
Monty Hall Problem - leeoniya
https://en.wikipedia.org/wiki/Monty_Hall_problem
======
drfuchs
I found the easiest way to convince non-mathematical types is by suggesting we
do the million-door version: “Ok, pick a door between 1 and 1,000,000. We’ll
call that door N. Now, I’ll open 999,998 bad doors. So, now only door N and
door 223,205 are still closed, and one of them has the good prize. Want to
keep door N, or swap?”

~~~
hinkley
You want to have an ugly argument with other developers, bring up the Monty
Hall problem. People who are wrong will _not_ let it go.

We had it beaten into us not to use truth tables to solve problems in college,
because it doesn't scale. But the Monty Hall problem has a finite set of
states you can easily write on a piece of paper. And then you'll see that it
jumps from 2/6 successes to 3/6.

I tried to replicate that table under duress once (see initial comment) and I
couldn't get it sorted. I should try it again and write it down this time.

Edit: Why do work when you can find someone else who already has:

[https://www.statisticshowto.datasciencecentral.com/probabili...](https://www.statisticshowto.datasciencecentral.com/probability-
and-statistics/monty-hall-problem/)

and particular to my statement, this table:

[https://www.statisticshowto.datasciencecentral.com/wp-
conten...](https://www.statisticshowto.datasciencecentral.com/wp-
content/uploads/2014/12/monty-hall-problem2.png)

When switching the number of lose scenarios is lower than a naive expectation
of outcomes.

~~~
daveFNbuck
You have to be careful with the truth table method. It's easy to get it wrong,
as your events might not have equal probability.

For example, we'll label the three doors P, L, R, for the prize door, the
leftmost goat door and the rightmost goat door. This gives possibilities

1\. You choose P, Monty chooses L - switching loses

2\. You choose P, Monty chooses R - switching loses

3\. You choose L, Monty chooses R - switching wins

4\. You choose R, Monty chooses L - switching wins

This seems to indicate that switching gives you a 50/50 likelihood of winning.
In fact, this is the correct truth table to use if Monty doesn't know which
door has the prize. In that case, these are all equally likely.

Instead, if he does know where the prize is and won't open that door, this
truth table is misleading because the cases where switching loses are less
likely. If you choose L, he'll definitely choose R, but if you choose P, he
has two options to split the probability.

Similarly, if you model Monty as not knowing what's behind the doors, then the
truth table you're thinking of doesn't work. Either way, the truth table on
its own isn't convincing.

The thing I find more clear is using Bayes' rule, as it shows exactly how the
model of Monty's behavior affects the probability that switching is
beneficial.

Let H be the event that your door has a prize and G be the event that Monty
shows you a goat door. Then P(H|G) = P(G|H) * P(H) / P(G). Obviously, P(H) =
1/3 and (assuming Monty must pick a door other than yours) P(G|H) = 1, so it
just comes down to what P(G) is.

If he always shows a goat, then P(G) 1 and you should switch. If he doesn't
know what's behind the doors, it's 2/3 and it doesn't matter whether you
switch. If he knows but prefers to spoil the prize, it's 1/3 and you
definitely have the prize door.

~~~
Sean1708
I've always preferred to draw it out as a tree of what your your possible
choices are at each stage, I find easier to get right _and_ I find that people
understand it more intuitively.

    
    
           ┌─────────────┼─────────────┐
           G             G             C
          ┌┴┐           ┌┴┐           ┌┴┐
      Stay│ │Switch Stay│ │Switch Stay│ │Switch
          G C           G C           C G

~~~
daveFNbuck
That has similar drawbacks of hiding the subtlety behind how Monty is being
modeled. The tree you've drawn will tend to lead people to the wrong answer if
you ask whether switching improves your odds when Monty doesn't know which
door has the car.

------
ivanb
This is only a paradox due to very poor definition.

