
Marilyn vos Savant and the Monty Hall Problem - ryan_j_naughton
http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/
======
bhc3
My crack at it...

1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance
it's behind it.

But there's a 2/3 chance it's behind one of the other doors.

So you've got two sets of outcomes at this point. Set A has 1/3 probability
(the door you chose). Set B has a 2/3 probability (the two doors you didn't
choose).

You then get this incredibly valuable information. The door in Set B that
doesn't have the prize. So now Set B _still_ has a 2/3 probability of having
the prize. But you know that higher probability now applies to only the _one_
door in Set B.

So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance

Make the switch every time.

~~~
dperny
I've known the answer to the Monty Hall problem for a long time, but this
particular explanation just so happens to be the first one that's brought me
closer to grokking the answer. I dunno what's different about this one, but it
makes sense to me. Congrats.

~~~
BerislavLopac
I think that the key sentence in his explanation is "So now Set B still has a
2/3 probability of having the prize." This really nails the whole thing down.

------
Strilanc
An intuition pump for the problem, especially if you know about conservation
of expected evidence [1], is that the asymmetry between the door you picked
and the other door that wasn't opened is that the other door _passed a test_.
The door we choose got a bye; we learned nothing about it's ability to win
car-containing contests. The other door was at risk, and survived. We gained
information about it.

1:
[http://lesswrong.com/lw/ii/conservation_of_expected_evidence...](http://lesswrong.com/lw/ii/conservation_of_expected_evidence/)

~~~
andresp
This was the clearest explanation in my opinion. I'm still a bit confused
regarding how much information have we actually got and what mechanism is
behind "passing a test" to influence the amount of information when the test
result is not definitive (as in this case). We were just given a complete
information about one of the options the user didn't choose, but not any
direct new information specific of the door the user didn't choose and that
remains closed.

Imagine that after opening the door everyone goes home and a new participant
who has no knowledge of the previous experiment is called to make a decision
between the two remaining doors. Obviously in this case the probably is 1:2,
as there is no information regarding the previous test. Actually, if we
imagine a similar situation happened before the initial problem with a
previous show with 4 doors, we may as well see that the initial probability of
1:3 is also wrong if we had access to the knowledge about the previous
experiment.

I still agree with the switch suggestion, but the mechanism behind it is still
not entirely clear.

A more intuitive example would also be to invert the problem: leave one of the
non picked doors (C) out of the random opening and use the door the contestant
picked (A) as a possible case in the random opening experiment. If we start by
opening A, then we already know enough to see if the user won or lost, if we
open B and there is the car behind it, we already know everything. If we open
B and there is a goat behind the door, consider whether you would switch your
choice from A to C. I guess you probably wouldn't and it would feel easier to
justify your position, as this doesn't involve the "psychological pain" of
changing a previous choice by reasons that you don't entirely understand
(associated with possible loss-aversion feelings).

------
sago
She went on later to talk about the Tuesday's child problem, which is even
more interesting and confounding.

 _I have a friend who has two children, he has a son born on a Tuesday, what
is the probability his other child is a girl?_

Compared with this problem

 _I have a friend who has two children, one I know is a boy, what is the
probability the other child is a girl?_

And this one

 _I have a friend who has two children, the oldest is a boy, what is the
probability the other child is a girl?_

Why do these three probabilities differ (in particular, since the last two may
be more obviously different, why is the first one equal to neither of the
others)?

I spent the best part of a week in grad school pondering this until I came up
with a deeper intuition. But the results are still surprising, I think.

~~~
javajosh
My intuition is totally failing me here, because I can't get over the 1/2\. If
I have a friend who has one child, and he's about to have another, what's the
probability that it will be a girl? Clearly 1/2\. What's the difference
between asking any of these questions and the one I just asked? I can't find
any difference.

~~~
sago
Right, that's the last case. The eldest is a boy (or a girl, it doesn't
matter), so the younger is equally likely to be either.

The middle case is 2/3 likely to be a girl. Think of the possibilities of the
sex of two children (sex given in birth order) M then M, M-F, F-M, F-F. I tell
you one is a boy, so we have MM, MF, FM, in two of the three cases, the other
is a girl.

In the first case the answer is 14/27 a girl, so almost 1/2, but not quite. To
figure this out you can consider all 196 combinations of two children with
birth-days-of-the-week, cross out all those that don't match the phrase "he
has a son born on a Tuesday", and you're left with 27 possibilities: 14 of
them are when the other is a girl (14 because she could be born first or
second, and on any day of the week 2x7=14), and 13 when the other is a boy (6
when the other is an older boy born on a non Tuesday, 6 when it is a younger
boy born on a non Tuesday, and 1 when both are both on a Tuesday).

So that's not the intuition, as much as the answer.

The intuition is that as you add more specific knowledge about which child you
mean, that question contributes less to the probability of sex, and is more an
irrelevant detail. Being more specific can be anything. Do it with birthday-
of-the-year, (more specific than day of the week) and it is even closer to
1/2\. Do it with something unique, like their phone number (or their birth
order, or 'the taller one is a boy') and it is exactly 1/2\. Be less specific,
make it so they both toss a coin and it is 'one of the children tossed heads
and is a boy' and it is 3/7.

