And mostly favorably, too. You might need a subscription to read the review, alas. The reviewer is Donald Granberg, a sociologist (now retired) at the University of Missouri. He published several papers on the MHP during the nineties.

I liked this part of the review:

The author does a masterful job of tracing the problem back to its origin.

And this part:

One difficulty with word problems is their ambiguity. Rosenhouse does a superb job of reducing, if not eliminating, this source of endless argumentation with his canonical version.

Not to mention this part:

The Monty Hall Problem is much more comprehensive and wide-ranging than the many articles on the subject that have dribbled out. Those, of necessity, are more sharply focused. Rosenhouse offers readers much to think about concerning the perplexing question of whether to stick or switch.

I also got a kick out of this:

The nonmathematician deserves to be warned that The Monty Hall Problem‘s contents are highly mathematical. What would one expect from a math whiz?

Math whiz? Goodness! Haven’t been called that since high school.

It’s interesting the difference in perspective between a mathematician and a sociologist on this issue. I tend to view the problem as an exercise in probability, whereas someone like Granberg views it as a problem in human decision making. I think that distinction makes itself felt in a number of places. Here’s one:

Of greatest relevance is the “three prisoners” problem first published by Martin Gardner in 1959. Three convicts, A, B, and C, are scheduled to be executed. The governor decides that one of them is to be pardoned and tells the warden his identity but asks the warden to keep it a secret. However, A prevails upon the warden to tell A the name of one of the other two who will be executed. The warden tells A that B is to be executed. A is encouraged, thinking that his chance of being pardoned has increased from 1/3 to 1/2. Is he justified in thinking that his chance of being pardoned had actually increased?

Rosenhouse claims that this “is the Monty Hall problem in all but name…. or at least of something formally equivalent to it.” This raises the question of what constitutes the essence of the dilemma. I contend that lies in Monty’s offering the contestant the opportunity to change a choice in light of new information. It is the second-stage decision to switch or stick that is so excruciating, engaging, and entertaining. If I’m right about this, primary credit for introducing the Monty Hall dilemma to academia goes to Steve Selvin and his two brief letters in The American Statistician

I’m not sure if that’s the sort of thing about which one can be right or wrong. The Three Prisoners Problem (TPP) and the MHP are mathematically identical. That the TPP is typically presented as a problem in probability while the MHP is typically presented as a problem in decision making is not relevant from a mathematical standpoint. But I can understand why a sociologist would have a different emphasis.

I would like to clarify one point:

And, in her initial response to her critics, she showed confusion over the concepts of odds and probability, writing, “The winning odds of 1/3 can’t go up to 1/2” where (judging from what she wrote elsewhere) she intended to say “odds of 1:2 can’t go up to 1:1.” For some reason, when Rosenhouse presents this section from her second column he corrects “odds” to “chances” and makes other small changes in the wording without marking them. Elsewhere he repeats part of the same section without altering the original wording. In her 1996 book (4), vos Savant corrected the errors herself.

I did not “correct” anything. I simply made a transcription error in one of the places I mentioned the vos Savant quote, and that is how “odds” got changed to “chances.” I would point out, though, that this small change had absolutely no relevance at all to any argument I was making.

A more serious error I made was in relying on vos Savant’s 1996 book for my quotes, rather than the original Parade magazine columns. It had not occurred to me that wordings would have been changed for the book from what was originally published. Live and learn. I would add, again, that absolutely nothing of relevance to the discussion was riding on these minor issues of phrasing.

At any rate, I’m glad Professor Granberg wrote such a generally positive review. I have already received one e-mail from a person telilng me he was going to buy the book based on the review. The first of many I hope!

Comments

  1. #1 Chris Bell
    October 9, 2009

    Jason,

    I was interested in buying the book. I have a Kindle and am interested in the Kindle version. I was wondering if there are illustrations or pictures in the book? If so, are they black and white?

  2. #2 Tom English
    October 12, 2009

    Predicting your fate and making up your mind just seem different, don’t they?

    This reminds me of people’s difficulty in equating “intelligence brings information into the world” and “love brings good to the world.” It’s just obvious that information is physical stuff and good is not. Information is processed by computing machines — right? I suppose there’s little hope of returning to the days when computers were data processors, and information was a matter of interpretation.

    Forgive my free association. Hope the book sells well.

  3. #3 Jason Rosenhouse
    October 12, 2009

    Chris –

    All of the illustrations are in black and white. My understanding is that Oxford University Press makes all of its books available in Kindle editions, but I have not specifically checked.