"You pick a door, say No. 1, and the host, who knows what's behind the doors,
opens another door, say No. 3, which has a goat." Where does it state that the
host always opens a door with a goat? From the statement it looks like the
goat was accidental. No wonder Paul Erdős got confused by this too. The
translation of the paradox to his language could have been even more
confusing.

~~~
lqet
100% this.

The first time I read about this problem, it was formulated in a way which
made it clear that the host opens a door with a goat behind it. If you learn
about it like this, you literally cannot believe how anyone (even Paul Erdős)
could believe that switching is _not_ better. After all, you choose a door
with a goat behind it with probability 2/3 on your first try, and if you do
this, than switching will _always_ win you the car. It is so trivial that the
only "paradox" here is that it is classified as a paradox.

E: the explanation on Wikipedia is unnecessarily elaborate and hard to follow:

"The given probabilities depend on specific assumptions about how the host and
contestant choose their doors. A key insight is that, under these standard
conditions, there is more information about doors 2 and 3 that was not
available at the beginning of the game, when door 1 was chosen by the player:
the host's deliberate action adds value to the door he did not choose to
eliminate, but not to the one chosen by the contestant originally. Another
insight is that switching doors is a different action than choosing between
the two remaining doors at random, as the first action uses the previous
information and the latter does not. Other possible behaviors than the one
described can reveal different additional information, or none at all, and
yield different probabilities. "

 _What?_

~~~
henrikschroder
There are plenty of people who keep getting tripped up by this problem even
when it's correctly formulated. Statistics and probabilities are sometimes
incredibly unintuitive and hard for a lot of people.

I only took a single class of statistics at university, but my biggest take-
away from that class is that you absolutely cannot trust your gut, because the
human bias is crazy strong.

------
alexhutcheson
This alternative formulation from a Mathematics StackExchange answer[1] makes
the decision to switch much more obvious:

"There are three boxers. Two of the boxers are evenly matched (i.e. 50-50, no
draws!); the other boxer will beat either them, always.

You blindly guess that Boxer A is the best and let the other two fight.

Boxer B beats Boxer C.

Do you want to stick with Boxer A in a match-up with Boxer B, or do you want
to switch?"

[1] [https://math.stackexchange.com/questions/96826/the-monty-
hal...](https://math.stackexchange.com/questions/96826/the-monty-hall-
problem/3360686#3360686)

~~~
jedimastert
I'm gonna have to do a lot of thinking to make sure this matches the problem
at hand, but if it does this is an incredible translation.

~~~
zaksoup
Here's the explanation I like to use every year when this makes it to HN:

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.

~~~
leeoniya
another way to think of it is:

monty has the car 2/3 of the time. when he has the car, he _must_ tell you
where it is (by showing you where it isn't).

that last part is key. he is not making a random selection. so if you go with
his remaining door, it will have the car the same 2/3 of the time.

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

this tidbit is pretty wild

~~~
turc1656
What's even wilder is that the people working for game shows utilized this
strategy long before it was a known, publicly verified thing. So someone who
wasn't a world famous mathematician figured it out and was confident in their
work - the same work that a renowned, industry-leading professional refused to
believe.

~~~
ghaff
Once you figure it out, it's pretty obvious though. And the game show people
don't need a rigorous mathematical proof. You can simulate it with some
playing cards and understand it in a hurry.

What made it very clear to me is if you just increase the number of incorrect
doors that you open by a lot. (Making the assumption that the host knows where
the prize is.) In that case, it's obvious that you're improving the odds by
showing where the prize isn't.

------
goto11
It is really important to make it clear that:

1) Monty always opens another door (regardless of what is behind the door
initially chosen by the contestant). He will never open the door the
contestant chose.

2) Always opens a door with a goat behind.

3) These rules are known to the contestant

These premises are not intuitive, and some versions of the problem does not
make them explicit, which is why it is more of a "gotcha" riddle.

In particular, the version quoted in Wikipedia does not make any of that
clear. It just mentions the particular case of the host opening another the
door with the goat. But without knowing the above "rules", this just means
there is 50% chance per door now.