~~~
javajosh
Maybe I begin to see the light. First, there is an important difference
between considering a random event on it's own and several random events
together: in essence, when your friend says they have a boy, he is telling you
a little something about _two_ ordered events. That's why you count the BF and
FB cases separately (which is the counter-intuitive part).

The actual Tuesday problem requires a little more thought. Somehow that extra
information makes you drop a case! I can't help but think that the ordering is
the critical factor, and that it's not a coincidence that the "extra
information" has to do with time. In other words, the dropped case happens
because we drop the distinctive ordering when the two boys are born on a
Tuesday.

~~~
sago
You picked up on a really important issue that is often assumed. Thank you for
not letting me just slip it under the radar without question.

To answer your question: we treat the MF and FM differently, just for ease of
calculation.

What we're doing is taking a set of possible situations (different 2-child
families), each with an attached probability (I'll come back to this),
throwing away those that don't match some known information (i.e. those
without a boy), taking those that do, grouping them into sets (the set of
families with the other being a girl, and the set where the other is a boy),
and calculating the probability the actual scenario was in each of those sets.

Now the 'attached probability' probably sounded odd in there, because that's
not what we've done, right? We've just been counting situations. True, but
that's because we carefully started with situations that are all equally
likely. So rather than bugger about with those attached probabilities, we
could just use raw counts, knowing that it would work.

So, if we start with four possible sexes of two children: MM, MF, FM, FF,
those are all equally likely. I'm using the birth order as a way of breaking
the two MF cases apart, so I've got two cases that are equally likely, letting
me do the counting trick. I can just count the 2 out of 3 remaining situations
that match the information, so 2/3\. Easy.

But I don't have to do it that way.

Let's say we follow your intuition and start with three possible sex
combinations: MM, MF, and FF. Now, those aren't equally likely, right? There
are twice as many families with MF children as with MM children. So I need to
use the attached probabilities.

MM (1/4 of 2-child families), MF (1/2), FF (1/4)

We can remove the FF as inconsistent with the information (as before), then
the probability the parent has an MF family (i.e. their other child is a girl)
is

1/2 over (1/2 + 1/4) = 2/3

as before.

In fact, for any reasonable calculation based on data, we're almost certainly
not going to be able to use the counting trick. In the case where I was using
actual birth rate statistics for sex (rather than assuming as many girls as
boys are born), I'd have to do it this way. The counting trick is really only
useful for toy problems and teaching basic probability: you were right to call
me out on it.

\---

The Tuesday's child doesn't drop a case. There are certainly seven cases
where, in an MM family, the eldest is a Tue child, and seven where the
youngest is, but one of those cases is shared: a case where both are Tue
children.

In terms of the intuition you can think of it this way, identifying by birth-
day-of-the-week is identifying which child is which, so we're almost at 1/2,
but there is one case (both Tue boys) where this information does nothing to
distinguish them, so that tiny bit of ambiguity remains, and it can't quite
reach the full 1/2, there's still a trace of the 1/3 result that came when the
ambiguity was total.

------
tmuir
None of the explanations in the article really convinced me either. But after
looking at the chart of all possible outcomes for a while, it finally clicked.
It makes more sense to think of it in reverse from the winning position.

Winning by switching doors requires your first choice to be incorrect.

You have a 2/3 chance of guessing incorrectly when all three doors are closed.

Thus, you have a 2/3 chance of putting yourself in the position from which
switching will win.

------
asdfologist
I felt pretty stupid when it took me a long time to convince myself of the
correct answer. But I felt better when I read that even "Paul Erdős, one of
the most prolific mathematicians in history, remained unconvinced until he was
shown a computer simulation confirming the predicted result"!

[http://en.wikipedia.org/wiki/Monty_Hall_problem](http://en.wikipedia.org/wiki/Monty_Hall_problem)

