Adventures in Ethics and Science

There’s a neat article [1] in the September-October 2008 issue of American Scientist (although sadly, this particular article seems not to be online) in which Brian Hayes discusses the Monty Hall problem and people’s strong resistance to the official solution to it.

Now, folks like Jason have discussed the actual puzzle about probabilities in great detail (on numerous occasions). It’s a cool problem, I believe the official solution, and I’m not personally inclined to raise skeptical doubts about it. What I really like about Hayes’s article is how he connects it to the larger ongoing discussion in which scientists engage:

The issue that concerns me here is not who is right and who is wrong about the odds of winning on Let’s Make a Deal. The issue is how I can persuade anyone that my answer — or any particular answer — is correct. I have been stewing about this for several weeks, frustrated that I am powerless to communicate what i take to be a simple truth. But I’ve finally decided that what the episode demonstrates is the vigorous good health of the scientific enterprise.

Making progress in the sciences requires that we reach agreement about answers to questions, and then move on. Endless debate (think of global warming) is fruitless debate. In the Monty Hall case, this social process has actually worked quite well. A consensus has indeed been reached; the mathematical community at large has made up its mind and considers the matter settled. But consensus is not the same as unanimity, and dissenters should not be stifled. The fact is, when it comes to matters like Monty Hall, I’m not sufficiently skeptical. I know what answer I’m supposed to get, and I allow that to bias my thinking. It should be welcome news that a few others are willing to think for themselves and challenge the received doctrine. Even though they’re wrong.

Very well put. One hopes, in the ongoing conversations, that all the parties prioritize the truth over their own biases and intuitions.

_______
[1] Brian Hayes, “Monty Hall Redux,” American Scientist (September-October 2008) Vol. 96, No. 5, 434-435.

Comments

  1. #1 RM
    September 18, 2008

    I think that one of the problems with the Monty Hall problem (and other such paradoxes) is the way they are explained. All to often I’ve seen explanations which belittle and dismiss the countering viewpoint. People doing the explaining don’t seem to realize that one of the main problems is differences in viewpoint, and that, as well as being correct, it matters that you can convince the other person you are correct. No matter how logical and clearly laid out your explanation is, you’re not going to convince them simply by bludgeoning them with the “correct” explanation. The people who stick to the wrong answer usually have some fundamental difference in the way they are interpreting the problem. (An axiom error rather than a logic error.) This axiom issue needs to be fixed before they’ll believe you. Ignoring the difference in the two axiom sets, or just telling them “you’re wrong” on the axiom won’t work. In the first case, even if you lay out your argument clearly and the person agrees that your argument is reasonable, you still haven’t disproved their original logic, so they’re left with two conflicting answers. In the case of a tie, status quo wins, so they stick with their old position. Just telling them “you’re wrong” isn’t going to work either – they’re not going to believe you (if it’s just your word against theirs, status quo wins again). In order for the explanation to work, they have to be guided to discard the incorrect axiom themselves. (Potential ways are showing the axiom intrinsically leads to a contradiction, or “simplifying” the problem to the point where the axiom is obviously wrong. – One I like for Monty Hall is increasing the number of doors to 1 prize and 999 goats, with Monty opening 998 doors after your pick.)

    This was recently brought home to me after perusing two “airplane on a treadmill” explanations. I thought http://www.airplaneonatreadmill.com/ was weak as a convincing argument, not because he’s incorrect, but becuse he’s condescending and ignores/casually dismisses the bone of contention the “no fly”s have. I got a “I’m right. And you’re just plain wrong.” vibe from it – as if just repeating assertions improves their veracity. In contrast I thought http://blag.xkcd.com/2008/09/09/the-goddamn-airplane-on-the-goddamn-treadmill/ was well written and likely to be convincing – mainly because he comes off as understanding of the opponent’s viewpoint, and takes time to carefully and sympathetically unravel the confusion.

    As a benefit, working through the other side sympathetically helps you to examine the biases in your own axiom set.