~~~
WilliamEdward
This ambiguity turned a really easy problem into one that people debated for
decades for some reason

------
mself
Here’s another version that helps clarify it...

There are three doors, one has a prize and the other two are empty. You pick a
door (say, Door 1). Monty then says that his assistant will move whatever is
behind Door 2 (if anything) to Door 3. He then opens Door 2 to show you it is
empty. Would you switch to Door 3? Of course you would.

But Monty’s assistant is lazy, so instead of moving the contents from behind
one of the closed doors to the other, he just asks Monty to open the one that
he knows is already empty. The result is the same.

------
DamnInteresting
I wrote about the Monte Hall problem back in 2005, and I made a simple
simulator to demonstrate the success rate of door-switchers versus non-
switchers: [https://www.damninteresting.com/appendices/monte-hall-
simula...](https://www.damninteresting.com/appendices/monte-hall-simulator/)

~~~
flashman
I have been reading your website since the heyday of Digg. Thanks for the link
- I thought you must have stopped writing around 2010.

------
Doubl
I think most people get it that prize is twice as likely to be behind one of
the two other doors rather than behind the door you pick. So one time out of
three you'll get it right while two times out of three the presenter
indirectly indicates the winning door.

Edit. Be nice if my downvoter would explain what's wrong with the above.

~~~
icotyl
This is probably the clearest explanation, sorry you were downvoted.

~~~
Doubl
Thank you! It's nice to know someone else feels that way.

------
teej
I saw an interesting perspective on the Monty Hall problem from judegomila
earlier this month -

[https://twitter.com/judegomila/status/1183425802399969282](https://twitter.com/judegomila/status/1183425802399969282)

"Your brain will compute monty hall problem in fast casual causal mode AND/OR
compute slow mode formally for the exact answer. Probability predictions that
accurately map reduce to efficient intuitive representations in our brain
dictate direction in our evolutionary biology."

"[since] we don't solve the monty hall problem effectively with our intuition,
an evolving life form would be poor at resource collection from such problem
sets without knowledge of axiom representation and formal probability."

~~~
godson_drafty
"Sire. Shall we hunt in the valley, in the forest, or in the mountains?"

We shall hunt in the forest.

"Sire! Our scouts have just returned from the mountain. They report that the
herds are not there at the present time! Shall we try the valley, or continue
on to the forest?"

A) Stick to the plan, and hunt in the forest B) Hunt in the valley instead

~~~
greenshackle2
The Monty Hall paradox hinges on the fact that the host knows in advance where
the car is. If your scouts don't know where the herds are, but just picked an
area at random to recon, then it is a different problem, and switching makes
no difference.

~~~
kortilla
But it doesn’t depend on Monty knowing ahead of time. Monty can learn after
you pick, but before he reveals (to be sure he doesn’t reveal a car).

The scouts can depart after you’ve chosen the valley, find nothing in the
mountain, give that information to you, and you _should still switch_ , even
if the scouts didn’t confirm where they are.

Monty(or the scouts), knowing if your original pick was right has no impact.
The critical piece is that they are forced to reveal to you a bad choice. 2/3
times it’s the only bad choice to reveal because you’ve picked the other.

~~~
dllthomas
If the herds might have been in the valley and don't happen to be, you learn
something. Your odds rise from 1/3 to 1/2\. If the scouts _know where the
herds are_ and specifically checked the valley because the herds weren't
there, you learn more - your odds rise to 2/3\. But not as much as if they
just told you where the herds are, so we're back to it being an unlikely
scenario.

~~~
kortilla
Nope. Monty still reveals a door even if your planned pick was correct.

The entire gain in odds is predicated on Monty being forced to reveal a bad
choice _after_ you have chosen.

Monty knowing ahead of time has absolutely no impact on the outcome.

~~~
dllthomas
Knowing that Monty _could not have_ revealed the car (vs _happened not to_ )
has an impact.

If you are objecting to how I described that, in terms of who knows what when,
then you may be right and I probably could have been more precise.

But if you are disagreeing that scouts sent to investigate an area and return
an answer (which this time happens to be negative) differs materially from the
original question then you are wrong.

