.

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And if you switch to two or
three, then you lose.

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But if you chose door one, the
prize is behind door two, then

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your friend has to open door
three, he is forced to do

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that, then staying with lose
but switching would win.

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And so on for the other cases.

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And so again, this is just an
exhaustive list of all the

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possible outcomes, from which
you can see that, in fact,

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staying wins, only if your
first choice was correct.

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And switching wins in
all the other cases.

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And so one third of the time,
staying would win, 2/3 of the

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time switching would win.

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OK, so now, we have
the answer.

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Let's try to figure out and
convince ourselves that it is

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actually right, because you
might think before going

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through this process that
maybe it doesn't matter

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whether you stay or you switch,
they both have the

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same probably of winning, or
maybe even staying is better.

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So why is staying worse
and switching better?

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Well, the first argument really
is something that we've

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already talked about.

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By staying, you're essentially
banking on your first choice

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being correct, which is a
relatively poor bet, because

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you have only one in three
chance of being right.

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But by switching, you're
actually banking on your first

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choice being wrong, which is a
relatively better bet, because

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you're more likely to be wrong
than right in your first

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choice, because you're
just picking blindly.

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OK, so that is one intuitive
explanation for why

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switching is better.

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Another slightly different way
to think about it is that

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instead of picking single doors,
you're actually picking

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groups of doors.

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So let's say that your first
pick was door one.

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Then you're actually really
deciding between door one or

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doors two and three combined.

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So why is that?

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It's because by staying
with door one, you're

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staying with door one.

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But by switching, you're
actually getting two doors for

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the price of one, because you
know that your friend will

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reveal one of these to be empty,
and the other one will

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stay closed.

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But switching really kind of
buys you both of these.

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And so because it buys you two
opportunities to win, you get

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2/3 chance of winning, versus
a one third chance.

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00:09:24,210 --> 00:09:26,930

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Another way of thinking about
this is to increase the scale

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of the problem, and maybe that
will help visualize the

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counterintuitive answer.

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So instead of having three
doors, imagine that you have

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1,000 doors that are closed.

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And again, one prize is behind
one of the doors.

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And the rules are similar-- you
pick one door first, and

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then your friend will open
998 other doors.

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And these doors are guaranteed
to be empty.

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And now you're left with your
initial door plus one other

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door that is unopened.

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So now the question is should
you stay with your first

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choice or switch to
your other choice?

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And it should be more
intuitively obvious now that

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the better decision would be
to switch, because you're

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overwhelmingly more likely to
have picked incorrectly for

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your first pick.

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You have only 1 in 1,000 chance
of getting it right.

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So that is kind of just taking
this to a bigger extreme and

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really driving home
the intuition.

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OK, so what we've really
discovered is that the fact

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that the rules of the game are
that your friend has to open

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one of the other two doors and
cannot reveal the prize plays

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a big role in this problem.

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And that is an important
assumption.

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OK, so now let's think
about a slightly

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different variation now.

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So a different strategy.

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Instead of just always staying
or always switching, we have a

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specific other strategy, which
is that you will choose door

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one first and then, depending on
what your friend does, you

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will act accordingly.

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So if your friend opens door
two, you will not switch.

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And if your friend opens door
three, you will switch.

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So let's draw out exactly
what happens here.

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So you have door one
that you've chosen.

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And the prize can be behind
doors one, two, or three.

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And again, it's equally
likely.

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So the probabilities of these
branches are one third, one

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third, and one third.

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And now given that, your friend
in this scenario has a

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choice between opening
doors two or three.

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And so because of doors, you
chose one, the prize actually

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is behind one, and so two and
three are both empty, so he

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can choose whichever one
he wants to open.

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00:12:08,030 --> 00:12:11,380

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00:12:11,380 --> 00:12:15,490
And the problem actually hasn't
specified how your

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friend actually decides
between this.

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00:12:16,660 --> 00:12:21,310
So we'll leave it in general.

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So we'll say that the
probability p, your friend

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will open two, door
two, in this case.

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And with the remaining
probability 1 minus p, he will

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open door three.

246
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What about in this case?

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00:12:34,340 --> 00:12:35,760
Well, you chose door one.

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The prize is actually
behind door two.

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00:12:37,490 --> 00:12:40,010
So following the rules of the
game, your friend is forced to

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00:12:40,010 --> 00:12:40,920
open door three.

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00:12:40,920 --> 00:12:43,530
So this happens with
probability 1.

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00:12:43,530 --> 00:12:46,710
And similarly, if the prize is
behind door three, your friend

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00:12:46,710 --> 00:12:48,890
is forced to open door two,
which, again, happens with

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00:12:48,890 --> 00:12:50,460
probably 1.

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00:12:50,460 --> 00:12:53,860
So now let's see how this
strategy works.

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When do you win?

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00:12:55,300 --> 00:13:02,300
You win when, according to the
strategy, your final choice is

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the right door.