  2. #2 Kenny Easwaran
    September 18, 2008

    I think the bigger problem with the Monty Hall problem is that the correct answer is so sensitive to the wording of the question. If Monty Hall doesn’t specifically aim for a door without the prize, but just happens to hit one without the prize, then there’s no gain from switching – the fact that he hit one without the prize is some slight evidence that you were already right to begin with. Another slight misreading of the problem that makes staying be the right answer is that people might just be conditionalizing on the fact that a certain door has no prize, rather than the fact that Monty picked that door because it had no prize – so they’re missing one slight piece of evidence that is relevant.

    But really, the issue is that Monty Hall isn’t a mathematical problem at all – it’s a problem in formal epistemology or decision theory. Maybe some people are trying to debate the correctness of the mathematics involved in calculating the probabilities, but I think the bigger worry tends to be whether we’ve got the right mathematical representation of the situation to begin with. And of course that’s a very common problem in science. But it suggests that consensus on the mathematics isn’t what’s relevant.

  3. #3 MartinB
    September 18, 2008

    I agree with Kenny on the wording of the problem.
    In most versions I have read it was not made explicit that the answer depends on Monty *always* opening one door. If Monty adapts his probability of opening a door to whether your initial choice was correct or not, you are no wiser at all.

  4. #4 Jeff Chamberlain
    September 18, 2008

    I’ve not read the whole article, but the excerpt does not seem quite correct. It’s relatively easy to convince most people of the correct solution to the Monty Hall problem. All you have to do is count. For example, run a computer simulation of switching vs. staying over a large number of tries, and the benefits of switching become apparent. It’s not, therefore, a matter of “persuading” someone that the correct solution is correct. It is, rather, a matter of “explaining why” that solution is correct.

    I rather like RM’s description of an “axiom error.” However, once again I suggest that the most interesting problem is not “convincing” people about what the correct answer is, but effectively “explaining why” that answer is correct. Most people (not all) will “believe” that switching is the way to go once they “run the numbers,” even if they still don’t understand why.

    The difference is between Monty Hall “deniers” (who don’t thing switching is beneficial) and those who accept the correct solution (and even understand at least some of the mathematics of it), but nonetheless still don’t “get it” (I’ll switch, I promise, but I still can’t quite “get my head around” why I should switch, beyond “I’ll have a better chance of winning if I do”).

    None of this is to challenge the underlying point: That effectively explaining the MHP (or other scientific/mathematical things) can be very difficult (when difficulty and effectiveness are measured by the likelihood of comprehension by the audience).

  5. #5 Tony Jeremiah
    September 18, 2008

    But really, the issue is that Monty Hall isn’t a mathematical problem at all – it’s a problem in formal epistemology or decision theory.

    I’d have to agree. In psychology (and undoubtedly other scientific enterprises that employ the scientific method), intro students are taught about the importance of operationally defining vague concepts such as intelligence, personality, and love. How one operationally defines such terms allows other scientists to scrutinize the validity of the subsequent empirical observations. For example, intelligence could be empirically observed as scores on a standard IQ test, grade point average, peer recognition as defined by number of academic accolades won, etc.

    In the instance of the Monty Hall problem, the issue seems to be how one operationally defines the Monty Hall problem. Via illustration of the most mathematically challenged presentation of the Monty Hall problem possible (i.e., no math involved–just a brute application of a decision making matrix), it does seem that the problem is epistemological.

    If we define the Monty Hall problem like this:

    you choose a door, then Monty Hall asks if you would like to switch to a different door (without revealing what is behind any of the doors). You have the option of keeping your original choice or switching to one of the two remaining doors:

    then switching is advantageous according to the decision matrix below. There are three possible situations where a prize can be, and outcome (Win or loss) depending on whether a person decides to stay or switch from a particular door.