~~~
kortilla
> Knowing that Monty could not have revealed the car (vs happened not to) has
> an impact.

Nope, it does not. The fact that he revealed a dud is what matters. The notion
of him maybe revealing a car is senseless anyway because you would just then
pick the car or the game would be over.

~~~
dllthomas
> Nope, it does not. The fact that he revealed a dud is what matters.

That's simply wrong. This has been discussed countless times on this forum.
The most effective way to proceed is for you to spend 5 minutes writing the
simulation that you think will prove you right. When I was on the other side
of this, that experience is what convinced me.

> The notion of him maybe revealing a car is senseless anyway because you
> would just then pick the car or the game would be over.

That's not a very strong objection - game shows are weird sometimes and could
often apparently be easily improved.

But more importantly, it's an objection to the wrong thing. My point was that
the presented "find the herds" problem _differed from_ the game show. "That
would have been a bullshit game show" is just confused.

------
hanoz
It's bedtime here so I'm just going to chuck this in and run away: when you've
got _Monty Hall_ sussed, try _Tuesday 's Child_, i.e. "I have two children,
one of them is a boy born on a Tuesday, what is the probability I have two
boys?"

~~~
chowells
You're going to say it's 1/3\. I'm going to say that the problem is not
described clearly enough to assert that.

For instance, let's say you tell me you have two children. One was born on a
Tuesday, the other on a Wednesday. I ask about the one born on Tuesday, and
you tell me it's a boy. With the information I have available to me at this
point, I am correct to judge that the odds you have two boys are 1/2.

This story differs from the one you provided in ways that are insignificant
for story telling.

The way you tell the story depends on the reader picking up on a rather tiny
detail to reach the conclusion you intend. Specifically, the reader must come
to the conclusion that you are someone playing games with them who has
intentionally used very careful wording to only eliminate the girl/girl entry
from the outcome chart.

That's not the way people communicate naturally. If someone says "I have two
kids. One's a boy...", they aren't playing probability games, they're thinking
of a child they'd like to tell you a story about. In that case, when they have
a distinguished child in mind before making that statement, the odds are in
fact 1/2 that the other child is a boy and hence you have two boys. (Modulo
all the legitimate criticisms that can be raised that there are more possible
outcomes than boy/girl and that even between those two the odds aren't equal.
This is a simplification of reality for the sake of a puzzle.)

In the end, problems like this mostly hinge on poor communication that abuses
common patterns to convey less information than the reader interprets from
their experiences communicating in the real world. As such, they're really
rather boring.

~~~
hanoz
> You're going to say it's 1/3.

I most certainly am not. I'm going to say it's 13/27.

I agree the imprecise nature of the English language muddies the water
somewhat, but the question can be re-framed to remove this aspect and yet
still have people give the wrong answer, and as such is not just a "really
rather boring" case of "intentional" "poor communication", but offers an
interesting insight into the shortcomings of our intuition for probability.

So let's set aside the original problem and put this one instead (which may or
may not be equivalent to the original depending on your interpretation, but
let's take this in isolation anyway):

    
    
      Woman: How may children do you have?
      Man: Two.
      Woman: Is at least one of them a boy who was born on a Tuesday?
      Man: Yes.
    

What now is the probability the man has two boys?

The answer is in fact 13/27, because of the unlikely extra information
revealed by the Tuesday condition, which I think is surprising and
interesting.

------
Myrmornis
Instead of asking yourself "should I switch?", ask yourself "At the time of
making my first guess, would I have bet that it got the car (1/3) or a goat
(2/3)?"

Obviously the answer is that your first guess is more likely to have been a
goat. So if you had to bet, you'd have bet on the car being in one of the ones
you didn't choose.

After the host reveals a goat, it's no longer "one of the ones you didn't
choose", but rather " _the one_ you didn't choose" (since the host has taken
the other out of consideration). So the choice is easy now (as is the
probability calculation).

------
melonkidney
You probably picked a goat. When the host shows you where the other goat is,
you switch to the car. Job done.