~~~
bmelton
The only explanation that convinced me that my original guess was incorrect
was this one (which is alluded to in the article):

If there are three doors, the odds that the car is behind any of them is 1 in
3, or 1/3\. Once you select a door, the odds are 1/3 for your door, 1/3 for
each of the other, unselected doors. Put more simply, 1/3 for your door, or
2/3 that it is _not_ in your door. Once another door is opened that does not
have the car, then the odds of it being your door are 1/3, and the odds of it
not being your door are still 2/3, so it makes sense to switch.

~~~
mjevans
Thank you. Adding the temporal domain and system states is what makes this
easy to understand.

At the time of selection you freeze the probability of the sets (as this is
the macroscopic world and the implications of the state in the other rooms
must have collapsed due to interactions with all sorts of radiation from the
rest of the universe).

Thus what you say is exactly correct.

* T0 - {1/3} {1/3} {1/3} * T1 - {1/3 Your pick} {2/3 = {1/3} + {1/3}} * T2 - {1/3 Your pick} {2/3 = {0/3 Host Pick} {2/3 Remaining door}}

------
mediocregopher
So I _believe_ the 2/3 answer to be true. I see the evidence in those tables.
But intuitively, I still don't get it, and the explanation in the article
isn't helping

> The short answer is that your initial odds of winning with door #1 (⅓) don’t
> change simply because the host reveals a goat behind door #3; instead,
> Hall’s action increases the odds to ⅔ that you’ll win by switching.

Well, why don't they change simply because the host reveals a goat? If you
show me a three sided coin, I'd say there's a 1/3 chance of getting a
particular side. If you then magically change gravity such that one of the
other sides will never ever be landed on, would the two remaining sides not
now have 50/50? I guess the article asserts that the "probability" doesn't get
redistributed amongst the remaining options evenly, and I want to know why it
doesn't and why it instead all "goes" to the one I didn't pick.

I don't mean to be confrontational in asking, and I don't doubt the answer, I
just don't get _why_. Could anyone explain it better, or point me somewhere
that does?

~~~
zaksoup
The way I like to think about it is as follows: You pick door 1. Out of three
doors you have a 1/3 chance of having a car behind your door. The chance that
the car is behind one of either of the other two doors is 2/3\. Let's pretend
that instead of opening a door and revealing a goat, Monty instead says to you
"You can switch to both of the remaining doors. If the car is behind either
one of them you get to keep it." Your likelyhood of getting the car by
switching is now 2/3.

Instead, Monty does the statistical equivalent: He allows you to know for sure
which of the other doors definitely has a goat.

~~~
choffstein
This is a great answer.

To answer the question about the 3-sided die: the question is flawed for 2
reasons.

First, the way you phrased it, you know the information a priori. It's like me
learning that the car isn't behind door #1 before I choose. In that case, I
have a 50/50 chance on the other doors.

But what if you didn't tell me before? This introduces our second problem.
Let's say I roll the die and you hide it under a cup. I tell you I chose side
A and you tell me A is physically impossible and ask me if I want to switch.
Well of course I want to switch ... but I didn't learn any more information
about sides B and C. It's a 50/50 toss-up. In terms of the game show, you
basically just opened the constant's door.

A critical element of the original problem is when the information is learned
and knowing that the contestant's door will not be opened. It is what allows
the phrasing from zaksoup above.

------
elihu
One thing that bothers me about the problem as it's stated in the article is
that it isn't explicit about whether the game show host is _required_ to open
one of the goat doors regardless of which door the contestant picks, or
whether the host is allowed to choose whether to open a door or not. If the
host were to only open a goat door if the contestant initially picks the
correct door, then the contestant would lose every single time if he/she
always switched doors.

~~~
mcphage
> whether the host is allowed to choose whether to open a door or not

The problem as stated is: "you select door #1. Then, the host, who is well-
aware of what’s going on behind the scenes, opens door #3". Why would you
think the host has the option of not opening a door? And if you think they
could do that, why couldn't the host also do other things, like offer the
contestant cash to walk away, or swap what's behind the doors, or add more
doors, or open all three doors, or sucker-punch them in the gut?

~~~
T-hawk
> Why would you think the host has the option of not opening a door?

Because the actual TV show of Let's Make A Deal worked that way. Here is
Monty's own take on it (click to page 2.)
[http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-
doo...](http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doors-puzzle-
debate-and-answer.html)

This is the source of all the confusion from vos Savant's original column. vos
Savant got the math right for her version of the problem where the host always
opens a door, but that isn't how the show worked. She presented a simplified
"spherical frictionless cows" version of the problem.

The actual show did not compel Monty with any rules on opening doors. He had
free rein to do whatever he wanted. He could reveal an incorrect guess right
away with no opportunity to switch. He could and would offer the contestant
cash to switch or not switch. Monty could try anything to trick or bluff or
deceive the contestant, all for the sake of producing entertaining television.

Of course, there is no correct mathematical answer at all to a game of
bluffing and outwitting. At best you might find a Nash equilibrium in game
theory, but even that is hard to pin down. You need to know Monty's tendencies
precisely, but even that is hard when his goal isn't to maximize expected
value but to produce entertainment.