    1. (P) (-) (-)

    Pick(1),Stay(1),Switch(-),Outcome(W)
    Pick(1),Stay(-),Switch(2),Outcome(L)
    Pick(1),Stay(-),Switch(3),Outcome(L)

    Pick(2),Stay(-),Switch(1),Outcome(W)
    Pick(2),Stay(2),Switch(-),Outcome(L)
    Pick(2),Stay(-),Switch(3),Outcome(L)

    Pick(3),Stay(-),Switch(1),Outcome(W)
    Pick(3),Stay(-),Switch(2),Outcome(L)
    Pick(3),Stay(3),Switch(-),Outcome(L)

    2. (-) (P) (-)

    Pick(1),Stay(1),Switch(-),Outcome(L)
    Pick(1),Stay(-),Switch(2),Outcome(W)
    Pick(1),Stay(-),Switch(3),Outcome(L)

    Pick(2),Stay(-),Switch(1),Outcome(L)
    Pick(2),Stay(2),Switch(-),Outcome(W)
    Pick(2),Stay(-),Switch(3),Outcome(L)

    Pick(3),Stay(-),Switch(1),Outcome(L)
    Pick(3),Stay(-),Switch(2),Outcome(W)
    Pick(3),Stay(3),Switch(-),Outcome(L)

    3. (-) (-) (P)

    Pick(1),Stay(1),Switch(-),Outcome(L)
    Pick(1),Stay(-),Switch(2),Outcome(L)
    Pick(1),Stay(-),Switch(3),Outcome(W)

    Pick(2),Stay(-),Switch(1),Outcome(L)
    Pick(2),Stay(2),Switch(-),Outcome(L)
    Pick(2),Stay(-),Switch(3),Outcome(W)

    Pick(3),Stay(-),Switch(1),Outcome(L)
    Pick(3),Stay(-),Switch(2),Outcome(L)
    Pick(3),Stay(3),Switch(-),Outcome(W)

    Total Wins: 9
    Wins due to switching: 6
    Wins due to staying: 3

    If we define it as it is written on Jason’s site:

    You choose a door, then Monty Hall (who knows where the prize is) opens one of the other doors, revealing that the prize is not there. Now you have the option of keeping your original choice or switching to the third door.

    it doesn’t appear to work out that switching or staying makes a difference. Monty Hall knowing where the prize is (which indicates that he will not reveal the door with the prize), changes the decision-making matrix:

    1. (P) (-) (-)

    Pick(1), Monty shows (2),Stay(1),Outcome(W)
    Pick(1), Monty shows(3), Stay(1),Outcome(W)
    Pick(1), Monty shows (2), Switch(3),Outcome(L)
    Pick(1), Monty shows (3), Switch(2),Outcome(L)

    Pick(2), Monty shows(3), Stay(2), Outcome(L)
    Pick(2), Monty shows(3),Switch(1),Outcome(W)

    Pick(3), Monty shows(2), Stay(3), Outcome(L)
    Pick(3), Monty shows(2), Switch(1), Outcome(W)

    2. (-) (P) (-)

    Pick(2), Monty shows (1),Stay(2),Outcome(W)
    Pick(2), Monty shows(3), Stay(2),Outcome(W)
    Pick(2), Monty shows (1), Switch(3),Outcome(L)
    Pick(2), Monty shows (3), Switch(1),Outcome(L)

    Pick(1), Monty shows(3), Stay(1), Outcome(L)
    Pick(1), Monty shows(3),Switch(2),Outcome(W)

    Pick(3), Monty shows(1), Stay(3), Outcome(L)
    Pick(3), Monty shows(1), Switch(2), Outcome(W)

    3. (-) (-) (P)

    Pick(3), Monty shows (2),Stay(3),Outcome(W)
    Pick(3), Monty shows(1), Stay(3),Outcome(W)
    Pick(3), Monty shows (2), Switch(1),Outcome(L)
    Pick(3), Monty shows (1), Switch(2),Outcome(L)

    Pick(1), Monty shows(2), Stay(1), Outcome(L)
    Pick(1), Monty shows(2),Switch(3),Outcome(W)

    Pick(2), Monty shows(1), Stay(2), Outcome(L)
    Pick(2), Monty shows(1), Switch(3), Outcome(W)

    Total Wins: 12
    Wins due to switching: 6
    Wins due to staying: 6

  6. #6 MartinB
    September 18, 2008

    Tony, I think you are miscounting choices here.
    Consider your first case (P)(-)(-)
    The fact that Monty has more options when you choose the winning door does not mean that you will do so in half of the cases, but this is what your numbering implies.