~~~
_def
This is the best explanation I've ever read. Thank you!

------
jhncls
Something that really adds to the confusion, is that "in a not mathematically
perfect" world, the observer doesn't know whether the host will always open a
door. When the candidate chooses the winning door, the host could always open
another door, and when the candidate chooses a loosing door, only open another
door with a reduced probability. Life teaches people to always be suspicious
when free money is promised.

~~~
mrob
The "Monty from Hell" scenario (Monty only offers the choice if you picked the
car, and otherwise does nothing), which the original statement of the problem
allows, is IMO the most likely in real life because it would make for the most
entertaining television.

~~~
kangnkodos
You have real world knowledge of what makes entertaining TV, and perhaps how
this particular show actually operated. And you are correct. In the real
world, Monty Hall always revealed the worst of the two prizes that the
contestant did not select. In the real world, you can use common sense and
assume that Monty Hall will use that algorithm.

But in a mathematical probability problem, you can't do that. For example, if
there's a slightly smaller object and a slightly larger object concealed in
two different sized boxes, common sense says that a person would put the
larger object in the larger box. But if it's a abstract mathematical
probability problem, and you have no information on the algorithm used, then
you have to assume the concealer flipped a coin when they decided which box to
use.

If I'm the one setting up the big-small game, and I use the random algorithm,
the smaller object is going to be in the bigger box about half the time. If my
mom is the one setting up the game, the smaller object is going to be in the
smaller box every time.

In abstract mathematical probability problems, if you don't know what
algorithm was used, you aren't allowed to bring human behavior in, unless it's
stated in the problem.

~~~
JadeNB
> But in a mathematical probability problem, you can't do that. For example,
> if there's a slightly smaller object and a slightly larger object concealed
> in two different sized boxes, common sense says that a person would put the
> larger object in the larger box. But if it's a abstract mathematical
> probability problem, and you have no information on the algorithm used, then
> you have to assume the concealer flipped a coin when they decided which box
> to use.

I think that this misrepresents mathematical reasoning.

For solving a problem in a textbook (which a mathematician does intensively at
the beginning of his or her career, and seldom thereafter), you must assume
neither—that the larger object goes in the larger box, nor that the concealer
flipped a coin; you must use only the information contained in the problem.
(From this point of view, many problems in most probability textbooks are ill
posed, because they force you to make some assumption while solving them; and
it then becomes a mind-reading exercise of whether you can successfully make
the same assumptions as the poser, rather than a mathematical exercise.)

For a mathematician who actually wishes to use theoretical probability to do
something useful, then you must again make assumptions that allow you to
translate the messy real world into an idealised mathematical object; but to
say that you must, or must not, assume any _particular_ thing is
otiose—different assumptions will lead to different results, and the only
guide to whether you made the 'right' assumptions is whether or not the
idealised mathematical results you get match the real world (to within
whatever bounds of error are tolerable).

------
Syzygies
As a math professor recalling reactions back in 1990, there were various forms
of "Marilyn vos Savant" bias in play. A good cover for gender bias was the
absurdity of the world's highest IQ writing newspaper columns. There are
plenty of other examples of narrow spikes in particular forms of intelligence
having a crippling affect, while the most creative people I know work to
change the world with less spectacular but more broad-spectrum cognitive
advantages. Still, it's a bit of a leap to conclude she was actually wrong
about this problem.

~~~
kangnkodos
The key was... did Marilyn vos Savant state in her version of the problem that
the host always uses an algorithm to open a door with a goat? (The alternative
was that the host randomly picked a door, and this one time it happened to be
a goat.) I read some of her articles on this topic, and my opinion was that in
some of her articles, she did not state what the host's algorithm was.

So, yes, even a person with the world's highest IQ can make a mistake.

~~~
mundo
Her column was written for an audience that was familiar with the show. Monty
Hall had, at that point, been opening doors and revealing goats-but-not-cars
to contestants five days a week for more than twenty years.

