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The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser

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If you are old enough, you remember the sensation that the Rubik's Cube caused all the world over in 1980. No one is still alive that remembers the 1880 fad for the analogous two-dimensional "Fifteen Puzzle", which had fifteen numbered blocks within a four by four container and you were supposed to arrange them numerically. Mechanical puzzles can make storms like these, maybe because you can solve them over and over again, but it isn't often that word puzzles produce such fads. True, the Zebra Puzzle, a reasoning exercise consisting of fifteen seemingly unconnected statements that if regarded together the right way make a logical whole, was popular in 1962. Once you solved it, however, that was that. The Monty Hall Problem entered the public consciousness in 1990 and has been completely solved, but because the solution is so counterintuitive, it is still on the minds of many. One of those minds is that of Jason Rosenhouse, an associate professor of mathematics who has written _The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser_ (Oxford University Press). "My original idea for this book," he writes, was that an entire first course in probability could be based on nothing more than variations of the Monty Hall problem." Indeed, some of the chapters here are full-power mathematics, with unknowns x, y, and z, summation or conditional probability symbols, and complicated equations choked with parentheses within brackets, and more. Math phobics won't get far with such stuff, but there is enough other material here, along with different explanations of the basic puzzle, that will be of interest to anyone who likes recreational mathematics in even the slightest degree.

People feel strongly that the answer the mathematicians have worked out is wrong and cannot be made right. Here is the problem: You are Monty's contestant, and he presents you with three identical doors. One hides a car, which you want, but the other two doors hide goats, neither of which you want. You pick a door, but instead of opening it, Monty opens one of the other two doors. Monty knows, of course, where the car is and where the goats are, and he only opens a door that shows you a goat; in the case where you happened to pick the door hiding the car, he chooses one of the two remaining doors randomly. So then you have one door open with a goat, and two doors unopened, including the one you picked. Monty now says he will give you a choice: you can stick to the unopened door you originally picked, or you can switch to the other unopened door. So, do you stick or switch? It is obviously a fifty-fifty chance, and like so many obvious things, it is also wrong. Rosenhouse goes on to show several ways of calculating the problem, and he is good at explaining why you are twice as likely to win if you switch. Essentially, Monty is giving you extra information when he opens that door with a goat behind it. You had a one third chance of picking the door with the car to begin with, and if you have picked that door and switch, you lose. But you also had a two thirds chance of picking a goat to begin with, and (under the conditions of the problem), if you picked a goat and switch, you can only switch to the door hiding the car.

Don't worry if the summary in this review isn't convincing. Many who first saw the problem in a _Parade_ magazine article by Marilyn vos Savant in 1990 weren't convinced, either. Rosenhouse prints some of the responses to her article, many of them from mathematicians and many of them withering in their disapproval of her correct analysis that switching is the best policy by a factor of two. He is embarrassed by the vituperative nature of some of the professional voices in opposition. If this puzzle were not puzzling enough, Rosenhouse goes through many variables of the problem and its effects in different schools of thought (including quantum dynamics), because there is a huge amount that has been written about it. Rosenhouse says he could write a second book with material he has reluctantly left out of this one, and this one covers: what if there are four doors, what if there are n doors, what if there is another player playing against you, or what if Monty opens any of the three doors randomly. It covers the history of the problem and the similar problems that went before it, and it covers the psychological causes of sticking or switching, and studies that show how people tend to stick in all the cultures so far tested. Best of all, for this reader anyway, it made the previously counterintuitive strategy of switching feel a little more sensible.

The "Monty Hall Problem" by Jason Rosenhouse is currently the best coverage of this important problem.

He covers the version of the problem as it was made famous in Parade by vos Savant, and also it numerous variations and generalizations, its history, its occurrence in various fields (psychology, philosophy, quantum theory), and he gives a rather extensive bibliography which will be of great use to the serious student. The depth of coverage varies depending on the topic. For example, the classical analysis is satisfyingly extensive, while the fringe areas (quantum Monty Hall, for example) are just touched upon, and then references are given in the bibliography.

The chatty tone of the text is such that it probably should be categorized as a "mathematics for entertainment" book. And as such, Rosenhouse has allowed himself literary license that one might not normally expect in a math book. For example, we have to wait to until page 42 before "probability basics" are actually discussed . Douglas Adams allusions aside, it might have been better to have given at least the classical definition of probability somewhat earlier. The definition is further developed in pages 84 - 88 when he excellently discusses the classical, frequentist, and Bayesian concepts of probability. This section I consider one of the best in the book.

Rosenhouse states that the book should be within the reach of any undergraduate math major. This is probably overkill. If you know what a binomial coefficient is and what it is used for, know the classical definition of a probability in terms of a sample space, and know how to sum simple series, then you should have no difficulties.