~~~
mcphage
> The actual show did not compel Monty with any rules on opening doors. He had
> free rein to do whatever he wanted. He could reveal an incorrect guess right
> away with no opportunity to switch. He could and would offer the contestant
> cash to switch or not switch. Monty could try anything to trick or bluff or
> deceive the contestant, all for the sake of producing entertaining
> television.

And really, the thing is, _none of that matters_. Maybe in this version he
could reveal an incorrect guess! Maybe in this version he could reveal the
door that the contestant picked! That's not what happened in the situation
you're asked to analyze. All that matters is that the host picked a door that
(a) was not the door the contestant picked, and (b) contains a goat. Nothing
beyond that is part of this problem, and nothing beyond that is necessary to
solve it. You're given a very specific situation, and asked what the best
decision to make is.

------
Terr_
I'd re-frame it this way:

"The host has zero incentive to reveal what's behind the door you already
picked (or the choice of switching would be a no-brainer) so the fact that
they left it shut doesn't tell you anything."

"However, the fact that they _didn 't_ (or couldn't) open the remaining shut-
door hints that it may contain the prize."

The key is that the host's choice _is not random_.

~~~
ninkendo
Exactly the way I like to think of it. When he opens the goat door there are
now two possibilities: you were right the first time, or the remaining door
has the car.

By selecting the other door your are now betting against your initial choice,
which _was_ a 1/3 chance. Thus your new odds are 2/3.

------
UnoriginalGuy
I just want to start off by saying that I am not, at all, trying to place
blame on Marilyn vos Savant for any of this.

When someone gets declared as the smartest/best in the world it first off sets
people's expectations so high that they expect perfection, and secondly it
places the individual onto a plateau for which others would love to knock them
down.

The article doesn't discuss this, but I'm sure that played a part in why her
detractors jumped on her so fast and so fiercely. They got told that this
person was smarter than them, and they were inferior, and now was their chance
at proving that that wasn't the case.

She was of course right, and she likely didn't declare herself the "world's
smartest woman" but unfortunately for her, that connotation was still attached
to her character, and no matter how modest she was it would have been
difficult for other people to take an ego knock by feeling like this person
was "better" than them.

Essentially the title of "world's smartest woman" set her up for something
like this to happen. Luckily for her her detractors jumped on something she
was correct about making themselves sound like morons. However it could have
happened when she made her first public mistake (it would have happened
eventually).

Fortunately this episode may have helped release the pressure somewhat that
was placed on her character. She saved face, but her detractors feel like they
took her down a peg so didn't continuously stalk her looking for mistakes.

PS - Just to reiterate I am in no way trying to blame Marilyn vos Savant or
deflect any of it onto her. Someone else likely declared her the "world's
smartest woman," that is outside of her control, and she dealt with this as
well as anyone could hope to.

~~~
mcphage
> The article doesn't discuss this, but I'm sure that played a part in why her
> detractors jumped on her so fast and so fiercely. They got told that this
> person was smarter than them, and they were inferior, and now was their
> chance at proving that that wasn't the case.

If only they had stopped to see _if she was even wrong_ first, they wouldn't
have proven the claim.

------
savanaly
I read about this problem and the fact that multiple educated writers falsely
called out vos Savant in an old Martin Gardner column, and it shines in my
memory as one of the problems that illuminated math for me and made me see
recreational math as worthwhile. Of course back when he wrote about it the
topic of sexism was mercifully left out, and I honestly hadn't given a
moment's thought to the fact that vos Savant was a woman until I read this new
take on it, but there you go.

Have standards for what makes something confusing and difficult to understand
risen significantly over the decades? When I brought this problem up with my
group of (admittedly geeky and mathy) friends no one was left unconvinced by
the end of the afternoon, and most even had the intuition as well as the
mathematical proof of the answer. Clearly we aren't any smarter than the
public figures who couldn't comprehend this puzzle a half a century ago-- so
it all in the framing of the problem?

~~~
sukilot
Vos Savant originally posed the problem with ambiguous wording.

