Three Strange Results in Probability: Cognitive States and the Principle of Indifference (Monty Hall, Flipping Coins, and Factory Boxes)

Estimated read time (minus contemplative pauses): 16 min.

Probability is known for its power to embarrass our intuitions. In most cases, math and careful observation bear out counterintuitive results. After many such experiences, one’s intuition improves (sometimes perhaps crossing into a kind of overcorrection—see the Optional Endnote for some inchoate thoughts on that). But some results stay strange, and it’s not always clear whether our rebelling intuitions signal a problem with formal probability, or simply confirm that human cognition has evolved to concoct tidy stories amounting to illusory—if sophisticated—representations of the world rather than to deal head on with complexity, chance, and uncertainty.

Here, briefly, are three results I find particularly interesting because, despite being strange (if not problematic), their solutions are simple within their given models. My favorite is (2).

(1) When the Monty Hall problem really is 1/2: The Monty Hall problem contains a strangeness I’ve not seen explicitly discussed, though it’s often implied by explanations of why the common-sense answer is wrong, including in my post, “Monty Hall Problem and Variations: Intuitive Solutions.” In case you need a refresher, the problem goes:

There are three closed doors. Behind one is a car. Behind each of the other two is a goat. You pick a door, hoping to find the car. The host, Monty Hall, then does what he always does in this game: he reveals a goat by opening one of the doors you didn’t pick. He then offers you the chance to switch to the door he didn’t open. (When you chose the car-concealing door, he chooses among the two goat-concealing doors with equal probability.) If you switch, what is the probability you find the car?

The answer is 2/3. Nothing strange about that. However, if we change the problem so that Monty Hall opens a door at random rather than always revealing a goat, it changes to 1/2. This is because 1/3 of the time he’ll reveal the car, thus ending the game.1

This strikes me as strange not because of the math, but because it allows for games that appear identical—or at least seem significantly similar—to yield different probabilities.

For example, imagine you guess Door #1, behind which stands a goat. Monty Hall—for the first and only time during his long hosting career—forgets which door conceals the car, so he acts naturally, hopes for the best, and randomly chooses Door #2, which happens to reveal a goat. You know that you had a 1/3 probability of having chosen the car. If you knew Hall had guessed, you would now update that to 1/2. But you don’t know Hall guessed, so you quietly continue to presume a 2/3 chance of winning by switching. You switch, you win, and nobody’s the wiser.

Now imagine the same scenario, but this time Monty Hall doesn’t forget, and knowingly reveals the goat. This time, the chance you chose the goat conditional on Hall’s revealing a goat stays 2/3 rather than updating to 1/2, because the probability that Hall reveals a goat is 1, rather than 1/2 as it was above. So, you’re in what externally appears to be the exact same scenario as above, but this time you’re correct to assign a 2/3 chance of winning by switching.

You’d see those probabilities borne out were you to run many instances of the game, but I’m especially interested with how we conceive of a single run of the game.

It strikes me as endlessly fascinating that two given scenarios could be externally identical, with the exact same promised outcome (i.e., switching will win), yet those final outcomes have different chances—or at least demand different subjective assignations—due to the mental state of one actor during the middle stage of the scenario. Once Hall makes and acts on the decision to open that door, the influence of his mental state passes.

Interestingly, my intuition tells me the opposite of what most claim to be the intuitive answer here. That is, it always feels like it should be 2/3 to me, once a goat is revealed. I figure that I most likely chose the wrong door in the first place, so getting the chance to switch now that one wrong answer has been removed seems smart. But this fails to account for the additional randomness introduced by Hall’s guessing (where “random” means he’s equally likely not to open the door with a goat, or, at the very least, that his opening that door could not have been predicted with certainty).

To jiggle that intuition around like a loose baby tooth ready to come out, I can think of the following. You play the game with a million doors. You pick a door. Chances are you chose a goat door. Hall, who’s forgotten where the car is, randomly opens 999,998 doors, leaving one door closed. Luckily, he only reveals goats. Do you switch? My intuition here is, overwhelmingly, that you should stay. Why? Well, you’d have to get really lucky to pick the car door on your first try, but not as lucky as Hall would have to be to pick a goat 999,998 times! In other words, chances are you chose the car door, thus making it easier for him to reveal all goats.

So, my intuition has changed from switch to stay. But there’s something very wrong with this intuition as well. Here’s how I’ll correct it.

Assume you chose a goat door. The first door Hall opens, he has a greater than 99% chance of revealing a goat, given that there are 999,999 doors to choose from, and only one hides a car. Once he’s down to ten doors, he’s still got a 9/10 = 90% chance of revealing a goat. Of course, that’s still a lot of opportunity for accidentally revealing the car. There’s only a 1/999999 chance he’ll reveal only goats; or, put another way, will choose the correct door to keep closed (he could have just chosen one door at the offset and then said, “Open all the others”). His probability for keeping the car door closed is low, but a little higher than yours was for picking that same door to begin with.

