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Good morning and Hugggggs to all who want them.

plf515 here, your associate professor of mathematical geekology. Thanks to NCrissieB for another opportunity to address the student body of Blogistan Polytechnic University

In a previous diary, I talked about the Kaplans and Math Circle; if you are interested in being trained as a math circle teacher, they have some spots available in their summer institute, July 5-11; more info here.

Today, I discuss the infamous Monty Hall problem - and review a book.

This is, by FAR, the most contentious math problem ever.  It depends on precise wording, but even then, is counter-intuitive.  

Join me below the fold, as we attempt to unravel it.

Here is how I will proceed:

  1.  Statement of the problem and brief notes on history
  1.  The answer
  1.  Intuitive approaches
  1.  Monte Carlo/computer programming approaches
  1.  A formal proof
  1.  A book review (beyond what's in the diary already)

and

  1.  A special bonus question.

I'll mention right up front that the book was sent to me by Oxford University Press, it's called the Monty Hall Problem, the author is Jason Rosenhouse, it's due out in early July, and I liked it a lot.  If you like this diary, I think you'll like the book

STATEMENT OF THE PROBLEM
You are on a game show.  You are presented with three doors.  Behind one of them is a car, behind the other two are goats.  You get to choose a door.  But before you open it, the host (Monty Hall), who knows where the car is, opens one of the other doors.  He always opens a door with a goat.  If both of the unchosen doors have goats, he picks one at random.  The car was placed randomly, with 1/3 chance for each door.

You are then offered a chance to change doors, or stick with your choice.

What should you do?

BRIEF HISTORY
This problem has been around a while, but it got famous when Marilyn vos Savant (self billed as the person with the world's highest IQ) wrote about it in Parade magazine.  She got it right, although her explanation wasn't 100% on the money.  She then got tons of mail.  Angry letters saying how stupid she was, bemoaning the state of education, and so on.  Some came from mathematicians, and some of the most vituperative ones were from mathematicians.  They were all wrong.

THE SOLUTION
You should switch.  If you switch, you have 2/3 chance of winning the car.  If you stick, you have 1/3 chance.

This is counter-intuitive to nearly everyone.  Even one of the greatest mathematicians (Paul Erdos) got it wrong.  But, I guarantee you, the solution is correct.

INTUITIVE APPROACHES
First approach - a video

elropsych was nice enough to forward me two videos that he uses with his AP Psychology students.
as he puts it

This one is animated, kind of childlike, but does a good job of presenting a graphic organizer that breaks down the probabilities, and their reversal, inherent in the problem.
http://www.youtube.com/...

This is a montage of 2 popular treatments of the problem, one from the tv show Numb3rs, the other from the movie 21. Neither clip explains the problem as well as the cartoon above, but I use them as entry points to begin the discussion as many of my students have either seen the show or movie, or at least heard of them.
http://www.youtube.com/...

Then, I get them up and we act it out, using students in the room to fulfill all the moving parts of the problem. At this point, most seem to "get" the reversal of probabilities in switching after the first card is revealed.

Second approach - the 100 doors
This one apparently convinces a lot of people.  It didn't help me, but maybe it will help you.  

Suppose there are 100 doors, and Monty reveals 98 of them.  Would you  switch?  

Third approach - added information - where does the information go?
It's clear that when Monty opens a door, he gives you added information.  But what information?  In this, the classic version, he cannot tell you  anything about YOUR door, only the other two.  Your door stays at 1/3, but all of the other 2/3 is now in one door.  OTOH, in an alternative version, where Monty does NOT know where the car is, he MIGHT give you information about your door.  If he shows a car, it tells you your door has a goat.

MONTE CARLO / COMPUTER methods

Statisticians use the term "Monte Carlo" methods for simulations that use random numbers to generate answers.  Often, this is done on computer.  Many people have programmed the problem, and they all get the answer that swapping is good, and all are at close to 2/3 vs. 1/3 ... as close as can be expected.  People have also done this using playing cards and simulating by hand.  They take 2 red aces (goats) 1 black ace (car) and then have pairs of people be Monty and the contestant.  This also gives the answer that swapping is right.

FORMAL PROOF

Here I rely heavily on the book (I rely on the book everywhere, but more so here).