  4. #4 eric
    October 14, 2009

    Chris and Jason,
    Its out in Kindle now. I’ve just started so I can’t say how the equations in the later chapters are, but the “1/2″ symbols in the first chapter aren’t great. Legible, yes; as good as the regular print, no. FYI I’m reading it on 2nd-to-smallest font, so maybe increasing text size solves the problem (I haven’t checked).

    None of which is Jason’s fault. Its a great book so far! Though I was always partial to the “Caesar breathes out his last breath; what are the odds your next inhalation includes some of that air” example, and was sad to see it didn’t make your top-three list. But I quibble. :)

  5. #5 eric
    October 16, 2009

    So I’ve just finished Chapter 1, and one point you made Jason was that some of the counterintuitiveness of the solution results from the fact that people don’t understand the critical value of how Monty picks which door to open. To most people, explaining that he flips for it is just an extra sentence that adds nothing to their understanding of the problem.

    For me, I get a better handle on the value of this information when I think about it from Monty’s perspective. For sake of brevity, say that the door initially chosen by the contestant is always considered door “A.” Off we go:

    I, Monty, am forced to put my cadillac on the line in this game. I don’t want the mark to take it. Every time I’m forced to play one of three results occur:
    1. A has a goat. I’m forced by the rules of the game to open door B (the other goat). If the mark switches, he takes my cadillac.
    2. A has a goat. I’m forced by the rules of the game to open door C (the other goat). If the mark switches, he takes my cadillac.
    3. A has a cadillac. I flip a coin to decide which door to open. If the mark sticks, he takes my cadillac.

    The game, from Monty’s perspective, is incredibly simple: switchers win every time I’m forced to choose a specific door. Given that there is a direct, one-to-one correlation between “forced choice” and “switcher wins,” I really, really don’t want the mark to know in which situations I flip and in which situations I’m forced to make a choice. Because if they know that, they can count how often (out of three options) I’m forced to choose a specific door. And that will tell them that switching wins 2 out of every 3 times.

  6. #6 Kim Burke
    February 7, 2010

    Jason

    I love your book. I have been a big fan of Monty for many years.

    One question. I can’t understand Section 5.6. This is the variation where there are six doors (2 cars, 2 motorcycles and 2 washing machines). After we select our door Monty reveals a door at random, without regard for its contents. We then have to choose whether to stick or switch.

    You point out that when we initially choose our door we have a 1/3 chance each of winning a car (worth $20,000), a motorcycle (worth $10,000) or a washer (worth $300).

    Then you say “we know these probablilities do not change when Monty opens a door.” That’s the part I don’t understand.

    You say that the expected value of sticking to our original choice after Monty opens a door remains at 1/3*$20,000 + 1/3*$10,000 + 1/3*$300 = $10,100 (this is a corollary of the probabilities not changing).

    Then you suppose Monty opens a door randomly and reveals a car.

    You then calculate the expected value of switching as $7,625 (1/3*$5150 + 1/3*$7650 + 1/3*$10,075) and conclude its better to stick. (The 1/3s represent your probablilities that our initial choice was a car, motorcycle or washer respectively, and the $values the weighted average value of the remaining 4 doors on each assumption.)

    Why is it that randomly opening a door to reveal a car does not cause us to revise our probability estimates for each item to be behind our door? After Monty reveals a car, isn’t the chance that we have a car now 1/5 (with P(motorcycle) = P(washer) = 2/5)? (What if he opened 2 doors and revealed 2 cars?)

    If I use the revised probabilities I get the value $8,120 rather than $7,625 for switching. This is also the value of sticking based on the same probabilities.

    Intuitively this sounds right – if Monty just reveals a door at random its analagous to your Version 2 where there is nothing to be gained (or lost) by switching.

    Where am I going wrong?

  7. #7 Tony Morton
    November 15, 2010

    One more vote of thanks Jason for a great read.

    This point about Section 5.6 is subtle. Like Kim, I wasn’t sure about the logic of keeping the same probabilities after Monty opens a door. The problem actually occurs first in Section 5.4, and AFAICT goes back to the original paper by Georges and Craine that Jason has cited. As Jason’s Chapter 3 makes clear, if Monty chooses a door at random, then the probabilities for the remaining doors do change, unlike in the original ‘classical’ problem.

    In the case of Section 5.6 with the different-valued prizes, I think you *can* use the original prior probabilities to calculate expected values when you stick with your first choice (because Monty’s choice of door doesn’t actually affect the contents of your own), but you have to use the posterior probabilities for your initial choice to work out the expected value of switching. In other words I agree with Kim that $8,120 is correct for the expected value of switching, but I agree with Jason that $10,100 is the expected value of sticking.

    I’ve tried to analyse things in detail in my own rather lengthy review of Jason’s book, which is available here.

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