Now let’s assume we don’t know whether you choose a goat or car door. You started with a 1/1000000 chance of having chosen the car. Every time he reveals a goat, you get evidence that you’ve made it easier for him (than had you chosen a goat) to not accidentally reveal the car (a little easier at first, a lot easier by the end). The upshot of this is that if he reveals all goats, the probability of winning by switching becomes, as in the two-door case, 50%. That’s a big if, but, in the unlikely-to-occur run of game you played, he did indeed make it that far without revealing a car, and you must update your probabilities to match that world. The challenge here (for me) is to not increase my confidence in having initially chosen the car to above 50%.

Still, this helps improve my intuitions about the two-door case (where I don’t feel tempted towards overconfidence). That is, if I learn he randomly revealed a goat, I feel my confidence rise a little for having made it easier for him to not reveal the car. If I’m appropriately careful about putting a number on that increase, I’d say it rises from 1/3 to 1/2.

(2) When it’s right to assign 1/2 to a 1/3 situation: While recently writing about the Two-Child problem, I noticed a variation on the above example. For a quick refresher on that problem:

A couple has two children, one of whom is a girl. What is the probability both children are girls?

The answer is 1/3 when it’s assumed one learned of the child through some deterministic means, and 1/2 when learned randomly. Most discussions of the problem assume a 1/3 circumstance. I won’t get into that here. You can read about it in my post on the topic, or you can see how those solutions play out by considering the following strange example (which also appears in the last section of that post).

If I flip two coins behind your back and then tell you one of them landed Heads, you should assign a 1/2 chance to both having landed Heads, due to the assumption that, in a Heads-Tails or Tails-Heads condition, I’m equally likely to reveal a Heads or Tails to you. So, for instance, in 120 runs of the game, 30 of the 60 times you hear me say “at least one landed Heads,” we’ll be in a Heads-Heads condition.

Despite this, the probability of Heads-Heads is really 1/3, because what I didn’t tell you is that it was my plan all along to tell you Heads if one landed that way. I would only have told you Tails had both landed Tails! This means that in 120 runs of the game, 30 of the 90 time you hear me say “at least one landed Heads,” we’ll be in a Heads-Heads condition.

What’s especially interesting to me here is that, in evaluating a single instance of this trial, you’d be justified in imagining a multi-trial simulation in which 1/2 of the outcomes are Heads-Heads. In a real series of such trials, however, a 1/3 result would emerge. You’d eventually catch on. For example, every time you’re told, “at least one landed Tails,” it would turn out to be in a Tails-Tails condition (a condition, interestingly, in which I’m unable to exercise the rule; so, if our first three runs were in that condition, it wouldn’t be of much help). But in the first instance of the game, you’d reasonably apply a principle of indifference that assumes even odds for my revealing a Heads or Tails. This is the same principle of indifference that leads folks to claim that the solution to the Two-Child problem is 1/2 (i.e., on the grounds that, for example, a father who mentions having a daughter might just as well have mentioned having a son).

This leads me to the third strange result…

(3) Factory Boxes and the Principle of Indifference: The principle of indifference, also known as the principle of insufficient reason, says that when you see no reason to weight competing outcomes differently, you should weight each of them as equally probable. I gave examples in (2) above. The most common application might be when we assume a given coin is fair.

Bas van Fraassen has produced a compelling paradox arising from this principle.2 Here I’ll quote Aidan Lyon’s discussion in his 2010 paper “Philosophy of Probability” (published as a chapter in Philosophies of the Sciences: A Guide):

Consider a factory that produces cubic boxes with edge lengths anywhere between (but not including) 0 and 1 meter, and consider two possible events: (a) the next box has an edge length between 0 and 1/2 meters or (b) it has an edge length between 1/2 and 1 meters. Given these considerations, there is no reason to think either (a) or (b) is more likely than the other, so by the Principle of Indifference we ought to assign them equal probability: 1/2 each. Now consider the following four events: (i) the next box has a face area between 0 and 1/4 square meters; (ii) it has a face area between 1/4 and 1/2 square meters; (iii) it has a face area between 1/2 and 3/4 square meters; or (iv) it has a face area between 3/4 and 1 square meters. It seems we have no reason to suppose any of these four events to be more probable than any other, so by the Principle of Indifference we ought to assign them all equal probability: 1/4 each. But this is in conflict with our earlier assignment, for (a) and (i) are different descriptions of the same event (a length of 1/2 meters corresponds to an area of 1/4 square meters). So the probability assignment that the Principle of Indifference tells us to assign depends on how we describe the box factory: we get one assignment for the “side length” description, and another for the “face area” description.

