
Coding Horror: Monty Hall, Monty Fall, Monty Crawl - Anon84
http://www.codinghorror.com/blog/archives/001278.html
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biotech
"Some objected to the way I asked the question...."

Here Atwood links to a very insightful blog post by Paul Buchheit:
[http://paulbuchheit.blogspot.com/2009/01/question-is-
wrong.h...](http://paulbuchheit.blogspot.com/2009/01/question-is-wrong.html)

To continue the quote:

"...but it was a simple question asked in simple language."

Yes, a simple question, with two perfectly valid interpretations, as described
by Buchheit. When I first read the question, I interpreted it the "wrong" way.
The simple language that is used to ask the question is _too_ simple, to the
point of not being specific.

"I think what they're really objecting to is how unintuitive the answer is."

No sir. I am in fact objecting to the wording.

~~~
3pt14159
I interpreted it the "wrong" way too. I thought about it like this: Instead of
telling me that there is a girl or a boy tell me only that "1 + 1 = 2" and
then ask me what the odds of the mothers child are for being a boy. I'll say
50%. I don't care what the state of some other child, I just care what the
state of the unknown is. Now, if he had said, "you ask a woman that has a
random assortment of children if she has at least 1 girl and she answers
'yes', what are the odds that the other is a boy" THEN I would be fine with
the problem.

------
michael_dorfman
I'm always surprised by how the Monty Hall problem (and other similar
problems, like the Bellhop and the missing dollar) messes people up. It's
really terribly simple when you consider that Monty is _adding information_ by
acting non-randomly. The 3-door problem seems counter-intuitive to most
people, so it is easier to imagine 10 doors-- after you choose, Monty opens 8
other empty doors, leaving one. It should be evident in this case that your
choices are either a) to keep the original door, with a 1-in-10 chance of
winning, or b) take the other 9 doors (8 of which are open) with a 9-in-10
chance of winning. Clearly, the winning move is to switch, and there's nothing
counter-intuitive about it. People tend to let the details of the set-up get
in the way of their seeing the structure of the problem. It's a cute puzzle,
but it really shouldn't have confused Erdős.

~~~
lacker
I doubt Erdős was really confused. The Monty Hall problem is complicated
because usually the person explaining it _tries_ to make it complicated by
leaving out necessary information. Specifically, they often leave out that
Monty _always_ opens an empty door and his choice of which empty door to open
is done at random. Without those facts (which some but not all people take for
granted) there is not enough information to solve the puzzle.

~~~
callahad
Why is Monty's choice of which empty door being random important? Isn't the
fact that he always opens an empty door alone sufficient?

If you select the correct door, switching will yield an empty door.

If you select an empty door, switching will yield a prize.

You have a 2/3 chance of selecting an empty door, thus, 2/3 of the time,
switching will grant you the prize.

~~~
anatoly
The reason it's important is that the problem, as usually phrased, is a
conditional probability problem. Say you picked door number 1. Given that
Monty opened door number 3 and it's empty, what's the probability you'll win
by switching?

Now suppose that Monty will always open the empty door as advertised, but he
will also always prefer to open a door with the smaller number, if he has a
choice of two empty doors to open. If that is how Monty always acts, then
given that he's opened door number 3, your probability to win by switching is
100%, and not the usual 2/3. If he's opened door number 2, then your
probability to win by switching is 1/2, not the usual 2/3.

If Monty will always choose which of the two empty door to open by random,
then the probability to win by switching is indeed 2/3.

There is a way to phrase the problem differently, so that it becomes an
unconditional probability problem, and then it doesn't matter how Monty
chooses the empty door to open. One way to enforce this interpretation is to
phrase it as follows: Suppose some player always chooses to switch. Over many
attempts, what will be the approximate proportion of his wins? The answer will
be 2/3. But that's not how the problem is usually phrased.

~~~
callahad
Oh, wow, you're completely right. That connection eluded me; thank you!

------
vinutheraj
_When told of this, Paul Erdos, one of the leading mathematicians of the 20th
century, said, "That's impossible." Then, when presented with a formal
mathematical proof of the correct answer, he still didn't believe it and grew
angry. Only after a colleague arranged for a computer simulation in which
Erdos watched hundreds of trials that came out 2-to-1 in favor of switching
did Erdos concede that he was wrong._

Can somebody please verify this from source, seriously !!