The great Hungarian mathematician Erdos, as Rosenhouse and others note, was a "victim" of the Monty Hall problem. As Erdos did work in the field of combinatorial mathematics and probability, this is significant. However, it must be emphasized that Erdos never actually attempted to solve the problem -- which takes all of about 1 minute to do if one writes down the sample space -- and which would admittedly would have been less than trivial for Erdos...No, Erdos was a victim because his intuition refused to accept the result, until somebody did a computer simulation and verified it for him. Hopefully, with the influence of this book (and others like it), this type of problem will find its way into high school textbooks so that future students of probability will develop a proper intuition.

Since this is a book that stresses math enjoyment, as mentioned earlier, Rosenhouse is allowed considerable license. Still I will point out a few things that bothered me:

On page 2, we are told that when physicist Paul Newman is interrupted in the 1966 Hitchcock move "Torn Curtain" by an impatient East German physicist who finishes off Newman's equations on a chalkboard that in reality "We don't finish off each other's equations." Well, physicists actually have done this, and rather famously so. So, quoting Max Born with regard to Oppenheimer: "In my ordinary seminar on quantum mechanics, he used to interrupt the speaker, whoever it was, not excluding myself, and to step to the blackboard, taking the chalk, and declaring: `This can be done much better in the following manner...' As Oppenheimer was a celebrity scientist during his lifetime, one can speculate that the script writer was aware of this anecdote.

On page 11, the famous "problem of points" of Pascal and Fermat is discussed. So the problem is, Alistair and Bernard are flipping a coin. Heads gives a point to Alistair, tails to Bernard, and the first person to 10 wins. The score is Alistair 8, and 7 for Bernard. If the game is stopped now, how should any prize be split? Rosenhouse correctly states that game will end after no more than 4 tosses...but it is a little bit too much license for my taste to claim, without explanation, that there are 16 possible scenarios. Drawing the tree diagram, we see there are only 10 real possibilities -- although each path of the tree is not equally probable, since the path lengths are different. Continuing this tree analysis, calculating .5 to the power of the length of a path gives a given paths probability and then summing each path's probability gives 11/16 for P(Alistair wins) and 5/16 for P(Bernard wins), and this agrees with Rosenhouse's result... However, Rosenhouse should point out that Fermat artificially allowed the game to continue even after a player had already won -- which is why he gets 16 possible scenarios instead of 10. Of course, Fermat included fictitious results in his calculation so that the paths would have the same length, and so by symmetry, the same probabilities.


An educational and entertaining read. Recommended.

Believe it or not, this entire book is on the Monty Hall problem! The author, a mathematics professor, has analyzed this fascinating brain teaser from a variety of angles. After discussing the problem's history, he presents various attempts that have been made to understand it. The earlier attempts, including those by Marilyn vos Savant, tend to focus on logical arguments in order to arrive at the correct solution. But in order to solve the problem with mathematical rigour, the author uses some of the tools of his trade such as conditional probability and Bayes' Theorem. But that's not all. He also discusses a series of variants to the problem and proceeds to solve those as well. Finally, psychological and philosophical issues are also presented, partly in an attempt to understand why the human mind has been shown to have so much difficulty in solving this problem. The writing style is clear, friendly and authoritative, although some of the unfortunate editorial errors that the book contains may contribute towards slowing down a reader's attempts at following some of the author's arguments. Regarding accessibility, general readers can learn much from a good part of the main text because of the many clear explanations; however, several sections are fairly heavy with mathematics, a few of which can be rather challenging. Consequently, although anyone with an interest in this problem can benefit greatly from reading this book, math and science buffs are likely to glean the most out of it.

Jason Rosenhouse has written what is a witty and informative book on the remarkable Monty Hall problem. Never has such a simply stated problem caused so much anger, confusion and irritation but so little understanding. Anyone who knows the problem will have gone through all the phases so amusingly described by Rosenhouse - denial, anger, bargaining, depression and acceptance.

The problem is named "the Monty Hall problem" because of its similarity to scenarios on the game show Let's Make a Deal, but was brought to nationwide attention by her column in Parade magazine. Vos Savant answered arguing that the selection should be switched to door #2 because it has a 2/3 chance of success, while door #1 has just 1/3. This response provoked letters of thousands of readers, nearly all arguing doors #1 and #2 each have an equal chance of success. A follow-up column reaffirming her position served only to intensify the debate and soon became a feature article on the front page of The New York Times. Among the ranks of dissenting arguments were hundreds of academics and mathematicians.