~~~
junar
But as Monty Hall himself notes, the game show rules did not actually require
him to open a door.

> Hall clarified that as a game show host he did not have to follow the rules
> of the puzzle in the vos Savant column and did not always have to allow a
> person the opportunity to switch (e.g., he might open their door immediately
> if it was a losing door, might offer them money to not switch from a losing
> door to a winning door, or might only allow them the opportunity to switch
> if they had a winning door).

If the host uses an adversarial strategy, such as only offers a switch when
the remaining door is a goat, switching is not necessarily wise.

------
RedComet
It is famous for a reason. One should neither be shamed nor feel ashamed for
not having intuition for this.

------
jaimex2
The key to explaining the Monty Hall problem to someone is to reverse the
logic.

When you picked a door at the start you may have excluded the winning door
from the host. This has a 1 in 3 chance of happening.

So that means the host has a 2 in 3 chance of being left with the winning
door.

The host who has knowledge can't pick the winning door, so those 2 in 3 odds
are passed on to you if you switch.

------
zaro
I think this solution can be very counter intuitive or natural depending how
you picture the problem in your head.

If you are thinking about you playing the game then of course it is very
counter intuitive, because you are playing the game and you have only one
chance. Doesn't matter that by switching your chances rise to 2/3 because you
play the game only once, and in the end you either win or lose.

On the other hand if you think about it from a statistic/probabilistic point
of view then it's all natural that the chance is 2/3 but there is the notable
exception that you don't play in this case and you neither win or lose.

------
DonHopkins
Not to be confused with the Monty Haul Problem from D&D.

[https://dungeonsdragons.fandom.com/wiki/Monty_Haul](https://dungeonsdragons.fandom.com/wiki/Monty_Haul)

------
raffy
Initial pick: 1/3 chance of winner

Question: did you pick the correct door?

    
    
      - If correct, both remaining doors are losers, eg. {loser, loser}.   
      - Otherwise, the remaining doors are {loser, winner}.
        And the loser must be revealed according to the game rules.
    

If you don't change your answer, you're stuck with the initial chance: 1/3

If you change your answer: (1/3) * 0 + (2/3) * 1 = 2/3

    
    
      - (1/3) you initially picked correct, then change to a guaranteed loser.
      - (2/3) you initially picked incorrect, then change to a guaranteed winner.

------
lucas_membrane
However, the details of the problem matter more than you might think. On the
original TV program, there were sometimes two contestants who picked doors for
the grand prize. In such cases Monty would sometimes reveal a small prize
behind the door selected by one contestant, and then offer the other
contestant the option to switch. In that situation, the frequentist logic
works out exactly the opposite, and the logic that says 'switch is better'
when there is only one contestant says 'switch is worse.'

------
dukoolio
I’ve had some luck explaining it by taking advantage of the sensitivity to
fairness that people seem to have.

You are the contestant. You choose a door. Monty reveals a goat from another
door.

Now, instead of giving you a switching choice, Monty picks [insert person you
find annoying] from the crowd. Monty gives them the option of choosing your
door (duplicating any prize) or choosing the other door.

The few times I’ve tried, people have seen it as unfair. But they don’t
necessarily see the 2/3\. They just see someone getting 1/2 vs their 1/3.

------
GistNoesis
Here is the scam version to make money out of math people who know the Monty-
Hall problem.

You look at the 3 cards then put them on the table and propose a fair triple
or nothing, if they can find the red queen.

You make them point toward the card they chose.

If they point a wrong card you go toward the card and reveal -> you won.

If they have pointed the right card, you return one of the card they didn't
pick and ask them if they are sure or want to change. Because they think they
are facing a Monty Hall problem they will switch. -> you won again. :)

~~~
twiceaday
This would only work on people who were made aware of the problem but spent no
time understanding it and just memorized 'switch.'

Even if you didn't cheat them, this game has zero expectation value... I guess
if you are working with people who wouldn't know this you are fine.