~~~
michael_dorfman
Sure.

It's the main subject of Chapter 6 of "The Man Who Loved Numbers: The Story of
Paul Erdős and the Search for Mathematical Truth", by Paul Hoffman. He quotes
Andrew Vázsonyi (who introduced the problem to Erdős) and Ronald Graham (who
provided the proof that finally convinced Erdős).

------
vinutheraj
Hey doesn't the Monty Fall problem also have the same answer as the Monty Hall
problem - since the host falls and opens that door which always contains the
goat, its same as saying that he knows which door contains the goat and opens
that only.

Please correct me if I am wrong !

EDIT: Saw the soln in the pdf and not able to digest it though. Well if Erdos
wasn't able to see it...

~~~
harpastum
Well, I think most of the confusion comes from the way that Rosenthal ignores
a key part of the Monty Fall problem.

In the monty fall problem, you first choose a random door. Assuming a random
placement of the car and a random guess, you should have a 1/3 chance of
choosing the correct door on your first guess, and therefore a 1/3 chance of
winning _no matter what else happens_.

I believe what's really happening in this example is that the Rosenthal is
allowing for a possibility that he doesn't explicitly mention: the chance that
the host opens the wrong door, nullifying the game and ruining the show.

Since there is a 1/3 chance that you picked the right door, it's a probability
of 2/3 that it is in one of the other two doors. The host accidentally opens
one of the two remaining doors. Since the host _randomly_ chooses one of the
remaining doors, the chance that _he_ picks the door with the car is 1/3. If
he does that, the game is invalidated, so Rosenthal ignores that result.

Therefore, the chance that you picked the correct door is 1/3, and the chance
that the remaining door is the correct one is 1/3 (the remaining third is the
chance that the host ruined the game). Since those are equal probabilities,
you end up with an even chance whether or not you switch.

~~~
rottencupcakes
What's actually interesting is linking this result back to Deal or No Deal.

This implies that if you got down to two briefcases on the show, with both the
$1,000,000 and $1 left, it's a 50/50 chance of picking the million dollar case
whether you switch or not.

------
asorbus
I'm confused about why people argue that it is not necessarily 2/3.

I assume that the sample space is: [(b,b),(b,g),(g,b),(g,g)] where the four
possible birth sequences are [(younger child is boy, older child is boy),
(younger child is boy, older child is girl), (younger child is girl, older
child is boy), (younger child is girl, older child is girl)] and the sequences
are all equally likely.

So now if you ask, what is the probability that the person has a girl and a
boy given that at least one of them is a girl, you can discard (b,b) from your
sample space because you know that there is at least one girl.

Now your sample space is [(b,g),(g,b),(g,g)] and the odds that the person has
a girl and a boy are 2/3.

~~~
anatoly
The way you phrase it - "given that one of them is a girl" - the answer is
2/3.

The people who argue that it is not 2/3 and yet aren't confused are choosing a
different interpretation of the problem. When your acquaintance tells you "one
of my children is a girl", they interpret it not as pure information, so to
speak, but as an event to condition on.

It's reasonable to suppose, for example, that your acquaintance could have
said "one of my children is a girl" or "one of my children is a boy", and used
the actual knowledge of his children's gender to pick one of the two
possibilities to say (choosing randomly if it's a boy or a girl). In that
case, the question becomes: given that he/she said "... a girl" rather than
"... a boy", what is the probability that it's a girl and a boy?

In that case, the sample space becomes larger: to (b,b) etc. add also B or G,
which is whether the acquaintance said "one of my children is a boy" or "... a
girl". The possibilities are:

b,b,B: 1/4 (b,b,G is impossible) g,g,G: 1/4 (g,g,B is impossible) b,g,B: 1/8
b,g,G: 1/8 g,b,B: 1/8 g,b,G: 1/8

Now if you condition this sample space on the event "G", you rule out exactly
half the possibilities, and the answer is 1/2.

------
olavk
A reframing which makes intuitive sense (to me):

There are two bucket containing a number of balls. A third of the balls
contain iPhones inside, two-thirds contain only navel-fluff. The balls are
mixed at random. You have to chose a single ball from one of the buckets.

Obviously the chance of any ball you pick to contain an iPhone is 1/3
regardless of which bucket you choose from - even if the buckets does not
contain the same number of balls.

But when you decide on one bucket to pick from, the host takes the other
bucket, and removes half of the balls. However, he doesn't remove balls at
random - he primarily removes balls containing fluff.

After this you are allowed to reconsider your choice of bucket. I think it is
pretty obvious now that there is a better chance of finding an iPhone if you
chose from the other bucket, where the host have removed some of the fluff-
balls.

The critical point in the Monty Hall problem is that the host eliminates a bad
choice, but he always eliminates it from the set of doors you didn't chose.
However, this is a subtle detail in how the store is told. If for example the
host randomly choose between opening any door containing a goat - including
potentially the one the protagonist chose if he happens to choose a door with
a goat behind - then it doesn't improve the odds to choose a different door.