    Call the door with the price behind it door A – you have a 1/3 chance of chosing it, not, as you get when you count up all your numbers above, a 1/2 chance.

  7. #7 Jim Thomerson
    September 18, 2008

    Tony, I’m a little confused on your first example where you find six wins for switching, vs three wins for staying. Seems to me that you are counting both switching possibilities, but, in fact the contestant has only one switching opportunity.

    The way I have understood the problem is that the contestant makes a choice of three doors, ie 1/3 chance of success. The contestant is then shown one of the non-picked doors is not good, and given the opportunity to switch. The odds are now 1/2. I think this is a new problem, thus the contestant should make a new pick, ignoring the previous pick.

  8. #8 Sven DiMilo
    September 18, 2008

    You (the contestant) make your choice. Then Monty opens a goat-door. Now, there are two possibilities: your choice is correct (the car) or incorrect (the other goat). If you were correct in the first place, then switching would always be a mistake because you’d get the other goat. However, if your initial choice was wrong, then Monty has just shown you the other goat, so switching pays off every time.
    Of course, you don’t know whether your initial choice is correct or not, but you know the odds: one in three. Therefore it will pay to switch 67% of the time.

  9. #9 Jason Rosenhouse
    September 18, 2008

    Hayes’ articles is available online here.

    I blogged about this article a while back,
    in this post.

    Janet –

    I’m afraid I have to disagree with your opinion that Hayes’s paragraph was well put. There is no received doctrine here. There is not a mainstream consensus challenged by a handful of plucky dissenters. There are only people who understand the problem and people who don’t. It is the great virtue of math problems over those of other disciplines that math problems can be solved definitively. Obviously I don’t think people should be stifled for expressing their opinions, but I do think people who deny the virtues of switching given the usual assumptions in the problem should be treated (gently) as people who are confused.

    RM –

    The confusion over the Monty Hall problem is not primarily about different starting assumptions. Psychologists and cognitive scientists have done a fair amount of research on precisely this point. It really is logic errors, as opposed to axiom errors (to use your terminology.) More precisely, it really is that people just have a hard time thinking clearly about probability. Most people, on hearing the problem for the first time, take the view that when Monty opens an empty door the only thing you learn is that that particular door is empty. In reality you learn that Monty, who makes his decisions according to strict rules, chose to open that particular door. This is a subtle point, and one that is easily overlooked.

    Concerning the rest of your comment, I can certainly understand your frustration with people who get very dogmatic in “explaining” things like the Monty Hall problem or the Airplane on a Treadmill problem. But I can also see things from the other side.

    In the final analysis these are not issues over which reasonable people can disagree. If there is some ambiguity in the problem statement, then people should be able to come to some agreement about precisely what needs to be clarified. But try having a reasonable debate with someone who claims that 2+2=5, and watch how quickly you lose control of your frustration. Arguing with someone who steadfastly denies that under the usual assumptions you should switch is a similarly frustrating experience.

    Kenny –

    You’re right that the Monty Hall problem is very sensitive to initial assumptions. I would add that most statements of the problem appearing outside of textbooks on probability are genuinely vague. But, again, that’s not what’s causing most of the confusion. For example, when the problem blew up in Marilyn vos Savant’s column, her critics were not challenging her for leaving crucial assumptions unstated. They were making the same assumptions she was, but were figuring their probabilities incorrectly.