~~~
GistNoesis
This is not cheating, you don't have to hide the fact that you know where the
good card is. It is entertainment teaching a valuable lesson.

It is, as I said before, a fair game so it has zero expectation value for both
players, better odds than a casino for the player and there is no reason for a
gambler to refuse this bet.

In expectation you won't lose money, but people who think they know better
will happily give you their money by making a bad move while being convinced
they are getting an edge.

Most math-type people don't know the hypothesis of the Monty-Hall Problem and
can't state it properly, but most have seen it played on TV-shows and know
they must switch.

In the heat of the moment, especially if you have played a few follow the red
queen games before, the brain is still in fast mode, and will jump to using
heuristics. Mind trick.

------
beefield
Monty Hall Problem is to me the single most bewildering and depressing thing
when I wonder how come mainstream/basic/MBA economics still keeps rational
agency as its central tenent.

I mean, if people can't sort _this_ out, how on earth someone that is not
either extremely stupid or extremely evil (or both) can seriously think that
people can figure out their optimal health insurance _even if_ the terms were
not incomprehensible legalese designed to trick you?

------
nkrisc
Here's how I finally understood it (I think):

Switching is only bad if you pick correctly the first time. Since you only
pick correctly 1/3 of the time, switching must be no worse or better 2/3 of
the time.

Since Monty will never open the door you picked nor the door with the prize,
the remaining door after he opens one is the prize if you initially picked
wrong, or a goat if you initially picked right. Since you initially pick right
only 1/3 of the time, you switch.

------
vladislav
It's simple: choosing again switches from a losing outcome to a winning
outcome and vice versa, thus the probability of winning by choosing again is
2/3.

~~~
Myrmornis
That's a clean explanation.

Maybe needs one introductory sentence: "Note that your first choice has
probability 2/3 of being a losing choice".

------
nstart
Since my brain could not accept this long ago, I wrote a tiny script to see
what the result would be if this was to be emulated. Works out perfectly!
(Might need to update script for python 3). It's weird. Still feels so
unintuitive. Which tells a lot about intuition I guess :)

[https://gist.github.com/kiriappeee/4316871](https://gist.github.com/kiriappeee/4316871)

------
peteforde
The one I have a hard time explaining to my friends is the Martian Potato
Paradox. It's hard because I actually cannot wrap my head around the truth,
even though I can explain why it's true.

[https://medium.com/i-math/matt-damon-s-martian-
potatoes-1bcd...](https://medium.com/i-math/matt-damon-s-martian-
potatoes-1bcde7c6f77d)

------
Hamuko
Did some academic rewrite the article or is there some explanation as to why
so many of the citations on that page are so un-Wikipeda-like?

~~~
stupidcar
[https://en.wikipedia.org/wiki/Wikipedia:Citing_sources#Citat...](https://en.wikipedia.org/wiki/Wikipedia:Citing_sources#Citation_style)

Basically, Wikipedia does not mandate a citation style, it just has to be
consistent throughout an article. And once set, another editor can't change it
simply for personal preference. Presumably the original author, or whichever
editor first added citations, was more familiar with this style, and so it has
stuck.

------
scoot
Here's a great explanation: [https://mathigon.org/course/probability/monty-
hall](https://mathigon.org/course/probability/monty-hall)

Had it in a browser reading list (that I'd forgotten was there), and came
across it by chance yesterday.

------
simplesleeper
I wrote a post that explains this as well, including a playable game with
cards (uses js)
[https://www.michalpaszkiewicz.co.uk/blog/montyhallexplained/...](https://www.michalpaszkiewicz.co.uk/blog/montyhallexplained/index.html)

------
cafard
There was a Tcl for Windows distribution that came with the simulation. A co-
worker ran it out to some number of tries, and was satisfied.

By the way, you can get old Monty Hall shows on satellite TV to this day (or
at least week)--a friend noticed it playing in a restaurant a couple of days
ago.