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00:13:02,900 --> 00:13:06,160
So according to the strategy,
in this case, your friend

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00:13:06,160 --> 00:13:07,650
opened door two.

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00:13:07,650 --> 00:13:10,560
And according to your strategy,
if door two is open,

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you don't switch.

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00:13:11,330 --> 00:13:13,660
So you stay with your
first choice of one.

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00:13:13,660 --> 00:13:17,380
And that happens to the right
one, so you win in this case.

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00:13:17,380 --> 00:13:18,500
But what about here?

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00:13:18,500 --> 00:13:22,110
Your friend opened door three,
and by your strategy, you do

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00:13:22,110 --> 00:13:26,080
switch, which is the wrong
choice here, so you lose.

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00:13:26,080 --> 00:13:28,750
Here, you switch, because you
open door three, and you

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00:13:28,750 --> 00:13:31,460
switch to the right door,
so that wins.

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00:13:31,460 --> 00:13:34,910
And this one, you don't
switch, and you lose.

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00:13:34,910 --> 00:13:39,110
All right, so what is the final
probability of winning?

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00:13:39,110 --> 00:13:42,130
And the final probably of
winning is the probability of

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00:13:42,130 --> 00:13:45,730
getting to these two outcomes,
which happens with probability

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00:13:45,730 --> 00:13:52,510
one third times p plus
one third times 1.

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00:13:52,510 --> 00:13:53,670
So one third.

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00:13:53,670 --> 00:13:57,010
So the final answer is one
third p plus one third.

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00:13:57,010 --> 00:14:00,520
And notice now that the answer
isn't just a number.

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Like in this case, the answer
was one third and 2/3.

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00:14:04,400 --> 00:14:07,930
And it didn't actually matter
how your friend chose between

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00:14:07,930 --> 00:14:10,220
these two doors when
he had a choice.

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00:14:10,220 --> 00:14:12,680
But in this case, it actually
doesn't matter, because p

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00:14:12,680 --> 00:14:14,090
stays in the answer.

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00:14:14,090 --> 00:14:17,420
But one thing that we can do
is we can compare this with

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00:14:17,420 --> 00:14:18,780
these strategies.

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00:14:18,780 --> 00:14:22,545
So what we see is that, well p
is a probability, so it has to

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00:14:22,545 --> 00:14:23,920
be between 0 and 1.

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00:14:23,920 --> 00:14:27,000
So this probability winning for
this strategy is somewhere

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00:14:27,000 --> 00:14:31,130
between one third times 0 plus
one third, which is one third.

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00:14:31,130 --> 00:14:33,460
And one third times 1 plus
one third, which is 2/3.

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00:14:33,460 --> 00:14:38,530
So the strategy is somewhere
between 2/3 and one third.

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00:14:38,530 --> 00:14:43,560
So what we see is that no matter
what, this strategy is

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00:14:43,560 --> 00:14:46,200
at least as good as staying all
the time, because that was

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00:14:46,200 --> 00:14:47,060
only one third.

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00:14:47,060 --> 00:14:51,610
And no matter what it can't be
any better than switching,

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00:14:51,610 --> 00:14:53,850
which was 2/3.

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00:14:53,850 --> 00:14:56,850
So you can also come up with
lots of other different

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00:14:56,850 --> 00:15:00,970
strategies and see what the
probabilities of winning are

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00:15:00,970 --> 00:15:02,580
in that case.

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00:15:02,580 --> 00:15:05,100
OK, so what have we learned
in this problem?

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00:15:05,100 --> 00:15:06,380
What are the key takeaways?

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00:15:06,380 --> 00:15:10,680
One important takeaway is that
it's important to really

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00:15:10,680 --> 00:15:13,500
understand a problem and arrive
at a concrete and

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00:15:13,500 --> 00:15:15,530
precise set of assumptions.

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00:15:15,530 --> 00:15:19,070
So really have a precise problem
that you're solving.

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00:15:19,070 --> 00:15:24,460
And another important takeaway
is to think about your final

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00:15:24,460 --> 00:15:27,230
answer, make sure that that
actually makes sense to you,

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00:15:27,230 --> 00:15:30,740
make sure that you can justify
it somehow intuitively.

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00:15:30,740 --> 00:15:35,030
In that case, you can actually
convince yourself that your

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00:15:35,030 --> 00:15:37,890
answer is actually correct,
because sometimes go through a

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00:15:37,890 --> 00:15:41,660
lot of formulas, and sometimes
your formula may have an error

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00:15:41,660 --> 00:15:43,040
in there somewhere.

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00:15:43,040 --> 00:15:45,040
But you could take the final
answer and ask yourself does

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00:15:45,040 --> 00:15:48,100
this actually makes
sense intuitively?

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00:15:48,100 --> 00:15:52,250
That's often a very good check
and sometimes you can catch

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00:15:52,250 --> 00:15:55,400
errors in your calculations
that way.

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00:15:55,400 --> 00:15:57,160
OK so we'll see next time.

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00:15:57,160 --> 00:16:03,000