    The Monty Hall problem can be viewed either as an exercise in pure mathematics or as an exercise in formal decision theory. In the pure math version all of the essential assumptions are (or at least ought to be) stated precisely and the only issue is whether you can apply probability theory (especially Bayes’ Theorem) correctly. This version causes ample confusion, for the reasons I have mentioned.

    Alternatively, you can imagine that you are actually on stage playing the game, and do not have enough information to do a proper probability calculation about what to do. That’s another interesting problem, but it’s not the one people usually have in mind in describing the Monty Hall problem as controversial.

    Jeff Chamberlain –

    I think you’re exactly right. A computer simulation very quickly convinces most people of the virtues of switching. But it does nothing to resolve the counter-intuitiion that it shouldn’t matter if you switch. Paul Erdos, the great twentieth century mathematician, did not beileve the correct answer. Upon being shown a computer simulation he became even more frustrated. On the one hand, he could not deny the results, but on the other he still did not understand the mathematics underlying them.

    Tony Jeremiah –

    Can you show me where I defined the problem in the way you described? It’s possible that I flubbed it somewhere, but that’s not the way I usually put thing. Viewed as a problem in pure mathematics, the key assumptions are these:

    1. Monty always opens an empty door.
    2. He never opens the door you initially chose.
    3. In those cases where your initial choice conceals the prize, Monty chooses randomly from among the other two doors.

    It’s that third assumption that frequently gets left out of casual statements of the problem (but which gets snuck back in during the solution). If you grant those assumptions then it is mathematically certain that switching is the way to go. If you start messing with those assumptions, then all bets are off.

    There’s much more to say, of course, but I think this comment has gone on long enough. I can’t resist mentioning, however, that I happen to know that sometime in the next few months Oxford University Press will be publishing a book on the Monty Hall problem that discusses all of these points in great (some might say nauseating) detail. The author? Gosh, I can’t seem to remember just now…

  10. #10 David DiBattista
    September 18, 2008

    I agree that a clear statement of the problem is essential. I have created a learning object for the Monty Hall Dilemma that I use in teaching an undergraduate statistics course, and in creating this object, I made very sure to state the problem very clearly. Anyone interested in viewing the learning object can go to my web page (www.psyc.brocku.ca/people/index.htm)and click on the link “Monty Hall Dilemma Learning Object.”

  11. #11 Tony Jeremiah
    September 18, 2008

    MartinB and Jim:

    My interpretation of the Monty Hall problem is based on discrete event probability rather than any one individual’s probability. That is, the statement that switching is better than staying makes more sense if one thinks about them in terms of adding together all possible discrete events that lead to a win (and a loss). So the scenarios above are based on the perspective of one person playing the same game multiple times (with each game representing one possible discrete event), or, different people playing one game each (with each player playing one of the possible discrete events assuming no psychological effects such as having a preference for a particular number and/or the social psychological literature that shows that people tend to stick with a decision after making them).

    The statement that any individual person has a better chance of winning by switching during any particular game (which seems to be the primary interpretation) does not make sense. For example, if the prize is in fact behind door 1 [represented as (P),(-),
    (-)] for a particular game, it would not be advantageous for the player to switch to either door 2 or 3 if their initial choice is door 1. It does make sense if one considers all possible discrete events associated with the prize being behind door 1 and the player plays multiple times.

    So what I’ve written above are two wordings of the Monty Hall problem, and all of the discrete events associated with each. As examples:

    1.(P),(-),(-) = The prize is behind door 1 and not doors 2 and 3.

    Possible discrete event associated with (P), (-), (-) (when Monty reveals what is behind the door the person ultimately decides to go with):

    Pick(1),Stay(1),Switch(-),Outcome(W) = The person
    initially picks door 1, decides to stay with 1 (after Monty Hall asks if s/he wants to switch), and decides not to switch. The outcome is that this person will win (W) the prize when Monty opens door 1.