~~~
mcphage
The original problem, from her website:

"Suppose you’re on a game show, and you’re given the choice of three doors.
Behind one door is a car, behind the others, goats. You pick a door, say #1,
and the host, who knows what’s behind the doors, opens another door, say #3,
which has a goat. He says to you, "Do you want to pick door #2?" Is it to your
advantage to switch your choice of doors?

Craig F. Whitaker

Columbia, Maryland"

(a) she didn't originally pose the problem, it was a question asked of her,
and (b) the wording of the problem is not ambiguous.

~~~
bztzt
the ambiguity is that it's presented as something the host does in this
particular instance, but the validity of the probability calculations depends
on it being a protocol the host always follows (in particular, that the host
will always reveal one of the goats _not_ behind the contestant's first choice
of doors).

~~~
mcphage
There _is no_ "always follows". This is an event which will not be repeated in
the history of the world. All there is is 3 doors: you pick one, the host
opens another door which contains a goat, and offers you a choice. Why he
picked the door that he chose is irrelevant. All that is relevant is that (a)
it's not the door you chose, and (b) it contains a goat. You don't need
anything beyond that to solve this problem. If you think you need something
beyond that to solve the problem, you're not solving it correctly.

~~~
PalmerEldritch
"Why he picked the door that he chose is irrelevant" I disagree. If he picked
the door at random and it just happened to contain a goat it's 50/50\. If he
picked the door because he was always going to pick that door if it contained
a goat then it's 50/50

------
vladislav
I don't buy this at all : "Despite its deceptive simplicity, some of the
world’s brightest minds -- MIT professors, renowned mathematicians, and
MacArthur “Genius” Fellows -- have had trouble grasping its answer."

The intuition is simple. Each time you initially pick a door with a goat, you
win the car by switching. This clearly happens 2/3 of the time.

~~~
diginet
That's not true, as Wikipedia says:

"Paul Erdős, one of the most prolific mathematicians in history, remained
unconvinced until he was shown a computer simulation confirming the predicted
result (Vazsonyi 1999)."

~~~
vladislav
Lol, Paul Erdos was 82 in 1995, which is when Vazsonyi claims they had this
conversation, a year before Erdos died. Beyond his advanced age, it's well
known that he did amphetamines his whole life and by that point his brain was
pretty fried. I guarantee you if you gave the Monty Hall problem to the great
Erdos in his primetime he would've thought it trivial.

------
coreyp_1
This was fascinating. I can't believe so many academics got it wrong, and then
attacked her gender over it!

~~~
bsdetector
I'm not surprised. At the time, almost every question every week in Parade
magazine was just pulp. Questions of the type "what's the world's smartest
woman's favorite color?". You could take anybody off the street and get the
same answers. Parade magazine was a gossip magazine.

Furthermore, at the time she wasn't listed in the _current_ Guinness Book so
anybody that looked it up would see the column as a fraud unless they checked
a specific old edition. Claiming somebody "is listed" when they had been
listed, for one year many years ago, is a lie.

So there were a lot of readers who thought of the column as a joke. If you
open 99 doors that are full of mindless drivel, what are the odds the 100 is
the one that has something interesting? This is why so many people were fooled
by it.

~~~
mcphage
> If you open 99 doors that are full of mindless drivel, what are the odds the
> 100 is the one that has something interesting? This is why so many people
> were fooled by it.

So they thought "ha, this is foolishly wrong, I'll write in and tell them
that—of course, without doing the 5 minute math check to make sure, first!"

------
QuantumGood
I'm a teacher, so I optimize for the result "understanding in my students." I
teach it like this:

You _don 't_ bring up Monty Hall first. It bends minds too much.

I give a parallel game, where basics are similar, but you can choose to reveal
3 doors, 2 doors, or just one door each time. Everyone (after making sure they
aren't being tricked) chooses to reveal three doors. Winner every time. Then I
change the rules: Can't open all three doors. Have to choose between 1 or two.
Everyone picks two.

Only _then_ do I give the Monty Hall problem, and show (as @bhc3 put it
succinctly) that it's essentially the same game they just learned: your choice
is between revealing two doors, or revealing one door.

The most interesting part of this, and why I do it this way, is that I
specifically optimize for my students to be able to explain what they know. I
can teach someone to win an argument with someone who is convinced it's a
50/50 chance this way. Other ways, people sorta, kinda get it, but can't
explain it to others who are confused.

EDIT:

The general principle here IMHO is teach something simple first, and show how
it can be applied to something more complicated. When people think something
is complicated, "breaking it down" into simpler concepts or making easy
analogies doesn't have that great a track record. Their brain seems to have
already classified it as "complex" and that has to be overcome —they have to
be convinced to drop the tag #complex— in addition to however you teach the
right understanding.

------
planckscnst
The way I see the problem is slightly different and I think quite intuitive.
If you've chosen a goat, then the host, by eliminating a door, is telling you
exactly where the car is. If you have not chosen a goat, then the host not
telling you where the car is. What are the chances that the host is telling
you the location of the car? It's the same as the chances of you having chosen
a goat the first time around.

------
MaysonL
I still think that the problem is wrongly cast as a probability problem,
rather than the game-theoretic problem that it is.

