
Monty Hall, Erdos, and Our Limited Minds - laacz
http://www.wired.com/2014/11/monty-hall-erdos-limited-minds/
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throwaway283719
The Monty Hall problem is actually a bit more subtle than this article makes
out. The exact wording of the problem can change the answer. For example, in
this version

> There are two doors with goats and one with a car. You choose one door from
> the three. The host selects one of the doors with a goat from the remaining
> two doors, and opens it. Should you switch doors if given the chance?

has a different answer than this one

> There are two doors with goats and one with a car. You choose one door from
> the three. The host chooses one of the remaining two doors at random and
> opens it, showing a goat. Should you switch doors if given the chance?

In the first example the hosts choice is either forced (if you chose a goat
initially) or doesn't make a difference (if you chose a car).

In the second example, the host _could have opened a door with a car_ because
he was choosing the doors at random. Given that he didn't, your estimate of
the probability that you have chosen the car should increase - in fact, it
should increase to 1/2, and switching doesn't make any difference.

To me, this is the really interesting thing about the Monty Hall problem - the
knowledge and intentions of the host are important in the correct solution to
the problem! So there are two ways to come up with the answer "it doesn't
matter if you switch" -

1\. You understand the problem in the first sense above, and your intuition
leads you to an incorrect answer.

2\. You understand the problem in the second sense above, and you are correct
that it doesn't matter whether you switch or not.

~~~
____a
My understanding is that in both situations as you described above you should
switch. You are still in a new conditional state. Telling you that the host
opened another door by chance and it is a goat removes the possibility that he
opened the door with the car behind it by chance. You have eliminated one of
the options and you have the 2/3 chance of winning by switching.

~~~
throwaway283719
No. In the first case, the fact that the host shows you a goat has _not_ put
you in any new state of the world (the host was going to show you a goat
regardless of what happened).

In the second case, you now know that you are not in one of the states of the
world where the host accidentally opened a door that had a car behind it. They
_could_ have opened a door with a car, but they didn't. That gives you some
information - it stands to reason that since they didn't open a door with a
car, it was probably a bit more likely that they would pick a door with a
goat, which makes it more likely that there are two goats than that there is
one car and one goat, which is equivalent to saying that it's a bit more
likely that you have already chosen the car.

~~~
____a
Okay, yes. Had to work it out on paper to convince myself that is works out to
1/2 in the second case. Thank you.

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oneeyedpigeon
My favourite bit of this article is right at the end:

"Pigeons repeatedly exposed to the problem show that they rapidly learn always
to switch, unlike humans (Herbranson and Schroeder, 2010)."

This just blows my mind and, as you might tell from my username, I'm a big fan
of the bird that some denigrate as 'rats with wings'; I prefer to think of
them as rat angels.

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simonh
The Monty Hall problem is a real classic, I remember puzzling over it for
hours and discussing it with family and friends for days after hearing of it.

It amazes me how many people believe that human beings are primarily logical
beings.

The best way I've come up to explain the Monty Hall problem is that in the
first place you have a 2/3 chance of choosing a Goat and in that case
switching guarantees you will win the car. Alternatively you have a 1/3 chance
you chose the car to start with and switching guarantees you will win a goat.
Therefore a strategy of always switching gives you a 2/3 chance of winning the
car.

~~~
oneeyedpigeon
Wow - I think that's the most straightforward explanation I've ever heard. I
think, to rephrase it, you're saying:

a) because a goat is revealed, swapping will always give you the opposite of
what you had

b) you're more likely to pick a goat in the first place, so you're more likely
to switch to a car

~~~
hamburglar
The notion of revealing the goat before you make your second choice is a bit
of a distraction. It becomes a lot more clear if you realize that in your
first choice, you're dividing the doors into two groups: the group you choose
has a 1/N chance of having a car in it, and the group you didn't choose has a
(N-1)/N chance of having a car in it. Now, imagine that without opening any
doors, Monty said: if you're willing to switch to the N-1 group, you win if
there's a car in any of them. Do you take it? Of course.

~~~
lukasb
Yeah this is it I think. If I can restate more simply for the 3-door case:

You pick. The doors are now in two groups: your door (1/3 chance) and the
other two doors (2/3 chance).