~~~
cruise02
Everyone agrees that your odds get better. Most people intuitively guess that
the odds go from 1/3 to 1/2. It's getting them to understand how you got to
2/3 that's the problem.

~~~
charltones
I found the easiest way was this: You pick door A, the host opens door B or C
and offers to let you switch to door C or B. He is effectively letting you
choose between either A alone (by sticking) or B & C together (by switching) -
i.e. by switching you open two doors instead of one, hence twice the
probability.

~~~
cruise02
I've found that the most enlightening explanation is to increase the number of
doors. What if the game is played with 100 doors, and the host opens 98 of
them after you've chosen yours? At the start you only have a 1% chance of
winning. Most people see right away that the host can't be increasing the odds
to just 50% by opening so many doors. They don't always see the logic
immediately, but at least they begin to question their intuition. Getting
people to admit that they might be wrong is often the first step toward
getting them to accept the correct answer. (That last sentence is a general
observation that applies to a much wider range of problems.)

------
tyn
From the Jeffrey Rosenthal paper that Jeff Atwood suggests
(<http://www.probability.ca/jeff/writing/montyfall.pdf>):

<Quote>

Nebulous Neighbours:

Your new neighbours have two children of unknown gender. From older to
younger, they are equally likely to be girl-girl, girl-boy, boy-girl, or boy-
boy. One day you catch a glimpse of a child through their window, and you see
that it is a girl. What is the probability that their other child is also a
girl?

Solution.

The probabilities that a glimpsed child will be a girl for each of the four
possi- bilities (girl-girl, girl-boy, boy-girl, and boy-boy) are respectively
1, 1/2, 1/2, and 0. Since the probabilities must add to 1, the probabilities
of these four possibilities are respectively 1/2, 1/4, 1/4, and 0. Hence, the
probability is 1/2 that the other child is also a girl.

</Quote>

Can someone explain how glimpsing to one child and seeing that is a girl is
different from hearing the parent saying that one child is a girl? How come
and getting the same information from your eyes instead of your ears changes
the probabilities? (1/2 for the other child to be a boy when seeing, 2/3 when
hearing).

~~~
anatoly
Seeing vs hearing is a red herring.

What really matters is whether information about one child being a girl fixes
a particular child. It doesn't matter how the child is fixed (a popular
variation is "I have two chilren and the older one is a girl"), but it does
matter that out of the two children, it is now known about a specific one of
them that it is a girl. In that case, out of 4 possibilities, 2 are ruled out,
and the answer is 1/2.

If all that is known is that one of the children is a girl, only one
possibility out of 4 is ruled out, and the answer is 2/3.

~~~
tyn
"If all that is known is that one of the children is a girl, only one
possibility out of 4 is ruled out, and the answer is 2/3."

Well, that's the case when you glimpse from the window. Probably I'm wrong but
I don't see why there is a fix, you still don't know if you see a G in GB or
in BG or in GG

------
edw519
"If all those Ph.D.s are wrong the country would be in serious trouble."

Yes. And yes.

------
TweedHeads
Monty Crawl: if you pick the 1st door and the host opens the 3rd door, you
have 100% chance to win if you switch.

~~~
georgeweinberg
Yes. But if picks the second door, it no longer matters whether you switch or
not.

Your final probability in Monty Crawl is still only winning 2/3, because you
can't do better than an always switch strategy, and you will pick the winning
door initially 1/3 of the time.

------
scscsc
Can anyone confirm this article by Atwood is worth looking at?

