

Monty Hall problem - Maro
http://en.wikipedia.org/wiki/Monty_Hall_problem

======
PanMan
I always find this the most interesting part of it: "A restated version of
Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-
answer column of Parade in September 1990 (vos Savant 1990). Though vos Savant
gave the correct answer that switching would win two-thirds of the time, she
estimates the magazine received 10,000 letters including close to 1,000 signed
by PhDs, many on letterheads of mathematics and science departments, declaring
that her solution was wrong (Tierney 1991). " Especially the science PhD's (I
have seen quotes from some of the letters..).

~~~
tel
Science PhD's are on roughly the same ground as those poor MDs who didn't know
what a positive test result does not imply high chances of having a rare
disease.

Statistics can be non-intuitive because common sense only ever approximates it
at best. You've got to either always run the math or build some good
heuristics for when to expect that common sense just failed you.

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cynicalkane
Similar problems arise in contract bridge, where players have a name for
it--"restricted choice". I've actually always wondered if mathematicians have
a name for it. Anyone know?

The most common example is this: say you and your partner have nine cards
combined in a suit, and you're missing four cards: the queen, the jack, and
two unimportant low cards.

Bridge is played like hearts or spades, where one person leads a card and
everyone else has to follow suit. It's your lead. You want to find the queen
and the jack. If you guess right, you win the contract (= win a lot of
points). If you guess wrong, you lose the contract.

Your opponents want to hide the queen and the jack. Assume that your
opponents, if one had both missing high cards, are devious enough to randomly
play one or the other.

You play the ace, and one defender drops the queen. What are the odds they
have the jack? Hint: they're not 50/50.

~~~
eru
It's similar in Skat and Doppelkopf. (Though even more devilish in Doppelkopf
since you do not know where your partner is at the beginning of the game.)

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code0
There is a certain subtlety to this problem which many people overlook. It is
that whether the host <em>always</em> gives the option to switch. Only if the
host always gives you the option to switch, there is a benefit in switching.
If the host acts randomly(say deciding on a coin flip whether to give the
option to switch), then the benefit of switching is nullified. Also, if the
host is malicious (gives the option to switch only if you have the door with
the prize), then of course switching decreases your chances of winning. I
believe that the reason several "PhDs" were confused over the answer has got
to do with the fact that they might have assumed a host which acts randomly.

~~~
tel
Not quite true.

If the host gives you the option to switch based on a coin flip, that flip
will be independent of the state of the game and just average the overall
probabilities between a regular game and a switch-style game.

Now, allowing the host to act totally randomly will remove the advantage. In
particular, the host needs to randomly choose which door to open from all
three (including your own). If he finds the goat, you reset and play again
perhaps, if he doesn't find the goat then you've got a 50/50 chance between
the remaining doors. The probabilities are simple because each door was
treated exchangeably and _there could be no possible information flow that is
mutual to, entangled with, the location of the goat._

~~~
code0
As you said, "average the overall probabilities between a regular game and a
switch-style game", which is 1/3 for a game with no-switching and 2/3 for a
game with always-switching. Averaging gives you a probability of 1/2. So if
you switch, you win 50 % of the time. But so do you if you do not switch. So
there is no benefit in switching, which was basically my point.

EDIT: In both the versions of the game, I am assuming an omniscience host who
always opens the door with a goat behind, if he does open a door at all.

~~~
tel
You're absolutely right and my argument fails, but it doesn't invalidate the
point I wanted to get at. In the long run, your coin flipping method does
cause the game to behave as though there were only two doors, but in any given
game you'll know whether or not the host is going to choose a door.

That means that, conditional on a good flip, you may still be able to act to
your advantage since there's information transfer from the host to you.

------
sprachspiel
Another similar problem from
<http://www.inference.phy.cam.ac.uk/mackay/itila/>

You visit a family whose three children are all at the local school. You don't
know anything about the sexes of the children. While walking clumsily round
the home, you stumble through one of the three unlabeled bedroom doors that
you know belong, one each, to the three children, and find that the bedroom
contains girlie stuff in sufficient quantities to convince you that the child
who lives in that bedroom is a girl. Later, you sneak a look at a letter
addressed to the parents, which reads `From the Headmaster: we are sending
this letter to all parents who have male children at the school to inform them
about the following boyish matters'.

These two sources of evidence establish that at least one of the three
children is a girl, and that at least one of the children is a boy. What are
the probabilities that there are (a) two girls and one boy; (b) two boys and
one girl?

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Roridge
Solution: Host doesn't know what is behind the doors either. Contestant then
has a true 1/3 chance. QED

I guess that defeats the game though.