    Possible discrete event associated with (P), (-), (-) (when Monty picks a non-prize door prior to the person making their final decision):

    Pick(1), Monty shows (2),Stay(1),Outcome(W) = Person initially chooses door 1, Monty shows door 2 (which does not contain a prize), person decides to stay with door 1 (which contains the prize), person Wins (W) because that is the location of the prize.

    Each situation represents a slightly different interpretation of the Monty Hall problem with the total wins due to staying or switching being the number of discrete events whereby switching results in a win, and staying results in a win.

  12. #12 Tony Jeremiah
    September 18, 2008

    Can you show me where I defined the problem in the way you described? It’s possible that I flubbed it somewhere, but that’s not the way I usually put thing. Viewed as a problem in pure mathematics, the key assumptions are these:

    Jason, I copied and pasted the description from here. It appears not to be your wording, but something that was quoted from Hayes.

    Monty always opens an empty door.

    He never opens the door you initially chose.

    In those cases where your initial choice conceals the prize, Monty chooses randomly from among the other two doors.

    Hmm,

    Even though the wording is different, when I run this through the (P),(-),(-) discrete event example that I’m using, I end up with a count that is the same for both staying and switching. Of course my operational definition could be completely different from the math that is actually used to solve the problem.

    Does anyone know if someone has actually looked at videotapes of the actual Monty Hall game show and counted how many times people have actually won and lost the prize, and, whether it is consistent with mathematical predictions and/or computer simulations (which I assume means that some kind of random number generator is involved rather than actual people)?

  13. #13 MartinB
    September 19, 2008

    “if the prize is in fact behind door 1 [represented as (P),(-),
    (-)] for a particular game, it would not be advantageous for the player to switch to either door 2 or 3 if their initial choice is door 1. It does make sense if one considers all possible discrete events associated with the prize being behind door 1 and the player plays multiple times.”

    I think you are wrong here (or I misunderstand your point).
    Let’s again consider the case (P)(-)(-) (In fact, since you don’t know which door the price is behind, this case alone suffices to understand what happens.)
    You have three choices: (1), (2), (3)
    Since you have no idea what door the price is behind, you will choose (1), (2) or (3) with equal probability: 1/3
    If you chose (1), you lose when switching.
    If you chose (2) or (3), you win when switching.

    I think you make the mistake of counting discrete events without looking at the propability of these events.
    To make it explicit: if you chose (2) (probability 1/3), Monty opens (3) with probability (1). Same if you chose (3)

    If you chose (1), Monty has two options – either open (2) (probability 1/2) or (3) (probability 1/2). The combinations:
    You chose (1), Monty opens (2) has an overall probability of 1/6,
    the same for: you (1), Monty (3).

    Here’s the overall table
    You (1) Monty (2): 1/6
    You (1) Monty (3): 1/6
    You (2) Monty (3): 1/3
    You (3) Monty (2): 1/3

    You must not simply count possibilities, you have to take into account the probability of each combination of cases.

  14. #14 Tony Jeremiah
    September 19, 2008

    (In fact, since you don’t know which door the price is behind, this case alone suffices to understand what happens.)

    This is probably where our mental models differ.

    I am thinking more like an experimentalist and not like a mathematician. In my thought experiment, there are three (thought) experimental conditions: the prize is behind door 1 [(P)(-)(-)]; door 2 [(-)(P)(-)]; or door 3 [(-)(-)(P)]. Those are the three situations possible concerning the location of the prize for any given game. Then for each one of those situations, I work out all the possible play permutations (discrete events) and whether each particular event will result in a win or loss (in terms of how the problem is defined). I assume that when the Monty Hall problem is set up like that, the exact solution space of the problem is operationally defined. So if one were to watch people actually playing the Monty Hall game, any particular game is defined by one of the play permutations written above. For example, for any game having situation (P)(-)(-), the only possible plays and their subsequent outcomes are the ones indicated by the play permutations [e.g., Pick(1),Stay(1),Switch(-),Outcome(W)].