------
dang
Previous threads for the curious:
[https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...](https://hn.algolia.com/?dateRange=all&page=0&prefix=false&query=Monty%20Hall%20Problem%20comment)

------
mhb
A Bridge from Monty Hall to the Hot Hand:The Principle of Restricted Choice

[https://pubs.aeaweb.org/doi/pdfplus/10.1257/jep.33.3.144](https://pubs.aeaweb.org/doi/pdfplus/10.1257/jep.33.3.144)

------
Vaslo
Not sure why you picked the Wikipedia entry - about 1000 more clearer
explanations out there.

------
injb
Even though I think I understand the maths behind this, I'm one of those
people who has never been able to fully accept it.

When you pick a door, the odds are that the prize is behind one of the other 2
doors (2/3 chance). Therefore when all but one of the other doors are
eliminated, the remaining one is twice as likely to have the prize as the one
you picked. Is that it?

I think my problem comes down to trying to reason about a past event using
probability. I just can't get my head around that. Yes you can predict what is
likely to happen if you repeat the experiment multiple times, but you're not
being asked about that. You're being asked about this particular instance of
the experiment. How can you know that you're better off switching _this_ time?

------
timtas
I wrote a simulation that proves it pretty convincingly.

[http://rubyfiddle.com/riddles/7066f/4](http://rubyfiddle.com/riddles/7066f/4)

------
overdrivetg
A fun way to play with people who won't believe you is to say OK then, let's
play 10 games @ $10/game:

Get 3 cards and pick one to be the prize. Your mathematically-challenged
partner gets to put them down each time.

You pick a card, they remove one of the other cards and then you always
switch.

I made like $50 from someone this way one time =)

I'm not totally heartless - I took him to lunch on me afterwards to lessen the
blow ;-)

If you _really_ want to drill the point home, offer to keep playing more &
more batches of 10 until they've had enough...

------
knzhou
I'm gonna pile on the hundreds of comments already existing and say that the
_real_ key to the Monty Hall problem isn't about pure probability, it's about
human interaction. The answer can be either 1/2 or 1/3 depending on Monty's
motivations, his knowledge, and the way he decides what information to give
you.

This is true for Bayesian updating on _any_ information, and is a good lesson
for life in general.

~~~
smohare
The probability only changes if Monty can remove a prize door and does not
reveal a goat. Monty’s motivations and knowledge are completely irrelevant in
the context of the original problem, because his actions are effectively
forced (he reveals a goat).

If he can remove an unchosen prize door with some unknown probability p and
does not reveal the contents of the removed door, then the probability of
getting the prize by switching is simply (2/3)(1-p).

His motivations and knowledge just amounts to some distribution of p, but
that’s a rather different problem.

When the expected value of p is 1/2, then switching is equally likely to win
you the prize as your original choice, but the probability of winning is still
only 1/3 in either case. Only when p=1/4 will the remaining door contain the
prize with probability 1/2.

~~~
knzhou
This is exactly demonstrating my point. When smart people disagree with you
over the answer, it's not because they're misunderstanding the question --
it's because they are not aware that Monty _must_ reveal a goat, which is
almost never properly specified in the problem statement. It's usually stated
as "you pick a door and then Monty shows a goat" which _does_ have an
ambiguous answer.

------
johnthescott
change the number of doors to 100 and the reasoning is obvious.

------
rq1
Change the 3 doors to 10^6 doors and it becomes obvious.

~~~
Myrmornis
No, it becomes less obvious, since the host's help has less influence the more
doors there are.

EDIT: My reply is based on a misunderstanding. I was assuming that the host
continues to open one door, whereas parent intended that the host would open
all-but-one doors.

~~~
rq1
How eliminating 10^6-2 doors is less obvious?

~~~
_vertigo
Clearly you're assuming that with more than 3 doors, the host will open all
but one, and they're assuming that the host will continue to open just one
door.

~~~
rq1
The idea is to understand why it's better to change, not to introduce a new
problem.