~~~
Terr_
A fair point, the hosts' choice of door is not random at all. He _could_ open
the door you'd already chosen, but that would make your choice to switch-or-
not a no-brainer.

------
WillNotDownvote
It's a testament to the brilliance of game show producers that they took
something like this that both the contestant and the audience will
misunderstand (but looks so simple!) and made a long running show out of it,
making millions along the way.

------
_pius
Quickest way I've seen to understand this result is referenced on Wikipedia:

 _An intuitive explanation is that if the contestant picks a goat (2 of 3
doors) the contestant will win the car by switching as the other goat can no
longer be picked, while if the contestant picks the car (1 of 3 doors) the
contestant will not win the car by switching. The fact that the host
subsequently reveals a goat in one of the unchosen doors changes nothing about
the initial probability._

[http://en.wikipedia.org/wiki/Monty_Hall_problem](http://en.wikipedia.org/wiki/Monty_Hall_problem)

------
transfire
There is a flaw in the chart. For the three listed Lose cases, it clumps two
distinct choices into one line item and counts them as such, and thus provides
an incorrect count of the possibilities.

Moreover the basic argument is flawed. The knowledge of Monty has no barring
on the knowledge of the contestant. If Monty simply presented you with three
doors that you were to pick from, but opened an incorrect door _before_ you
choose your first door, then what would your odds be? The fact that he opens
one after you choose does not matter. Monty has total knowledge of the setup
and you still have zero.

~~~
erroneousfunk
After you choose a door, Monty cannot choose it -- he's stuck choosing one of
the two doors that you did not. So, the order of choosing the doors absolutely
matters.

~~~
transfire
OK. You are right. I worked through the whole problem meticulously and figured
it out. It's actually rather easy to explain once worked through, as follows:

The odds of initially choosing correctly are 1 in 3. In that case if you
switch you will loose, but if you stay you will win. The odds of choosing
wrong initially are 2 in 3. In which case if you stay you will loose, but if
you switch you will win. So it is actually that simple. If you always stay you
will win 1/3rd of the time and if you always switch you will win 2/3rds of the
time.

It only seems like it should be 50/50 at first blush b/c that is what the odds
would be if you choose to stay or switch arbitrarily. That is also why if you
write down all the possibilities and simply add them up it will appear to be
50/50 too (which is what I did).

The only thing I still can't quite figure out is exactly how the exposure of
one of the renaming two doors increases one's knowledge. I thought I could
take the list of possibilities and eliminate those that no longer apply and
then add them up, but that doesn't seem to work.

------
simple1
One way to intuitively believe the answer is to imagine there were 100 doors,
you picked one, then the host opened up 98 of the other doors, revealing
goats. Switching would be the obvious answer in that case.

~~~
cdelsolar
That's still not obvious to me

~~~
peripitea
I'm a little intoxicated so this may be way over doing it or just not make
sense, but...

Let's make the number even bigger, so we don't even need to deal in
probabilities -- a sextillion (10^21) doors. That's more doors than there are
grains of sand on the planet. Say you're given a sextillion doors, and you
pick one lonesome door, call it door A. What are the odds that door A is the
right one? Essentially zero, right? I'm going to say straight up zero because
now is not the time for you to be a dreamer.

Ok so now that you've made your futile pick, poor Monty has to go and open
every other door, all sextillion of them, except for yours and one other
(let's call it door B) which he leaves closed. Every door he opens contains a
goat.

So now out of this vast universe of doors, there are (10^21-2) doors opened
with goats bleating away, and only two doors left unopened: door A, aka yours;
and door B, the one that Monty mysteriously left unopened. One of them must
contain the car.