The host picks. The second group is now reduced to a single door, still with
2/3 chance.

Now you decide - stick with the door with 1/3 chance, or switch to the door
with 2/3 chance?

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Avernar
More intuitive explanation.

Let's play a similar game with three doors, two goats and a car. But you get
to pick one of two rules to play by:

A) You get to pick one door you want and get what's behind that door.

B) You get to pick one door you don't want and get the better prize behind the
other two doors.

Pretty much everyone is going to pick rule B. So after picking the door you
don't want I will open the other doors. If the car is behind one of those
doors I will always open that door last. Does the last sentence make you wish
you picked rule A to play by? Heck no. We already know that one of those two
will be a goat. The order I open the doors won't change anything.

Even if I let you retroactively switch to rule A after opening the first door?
This is what trips people up. They think they're getting new information and
that their original door choice may be incorrect.

But in this version of the game it's easy to see that we're not changing our
choice of door, we're changing from the better game rule B to the worse rule
A.

So the correct choice is to always stick with rule B.

Now we map the rules onto the real Monty Hall game. Rule A is pick the door
you want and stick with it. Rule B is pick the door you don't want and switch.

So the correct choice is to always switch.

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GregBuchholz
My criticism of the problem is that the presentation is usually poor, and it
isn't always noted that the host knows what door the prize is behind, and
always opens a goat door. I wonder what the responses would be like if it was
presented as examples, in the same manner as was presented to the pigeons.

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madsushi
The way that I've tried to explain/understand the Monty Hall problem is to
expand it to 100 doors.

You're allowed to pick 1 door out of 100. Your chance of being correct is 1%.
Then, the host gives you the option of swapping to the _entire other set_ of
99 doors. Of course, your chance of hitting the car would be 99%, so you swap.

Opening 98 doors with goats behind them is _expected_ , and so just because
the host chooses to open 98 goats in a row doesn't mean that the chance of the
last 2 doors (your final door out of the 99, and your original 1 door) both
have an equal chance. Your set of 99 doors was opened in a _specific order_
and that's what actually matters. The 99th door still has a 99% chance to
contain the car, because your set had a 99% chance to contain the car. The
order is everything!

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ajvdarlington
Man the number of people suggesting it matters whether the host knows or not
makes my head hurt. It doesn't. The problem is equivalent to "given that the
host reveals a goat and you swap doors, what is the probability you get the
prize". This will be 2/3\. Whether the host knows or not is irrelevant.

There are only 2 possible paths in the problem: Path 1: you originally picked
a goat (p = 2/3), now you swap and get the prize

Path 2: you orinally picked the prize (p = 1/3), now you swap and get nothing

This depends in no way on the host having knowledge.

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barbudorojo
If you imagine the same game but with a million doors and in the end there are
only two doors, then you realize that there is a strong asymmetry and that the
initial door is special. Going to the limit, with infinite doors to choose,
the best thing you can do is choose the initial door, all the other doors are
just to confuse you.

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mindvirus
Another way to think about it: Imagine instead of opening the door, Monty
points to a door and says that the prize is either behind this door, or the
one you chose, and the question is, who is right? This is equivalent, but
seems easier to think about. Also imagine this with 5 doors - then it seems
much more clear.

~~~
dllthomas
Could you elaborate? This seems less clear, to me.

~~~
mindvirus
The variant works like this: A player picks a door, and Monty picks a door,
and whoever gets the right door gets the prize. Monty knows what is behind
every door, so the prize is either behind the player's door or Monty's.

So now, suppose you get to bet on who will win.

Its clear that the chance of the player winning is 1/3\. Thus the chance of
Monty winning is 2/3\. So bet on Monty.

To change it back to the original problem, you are also the player in the
game. So now, do you bet on yourself or on Monty?

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jonsen
This thread needs an argument for the case where the host picks a door at
random.

The chance for the host getting the car is of course also 1/3\. Then the
complement 2/3 chance of the host getting a goat splits the car chance evenly
between the two remaining doors as 1/3 for each door.

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nnq
...am I the only person for whom the intuition about this problem has always
yelled: "neither! choose whether to keep the choice or switch it with 50%
probability - in other words, flip a coin." ?

~~~
dllthomas
That's a good way to make _sure_ your odds are 50/50...