First, we need to define a sample space.  Whenever you do an "experiment" broadly defined, the sample space is everything that could happen.

if you flip one coin, the sample space is {H, T}
if you flip two coins the sample space is {HH, HT, TH, TT}
if you  flip one coin and roll one die, the sample space is
  {H1, H2, H3, H4, H5, H6
   T1, T2, T3, T4, T5, T6}

the various outcomes do not have to be equally weighted.  If you ask two kossacks who they voted for, and use Mc, Ob and Ot for McCain, Obama, Other, then the sample space is {McMc, McOb, McOt,
                                ObMc, ObOb, ObOt,
                                OtMc, OtOb, OtOt}

but I know which one I'd bet on happening! :-)

in the Monty Hall problem, three things happen: You choose a door, Monty opens a door, and the car is behind a door.  We can represent each of these with A, B, and C for the doors.  So, if you choose door A, Monty opens door B and the car is behind door C, we would write {A, B, C}

The sample space in the classic Monty Hall problem is
{ABC, ACB, ABA, ACA,
BAC, BCA, BAB, BCB,
CAB, CBA, CAC, CBC}

We can make things a little simpler by assuming you choose door A to start with.  Now the sample space is
{ABC, ACB, ABA, ACA}

note that some triples are impossible.  e.g AAB is impossible, because Monty never opens your door, and ABB is impossible because Monty never opens the door with the car.

Remember that the four outcomes do NOT need to be equally likely; in fact, here, they are not.

We are told that the car is equally likely for the car to be behind any door. The location of the car is the third item in the triple, so this means
P(ABC) = P(ACB) = P(ABA) + P(ACA)

each being 1/3.  Note that there are TWO ways for the car to be behind door A.  But the total probability for door A is 1/3.

We are also told that, when Monty can open either door, he chooses at random so:
P(ABA) = P(ACA)
and, since the total is 1/3, each of these is 1/6.

OK, so we have
P(ABC) = 1/3
P(ACB) = 1/3
P(ABA) = 1/6
P(ACA) = 1/6

when do you win by switching?  In the first two cases, total probability 2/3.  When do you win by sticking?  In the last two cases, total probability = 1/3.

Alternatively, we could look at the sample space after Monty opens a door.  Say he opens door B. Since he does this half the time, we halve the sample space, but we have to double the probabilities associated with the outcomes.  The sample space is now

{ABC, ABA}

P(ABC) = 1/3*2 = 2/3
P(ABA) = 1/6*2 = 1/3

you win more often by switching.

BOOK REVIEW

This is one diary and the book is 200 pages.  So, what else is in the book?  There's considerable detail about the origins of the problem and the huge outcry when vos Savant published the right answer.  There's extensive coverage of a lot of variations of the Monty Hall game (e.g. different probabilities, more doors etc).  Much more important, though, this is a (mostly successful) attempt to teach a course in probability theory through the use of the MH problem.  

Who should read the book?

I think it has a couple audiences.  First, if you are taking a formal probability course at university, this could be a good backup to your text.  OTOH, if you  are teaching such  a course, you could use this as a main text (I've never seen a probability text that is this much fun to read).  A course based on this book would cover a lot of the ground of a one-semester intro to probability course.

Probably few at daily Kos are in either of those groups.  Among the general population, I think this book could be read in two ways: First, you could read chapters 1, 2, 6, 7, and 8, and either skip 3, 4, and 5 or skim them.  (Chapter 4, in particular, will be heavy going).  Second, if you want to learn probability theory, you could read the whole book.  In this case, you'll want to read it more like a text book.

Speaking of chapters, here's the table of contents:

  1. Ancestral Monty
  1. Classical Monty
  1. Bayesian Monty
  1. Progressive Monty
  1. Miscellaneous Monty
  1. Cognitive Monty
  1. Philosophical Monty
  1. Final Monty

BONUS QUESTION
Suppose 500 mathematicians wrote to Marilyn vos Savant, and 450 of them said she was wrong.  What proportion of mathematicians in the US are mixed up about this?  (take the poll)

OK, I had bonus question for you, so I will try to give bonus answers, by throwing the floor open for statistics questions.  If I can answer briefly, I'll do so in comments; if it requires a long answer, I'll write a diary.  If I can't answer, I'll try to say why, and provide a reference.

UPDATE
BONUS from the comments, where Merrily1000 tells more about how vos Savant argued

I even wrote to her. She got the answer correct, but really botched the explanation. She didn't explain that the probabilities changed because opening the door was not a random event, so people were confused. That's why a lot of people thought the answer should be 50 percent probability on each unopened door after Monty opened one. It really would be 50 percent if the choice of what door to open was random, and therefore if the opened door sometimes disclosed the car. Monty knew where the car was, and he never opened that door, and that's the point.

If you have the car behind your door, which is one third of the time, it doesn't matter which door Monty opens, you still have it.

But two-thirds of the time you don't have the car, and Monty simply opens whichever of the other two doors doesn't have the car, so two-thirds of the time the unopened door holds the car.

After her column, when a lot of people wrote her, Marilyn wrote a second column supporting her first one, and then asked people to run the experiment, which she presented as what would be the true proof.

Originally posted to plf515 on Tue May 05, 2009 at 03:52 AM PDT.

Poll

see question in diary

11%23 votes
8%17 votes
3%8 votes
8%17 votes
68%143 votes

| 208 votes | Vote | Results

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