(I'm guessing at this point it will be obvious for many people that door B
contains the car, but I'll elaborate further in case it isn't clear yet.)

So we're left with your door A and Monty's door B. At this point there are
only two scenarios that matter. Either you picked the right door originally,
in which case staying put is the winning move. OR, you picked the wrong door
originally, in which case switching is the right move. And we already know you
picked the wrong door (remember, no dreamers allowed here), so you should for
sure switch to this by-now-extremely-conspicuous-looking door B.

Door B isn't just some random door; the fact that it was chosen as the only
door to not be opened tells you a tremendous amount of information about it.
Think about the game from Monty's perspective. Either you pick the correct
door initially and he gets to make door B a random goat door, or you picked
the wrong door and he has no choice but to make door B the door with the car
behind it. Since we know you picked the wrong door, we know what Monty's gonna
have to do.

------
transfire
It didn't seem right to me at first, so I worked through it. The trick was to
realize that switching always gives the opposite result of the initial choice.
So if the initial choice is correct, switching will make it wrong. On the
other hand, if the initial choice is wrong, switching will make it right.
(Recall that Monty eliminated one wrong choice, so there is only one other
choice from which to choose.) Since you have 1/3 chance of being right and 2/3
chance of being wrong on the initial pick, switching will reverse your odds.

------
whoopdedo
From Monty Hall's interview:

> He retained the authority to offer the contestant cash NOT to switch. "I
> wanted to con [them] into switching there."

So he wanted them to win the car. That seems interesting. I suppose it makes
sense as this was an entertainment program. Viewers like a happy ending. Or if
the guest accepts the cash and stays with the lower probability door, it's
okay that they get the goat because they have cash. So win-win just for
appearing. Also prevents unhappy guests from saying bad things about the show.

------
JohnTHaller
It always amazed me that so many academics got it wrong. Once you understand
the way the actual game works, it's pretty obvious that you gain a statistical
advantage if you switch.

~~~
savanaly
I know right? I don't blame them for getting it wrong at first glance-- I'm
pretty sure 999 out of 1000 humans would. But the mathematics of the
probabilities can be written down on a napkin they are so simple, so how was
no one able to convince these scientists with a clear, concise technical
proof? They really needed a computer simulation, as I've seen it claimed?

~~~
function_seven
> But the mathematics of the probabilities can be written down on a napkin
> they are so simple, so how was no one able to convince these scientists with
> a clear, concise technical proof?

I'm even more appalled that no one tried, you know, actually running the game.
Have someone take a quarter and two pennies, mix them up under cups, let you
choose one, show you a penny, then switch or not. Repeat several times. Switch
places. When you're playing the role of Monty, the truth will reveal itself...

~~~
MichaelGG
I have a close friend that was incensed about my explanation about how he was
misunderstand. He even ran a broken simulation that showed he was right. When
I corrected his source code with detailed comments, he further accused me of
trying to fool him and make fun of him. Didn't speak to me for a month.

Bizarre. But I suppose this kind of thing that goes against instincts can
really upset people.

------
pieguy
What many descriptions of the problem leave out is that it is critical that
Monty knowingly picks a door with a goat. If Monty chooses at random, then the
winning odds do indeed increase to 50% if he happens to open a door with a
goat behind it.

The randomized variant, by the way, tends to garner roughly the same degree of
"you're completely wrong" responses when given to people who know just enough
about the normal Monty-hall problem not to be fooled by it.

------
rhino42
One interesting variant, is the case where the Host opens a door without
knowing if that door had a goat. Expressing relief that he opened a goat door,
he asks you to consider switching doors.

IIRC, you get less information in this case, since there was a 1/3 chance that
the host would accidentally pick the "car" door, but you still have a 1/2
chance of winning by switching.

~~~
rando3826
> still have a 1/2 chance of winning by switching

It seems that you are referring to the other variant, in which case this is
false, because that one has a 2/3 chance of winning by switching.

~~~
taeric
No, you have a 1/2 chance of winning _after a door has been opened._ However,
you also have a chance of losing on that reveal. Something you don't have in
the normal variants. (That is, in normal variants, you only lose at the end of
the game. This one, you could effectively lose when the host opens the actual
winning door.)

So, your overall odds aren't as different as it seems. Just you have a much
better odds of winning by switching, if you are still in the game.

------
Orangeair
"You made a mistake, but look at the positive side. If all those Ph.D.’s were
wrong, the country would be in some very serious trouble."

Best quote in the entire article.

And if I had to explain the problem in one sentence: Opening the first door
obviously gives you a 1/3 chance, which remains unaffected by the host opening
another door.