~~~
sprachspiel
This is actually an interesting variation of the problem. If a random door
opens but by chance happens to not have the prize, then the result is
different.

~~~
cynicalkane
But the GP is wrong. If a random door opens, then the odds are 1/2, just like
the intuitionistic answer predicts.

There's a 1/3 chance you picked right, a 1/3 chance you picked wrong and a
wrong door was revealed, and a 1/3 chance you picked wrong and the right door
was revealed. Throw the last third out of your sample space and you get that
the odds are 1/2.

This is what that forced Marilyn to admit the odds can be 1/2 on some readings
of the problem, which was ambiguously phrased the first time it was printed.
(But I think most people read the problem for the 1/3 solution and still came
up with 1/2.)

~~~
Splines
That doesn't make any sense. The intentions of the host prior to opening the
second door shouldn't have any impact on the odds of picking a winning door.

I could be wrong, but I think of it this way: You have a choice between
picking one door, or picking two doors. If you choose to pick two doors, the
host will happen to open one of them to reveal the non-prize. Irrespective of
the differences between the wording of my problem and the MH problem, how are
the actions different between the two?

Here's the flaw in your logic:

There's a 1/3 chance you picked right, and a 2/3 chance you picked wrong. In
the 2/3 chance you picked wrong, there is a 1/2 chance of the correct door
being revealed. No matter if the correct door is shown or not (in this case,
it isn't shown), it doesn't change your original odds.

~~~
cynicalkane
The intentions of the host pass you information that you can use to choose the
correct door. The information is this: "There's a 2/3 chance I was forced to
reveal this door; ergo, there's a 2/3 chance the other door is the correct
one."

If the host is choosing randomly, no information is passed.

With regards to your last paragraph, the odds of the host picking wrong are
either 1/2 or 0 depending on the intentions of the host. I'm a tiny bit
annoyed that you think you're pointing out a "flaw" in my logic when you
missed something so simple, but whatever.

If you don't like my explanation, direct your attention to the OP's link to
Wikipedia, where there are many more variations and explanations that you
might like.

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tome
A recent paper with a thorough analysis:

<http://arxiv.org/pdf/1002.0651v1>

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nazgulnarsil
I've heard secondhand of this being used as an interview question.

~~~
anatoly
Sounds like a really bad idea to me.

~~~
nazgulnarsil
why? for a job that involves work with probability modelling it would
certainly be relevant.

~~~
anatoly
1\. It's notorious for clashing with many people's intuitions and being very
confusing. Many smart people who are experts in probability theory answer it
incorrectly at first and then it takes them a while to convince themselves
they were wrong.

2\. On the other hand, if you read about it and understood it once, it's
trivial, and you can give a correct answer confidently while still being very
bad at probability in general. So you'd be selecting for knowing a piece of
trivia rather than any real knowledge or experience.

3\. Interviewers are not flawless mechanisms, they're people. They have biases
that are different to control for even when they're aware of them. You want to
steer clear of questions which a) make it easy for the interviewer to feel
smugly superior to the candidate; b) the answer to which looks very easy and
natural once you know it well, and it's difficult to remember how non-
intuitive it had been before. This question fails badly on both counts.

~~~
Maro
In an interview scenario, I'd give the problem and the correct solution, and
just ask the interviewee to tell me 'why?'.

------
ARR
Wasn't this in the movie 21?