Now the heat that that this problem has generated is itself an intriguing phenomenon. According to Wikipedia Vos Savant is apparently a relative of physicist Ernst Mach and wife to the inventor of the artificial heart (Jarvik). She was listed in the Guinness Book of Records at some stage as the person with the highest IQ. She is a long standing columnist for Parade magazine which, with all due respect to that august magazine, is not exactly something you'd associate serious mathematical discourse. Check it out for yourself: [...]With that background and being a woman I think some people in the mathematical community actually played the woman rather than the problem and ultimately paid the price of being proved wrong in a rather ignominious fashion. I bet the people concerned don't put it on their CVs.

Because this problem is so simple to pose it seems accessible to professionals and amateurs alike. An amateur does not attempt to venture a public opinion on the Riemann hypothesis but the Monty Hall problem is something the man in the street feels he can offer an opinion on. Suffice it to say that probability is hard, and it seems we may not be built to intuitively perform accurate calculations be they classical or Bayesian. When I first came across this problem probably over 10 years ago I solved it along the lines of page 54 of the book (an exhaustive enumeration of cases where the goats are distinguished ) which seemed to me the most basic way to do it given that many combinatorial problems can be solved by applying appropriate labeling. I was surprised by Rosenhouse's classification of this approach in "other approaches" as I thought it was the obvious way to do it. Just goes to show how wrong you can be.

The book does a good job of teasing out Bayesian theory and much more. Rosenhouse takes the basic Monty and like all good mathematicians he fiddles with the assumptions to develop more and more general problems. That type of approach is useful training for mathematically inclined people. On page 62 Rosenhouse does a basic set of calculations which he says can be "quickly" determined. He's right but I found it odd that he did not footnote the intermediate steps given the amount of time he spends on other steps which are no more and no less difficult. Given the sort of heat this problem has caused one would not want to lose someone for the want of two lines of explanation.

These are minor quibbles. The book is very good and the generalizations particularly interesting.
[...]

THE MONTY HALL PROBLEM: THE REMARKABLE STORY OF MATH'S MOST CONTENTIOUS BRAIN TEASER follows one of the most interesting mathematical brain teasers of modern times, and uses a surprising minimum of math concepts in the process. Any interested in puzzle challenges - which ranges from college-level libraries strong in math to general-interest holdings - will find this a lively review of basic math problems and philosophy alike.

Jason Rosenhouse's witty and most enlightening perspective into the otherwise dry field of mathematical probability will help even the novice (for example, me) get a better understanding on why so few cars were doled out on "Let's Make a Deal", and why so many goats were distributed by the devious Monty Hall. He knew what he was doing, and the poor studio audience didn't have a clue; neither did I.

Thanks to people like Marilyn Vos Savant and now Jason Rosenhouse, the masses have a slightly better understanding of the laws of probability; at least we've been presented with the evidence. You don't have to be a genius to understand the findings; but you do need to keep an open mind and at least realize you'd have a very good chance of going home with some stupid goat at the end of that show if you didn't take the chance to switch to a different door when Monty taunted you with that chance.

This is a wonderful book and will make anyone feel like a genius once they've perused it. Too bad Monty's not still wheeling and dealing. Now we're ready for him.

monty-hall is a great probability problem, this book does a good job at reviewing it, describing it and relating its history and other details

I applaud the support for Marilyn (page 29) against the attack on her by Morgan, Chaganty, Dahiya, and Doviak (MCDD) in 1991. As Marilyn stated in her reply, her original statement of the problem "was not intended to be subject to strong attempts at misinterpretation." I believe that that's just what MCDD did, that they went out of their way to misinterpret her statement of the problem, seeing it as another problem, and that they then called her solutions "false" because they did not apply to their own problem. It appalls me that the MCDD paper was published by The American Statistician.

This book agrees with MCDD that Marilyn shifted from one problem to another. In my reading, she did not shift. I see Marilyn's initial statement (page 23) as describing the same problem that she offered solutions for and that she later described, repeatedly, in other ways.

This is from her original statement of the problem: "You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat." The use of "say" in the appositives ("say number 1" and "say number 3") marks them as exemplary, not as providing essential or definitive data. The problem is defined by what is left when the exemplary appositives are omitted: "You pick a door, and the host, who knows what's behind the door, opens another door, which has a goat." This is an adequate statement of what Marilyn intended, and it took a deliberate distortion for MCDD to turn it into another problem by treating "number 1" and "number 3" as essential data that define the problem.

Regarding mathematics, I find no fault with MCDD.

This book correctly describes the difference between Marilyn's and MCDD's problems on page 29. Marilyn: "... after the contestant chooses a door, the host opens one of the goat-concealing doors." MCDD: "... the contestant always chooses door one and he host always opens door three ..."