~~~
danellis
> Opening the first door obviously gives you a 1/3 chance, which remains
> unaffected by the host opening another door.

You had a 1/3 chance if you opened the second door too. Why does the second
door's chance change, but the first door's chance is somehow locked by you
having chosen it?

~~~
erroneousfunk
The chance is locked in because, by the rules of the game, the host can't open
_that one_ and show you anything -- regardless of what it contains. You know
that there's at least one door that the host can open (of the two remaining
doors), so them opening a door changes nothing about the door you've chosen.

------
ucha
The problem is not well-posed. The debate should be about its formulation, not
about the math...

The key here is to understand that no matter what you do, the game organizer
will ALWAYS open one of the two others doors.

Now, what if the organizer was to open a door only in some cases? He could
open the door only when you choose the correct door. In that case, you have 0
chance of winning when you switch doors when the organizer opens another one.

What makes the 1/2-1/2 probability - that so may math PhDs defended -
acceptable is this: the organizer is selecting a door at random and opening
it. Then if I choose a door and the organizer opens another one, I know that I
have a 50% chance of having selected the correct one. Of course, he could also
have opened my door in which case the game ends.

This is a good example of how an ill-posed problem have different solutions
depending on its implicit interpretation.

~~~
mcphage
In the problem as stated, the organizer _did_ open a door, which _did_ contain
a goat.

~~~
ucha
That's not enough. You need the knowledge of "the organizer will ALWAYS open a
door that contains a goat" to draw the conclusion of the article. That
knowledge is implicitly assumed. I have shown that by assuming a different
knowledge - in this case, the organizer only opens the door that contains a
goat when you choose the door that contains the car - you can end up with
different probabilities.

I don't have an IQ of 200+ nor do I have a PhD in math, but I'm pretty certain
that the probabilities in this problem are unknown unless you know the
behavior of the organizer in EVERY situation, not in one particular case.

~~~
mcphage
> That's not enough. You need the knowledge of "the organizer will ALWAYS open
> a door that contains a goat" to draw the conclusion of the article. That
> knowledge is implicitly assumed.

It's not implicitly assumed, it's stated in the original question: "You pick a
door, say #1, and the host, who knows what’s behind the doors, opens another
door, say #3, which has a goat." You pick a door, the hosts opens another door
which contains a goat. He can't open the door you chose, he can't open a door
with the prize, he can't not open a door. Introducing a possibility which is
not a part of the specified problem turns it into a different problem.

> I have shown that by assuming a different knowledge - in this case, the
> organizer only opens the door that contains a goat when you choose the door
> that contains the car - you can end up with different probabilities.

For every "the organizer only the opens a door that contains a goat when you
choose the door that contains a car", there's an "the organizer only opens a
door that contains a goat when you open a door that doesn't contain a car".
Not knowing the host's behavior means that it isn't part of your analysis. If
you don't know why he does what he does, all you can use are the facts in
front of you: there are three doors, you pick one, you are shown a second door
which doesn't contain the prize, and are offered a chance to choose.

> I'm pretty certain that the probabilities in this problem are unknown unless
> you know the behavior of the organizer in EVERY situation, not in one
> particular case

Not so. If you knew the host's behavior (which the original problem does
specify precisely), you can use it in your analysis. But not knowing it, means
the best decision has to be independent of it. And if you knew the door that
contained the prize, you could use that in your analysis. But not knowing it,
means the best decision has to be made independent of it.

~~~
ucha
You are making an assumption on the host behavior, the assumption being that
if you repeat the experiment the host will open a door containing a goat. If
you think you're not and that the 1/3 - 2/3 probabilities are correct based on
the single observation you made, consider this:

you have 1 black ball and a 2 white balls in 3 different urns each containing
a ball. You select one urn A. Some external agent randomly overturns an urn.
That ball it contains is white. What's the probability that the ball in urn A
is black? The answer is 1/2\. Not 1/3\. It's 1/2 because you the assumption
you make on the external agent's behavior is that it's random. If you don't
make an assumption on its behavior, you can't calculate probabilities. If you
make the assumption that the external agent picked up the white ball on
purpose and will repeat that behavior when faced with similar circumstances,
the probability is 2/3.

My point is that if you don't make an assumption on the host's behavior, you
can't calculate probabilities.

------
pbreit
There's an important rule that isn't always communicated correctly: the
participant must know ahead of time that the host is always going to open a
door.

------
bwy
Impossible to graduate from Berkeley with a CS degree without hearing this
problem :)