    So, the fact that the actual game involves a contestant not knowing the location of the prize does not seem relevant, if one defines all of the possible play permutations per prize location.

    Further, based on working out the Monty Hall solution space (depending on how the problem is defined), there seems to be one situation where switching is beneficial, and another where it makes no difference.

    The situation where it is beneficial, appears to be the one where I define the problem in terms of Monty not opening any door. The play details is such that the player initially chooses a door; Monty Hall asks if the player wants to switch; the person either switches doors or does not; Monty Hall reveals what is behind the door the person chooses. When the solution space is worked out for this scenario, there are 9 play permuations that result in a win: 6 situations involve wins that are due to switching, and 3 involve wins that are due to staying. So this works out to 67% (6/9) wins due to switching and 33% (3/9) wins due to staying.

    If we look at the entire solution space of this particular version of the Monty Hall problem, there are 27 possible play permutations. So if one were to observe many Monty Hall games defined as it is defined here, 9/27 (33%) of the games should result in wins, for which 6/27 (22%) are due to switching and 3/27(11%) are due to staying. So in this version of the game, the majority, 18/27 (67%) should be observed to result in a loss.

    In the standard description of the game (as stated by Jason):

    Monty always opens an empty door, He never opens the door you initially chose, In those cases where your initial choice conceals the prize, Monty chooses randomly from among the other two doors

    there does not seem to be an advantage for switching. And from my mental model, I’m defining it in terms of the situation where 12 wins are observed, for which 6/12 (50%) are due to switching and 6/12 (50%) are due to staying. And if you observe each of the 24 permutations that result from this version of the game, they conform to the stated rules of the game. When you consider the entire solution space (i.e., the losses), 6/24 (25%) of the games won should be observed to be due to switching; 6/24(25%) of the games won should be observed to be due to staying; and 12/24 (50%) of the games should be observed to result in a loss.

    The difference between the two games (which, technically, would represent experimental and control conditions), appears to be whether Monty eliminates a door prior to the contestants final decision.

  15. #15 Janet C
    September 19, 2008

    The problem I always have with the “standard answer” is that a basic concept of probability theory is “In a fair game of chance, no fixed strategy beats a random throw of the dice.” The “always switch” strategy seems to contradict that.

    But, in the beginning, you randomly choose from 3 boxes. (Probability of getting prize is 1/3.)

    Then, Monty shows you one box, that you didn’t pick, that doesn’t have the prize. Monty is not a random event. Monty has more knowledge than the game player, and is always showing you a door that does not have a prize, so you cannot model him mathematically as a random event. The outcome is always the same after Monty does his thing — one box with a prize and one without, and your current pick resting randomly on one of those two boxes.

    The second game, where you get a chance to change your pick, is mathematically independent from the first game, where you made your original pick.

    Your choices are really “stick with your win or loss the first game” (keep your choice) or “play the second game” (which turns out to have better odds).

    You get a chance to pick between the two remaining boxes. If you flip a coin here, probability of getting the prize is 1 in 2 instead of 1 in 3. You should flip that coin, because your odds of winning have improved.

    I think if the “official answer” were “flip a coin” between the two remaining doors, instead of “always switch,” it would make perfect sense to anyone. “Always switch” cannot do better than “flip a coin,” because you can’t beat a fair pair of dice.

    So, even though “always switch” is a correct answer, it is not the only correct answer. It just turns out, that if your original pick was random, that “always switch” is as good as “flip a coin.”

    And, even more counter-intuitively, “always stick” is not just as good as picking randomly, because that is equivalent to keeping your result from the mathematically-independent first part of the game, where the odds were less in your favor.

    –JanetC

  16. #16 abb3w
    September 19, 2008

    I suspect this ties to a more general problem that I has piqued my (amateur) curiosity for some time.

    Presumably, perception of evidence can have some effect on what we believe; sometimes, however, seeing is not believing. My understanding is this is because of ingrained thought patterns. So, what determines if evidence is sufficient to change an ingrained idea?