The rest of the book ignores this distinction. It offers a series of solutions to "Version One" (pp 35-36), which is Marilyn's problem plus the irrelevant stipulation that the host "randomly chooses which door to open when he has more than one option." In Marilyn's problem, it doesn't matter how the host makes a choice when there are two doors he could open to reveal a goat. In MCDD's problem, it does matter.

The confusion begins on p 46 and continues throughout. The solution present on pp 46-49 is for Marilyn's problem. The stipulation about randomly choosing is used there, but it is neither essential nor relevant to the solution. The same is true for the "Monte Carlo" method in the following section: a randomly choosing host is used but it need not have been. The same thing, use of the same irrelevant stipulation, is found in the probability tree on p 53 and in the assignment of equal probabilities to the lines of Table 2.4, p 54. All of these treatments of Marilyn's problem are incorrect in the sense that they include an unnecessary stipulation in the statement of the problem and proceed to use that stipulation in describing the solutions.

On pp 73-74 Bayes' theorem is applied to "Version One," but this time the problem that is solved is MCDD's, not Marilyn's. The book is silent on the shift from one problem to another.

The reader can sort things out by asking, for every solution that is offered for "Version One," "Does it really matter whether or not the host chooses randomly when he has a choice of goats to show?" If it does not matter, then the solution is for Marilyn's problem. If it does matter, then the solution is for MCDD's problem.

Full disclosure up front: I have not read the book, but have a point to make about the way it is being presented here.

Amazon is in business to sell books (and other things) and I have been a satisfied customer for almost the entire life of the company, but there is a truth in advertising issue here:

You do not need to buy this book in order to understand
(a) whether you would be better off, in the three doors game it discusses, by "sticking with your first choice, or switching to the other remaining door" and
(b) why.

You can go, as I just did, to Wikipedia and get a plethora of short or shortish explanations. The approach I found easiest to grasp was in a footnoted article there.

Suppose there were 100 doors instead of 3. You'd then have a 1% chance of picking the right one initially. The host "Monty" (the real Monty is still alive, by the way, and I wonder what he thinks of all this!) then shows you 98 doors with no prize. Clearly, you have a very good chance of getting the prize if you then switch and pick the one other remaining door rather than sticking with your first choice. But, it is actually a 99 out of a 100 chance, not a 98 out of 99 chance. This is because originally there was a 99 out of a 100 chance of the prize being behind a door you didn't pick, and that chance wasn't altered by your pick or the host then opening and showing you all but one of the remaining doors (and having doors HE opens always have no prize, at least that is they way this "paradox" is usually presented).

The same logic applies when there are only three doors instead of a hundred. You have a 1 in 3 chance of picking the right one originally, which means there is a 2 in 3 chance of the prize being behind one of the two doors you didn't pick. After "Monty" then opens one of those two and shows you that there is no prize there, the chances of the remaining door having the prize are also 2 in 3. You are better off switching.

The odds of getting the prize by switching are not as good when there are a total of 3 doors rather than 100, but you still have a 2/3 chance of getting the prize by switching and only a 1/3 chance by sticking with your original choice.

It is a tease to imagine playing the game, and pondering why there is any paradox at all.
You might wonder, as I did: Isn't there a 50-50 chance on the last go-round, no matter what, so how can it matter whether you switch or stick? Then you can figure out why this is actually not correct; actually, given that you were more likely to be wrong on first round, you are better off on the second round betting that you WERE wrong originally by switching to the last of the remaining doors you did not pick initially.
Or you can get stumped, as I did, and look up the answer.
This is a tease.

Having to decide if NOT figuring out the paradox has become so frustrating that you would pay $17.96 plus postage to end the frustration, and whether you want to blow that much of your Christmas budget and get the frustration-releasing answer only after waiting another week or two later for the book to come, is more like torture than tease.

For all I know this may be a wonderful book. Some other reviewers seem quite happy with it. But if so, it is not wonderful because you spend the whole book having fun, ala Agatha Christie, trying to find the solution yourself to one of the all-time "most interesting mathematical brain teasers before it is revealed to you."

I cannot believe there is a book on this
"People feel strongly that the answer the mathematicians have worked out is wrong" (because it is)
On your first choice your probability of winning is Zero, because regardless of your choice Monty does not open your door. Monty by eliminating a wrong door has invalidated your first choice entirely for purely theatrical extension of the suspense.
Probability works when things are statistically random, when someone has special knowledge (knowing the empty door) it does not. You have a 50-50 chance, because that is the only random choice you are allowed to make.
If the question is wrong the answer wrong, it is not always about just checking the math.