    Whether it’s done rationally or not, how and why do people “really” change their minds?

  17. #17 RM
    September 19, 2008

    (On the “making unstated assumptions explicit” issue, note that I’m not really talking about which answer to the problem is correct – I’m actually talking about how best to convince someone else that the correct answer is really correct.)

    Jason Rosenhouse: It really is logic errors … Most people … take the view that when Monty opens an empty door the only thing you learn is that that particular door is empty. In reality you learn that Monty … chose to open that particular door.

    What I meant by “axiom error” (perhaps not the best term) was an error in the basic assumptions used in approaching the problem, as opposed to an error in carrying out the logical deduction from those assumptions. I would view the error as you stated as such an “axiom error” – people are mistaken as to the basic rules of probability, and what one can learn from Monty’s selection.

    But I can also see things from the other side. In the final analysis these are not issues over which reasonable people can disagree.

    Oh, certainly. I never intended to imply that the no-switch/no-fly people were correct, or that their answer somehow had “equal truth” to the switch/fly one. I was simply pointing out that until they accept that their assumptions are incorrect, they’ll never agree with you. (Convincing someone isn’t an epistemological issue, it’s a sociological one.) They think they are right, and when they work through things (starting with their incorrect assumptions as a given) all the logic checks out. Simply telling them that they’re wrong isn’t going to change what they believe (i.e. that their assumptions are correct), especially if you start belittling them.

    Arguing with someone who steadfastly denies that under the usual assumptions you should switch is a similarly frustrating experience.

    I’d maintain that the real issue is that you have different (unstated) assumptions, like what Monty’s selection tells you. Ferret those out and clarify to all’s satisfaction, and the disagreement goes away. (Just because you think all the assumptions are stated, doesn’t mean that they are.)

    Tony Jeremiah: Does anyone know if someone has actually looked at videotapes of the actual Monty Hall game show

    Interestingly, “Lets Make A Deal” never actually had a situation set up like the “Monty Hall Problem”. Sure, Monty sometimes gave people opportunities to exchange their prizes, but never under the three-doors/two-goats situation.

  18. #18 Tony Jeremiah
    September 20, 2008

    Interestingly, “Lets Make A Deal” never actually had a situation set up like the “Monty Hall Problem”. Sure, Monty sometimes gave people opportunities to exchange their prizes, but never under the three-doors/two-goats situation.

    Hmm,

    Well this does seem to explain the nature of the controversy; an artificially created construct never tested outside the boundaries of an artifically created reality.

    perception of evidence can have some effect on what we believe

    I would contend that the reverse is true; what we believe effects our perception of the evidence. So the issue boils down to how does one alter engrained beliefs, which is related to a couple of your questions…

    So, what determines if evidence is sufficient to change an ingrained idea? Whether it’s done rationally or not, how and why do people “really” change their minds?

    These are fundamental questions in social psychology, particularly research concerning cognitive dissonance. One of the primary findings is that if you make people behave in a way inconsistent with a pre-existing attitude/belief, they will tend to change their belief so that it is consistent with the counterattidunal behavior. An example related to the Monty Hall problem indicated by previous commenters, is where Monty Hall deniers are asked to participate in a computer simulation that shows them that switching works. Since the act of engaging in a computer simulation (the behavior) involves convincing people about switching (which is inconsistent with the pre-existing attitude that switching makes no difference), the attidue should change towards believing that switching works.

    But in terms of actually understanding the problem, doing so by computer simulation still does not really result in understanding; that is still just elaborate persuasion without comprehension.

    To really do the job of comprehension, an empirical verification of the problem in a real-world setting (not a computer simulation–which is probably only as good as the hidden assumptions behind the construction of the program that is the basis of the simulation) would be required.

  19. #19 Anonymous
    June 15, 2010

    1 0 0
    0 1 0
    0 0 1
    – – –
    1/3 1/3 1/3

The site is currently under maintenance and will be back shortly. New comments have been disabled during this time, please check back soon.