There have been several attempts to save the classical interpretation and the Principle of Indifference from paradoxes like the one above, but many authors consider the paradoxes to be decisive. See Keynes [1921]3 and van Fraassen [1989]4 for a detailed discussion of the various paradoxes, and see Jaynes [1973]5, Marinoff [1994]6, and Mikkelson [2004]7 for a defense of the principle. Also see Shackel [2007]8 for a contemporary overview of the debate. The existence of paradoxes like the one above were one source of motivation for many authors to abandon the classical interpretation and adopt the frequency interpretation of probability.

Lyon then goes on to discuss the frequency interpretation, which comes with its own problems. Here’s a taste, which ends with a segue into what’s known as the reference class problem:

Ask any random scientist or mathematician what the definition of probability is and they will probably respond to you with an incredulous stare or, after they have regained their composure, with some version of the frequency interpretation. The frequency interpretation says that the probability of an outcome is the number of experiments in which the outcome occurs divided by the number of experiments performed (where the notion of an “experiment” is understood very broadly). This interpretation has the advantage that it makes probability empirically respectable, for it is very easy to measure probabilities: we just go out into the world and measure frequencies. For example, to say that the probability of an even number coming up on a fair roll of a fair die is 1/2 just means that out of all the fair rolls of that die, 50% of them were rolls in which an even number came up. Or to say that there is a 1/100 chance that John Smith, a consumptive Englishman aged fifty, will live to sixty-one is to say that out of all the people like John, 1% of them live to the age of sixty-one. But which people are like John? If we consider all those Englishman aged fifty, then we will include consumptive Englishman aged fifty and all the healthy ones too. Intuitively, the fact that John is sickly should mean we only consider consumptive Englishman aged fifty, but where do we draw the line?

To summarize the factory box dilemma: in “side length description, there’s a 1/2 chance of getting a box with an edge length in the 0 to 1/2 range; in the “face area” description that same range has a 1/4 chance. Rather than merely a strange result, perhaps this demonstrates a technical or theoretical problem (if so, maybe it’s similarly wrong in the Two-Child problem to assume a parent could mention either gender in a seemingly neutral context unless that’s explicitly built into the problem).

But perhaps this result isn’t so worrisome, depending on what we expect from probability as a formal tool for doing better under uncertainty than we otherwise would. Maybe both those probabilities are equally recommendable so long as we are ignorant of actual data about the frequency of edge lengths produced—just as you are correct in assigning 1/2 rather than 1/3 in the coin example of (2) above, until your observations improve that assignment. And perhaps either assignment would still be better than what intuition alone would likely provide.

That an initially correct assignment can be better aligned with the world suggests to me that uncertainty is cognitive. Events in the world do not seem to entail paradox, but rather paradoxes seem to arise due to what we don’t—and perhaps cannot—know about the world (even should some of those events turn out to be knowable only probabilistically, even to an omniscient observer).

Optional Endnote: Probability and Intuition
I conclude with some rough thoughts I’ve been mulling over in recent months, other strains of which may be found in various recent writings I’ve posted on probability.

Probability isn’t just a technical discipline. It also asks us to systematically adjust the informal probabilistic worldview each of us inevitably develops, usually unawares, as we go through daily life guided by intuition and common sense. It’s in large part this relationship to intuition and the real world that makes the subject so captivating to me. Indeed, even which probability model to adopt can be a matter of disagreement between intuitions (for elaboration, see Lyon’s excerpted paper above).

The tuning of one’s intuition usually starts with coins and dice and jars of marbles and such, and gets rougher as increasing doses of complexity, interdependence, and uncertainty are injected.

Nassim Taleb has pointed out the error—what he calls the “ludic fallacy”—of turning one’s game-derived probability sense towards the complex real world. It seems that many of the results I find most intriguing rest in the hazy penumbral area between games, with their readily calculable expectations, and the real world, which consists of events lacking clearly shared features and, most inscrutably, the human mind. In other words, we may even be committing the ludic fallacy when we apply game probability to games!

And so I resist fully surrendering my intuition to formal disciplines (though by “I” here I really denote my stubborn intuition itself, rather than some central executive who gets to decide what my intuitions do from moment to moment). Rather, I think we should pay attention when a result carries a whiff of strangeness about it—especially when it endures the hard work of understanding that result’s technical dimensions. I often observe people rehearsing counterintuitive probability results as though they’re obvious. But clearly they’re not obvious9, or else it wouldn’t have taken as long as it did for the field of probability to develop, and there wouldn’t be a history of brilliant mathematicians getting such problems wrong—e.g., when Paul Erdős rejected the 2/3 solution to the Monty Hall problem.

Presumably Erdős rejected that solution until he understood it for himself—until it felt right—rather than accept it because other mathematicians said so. Maybe many results these days are accepted on the grounds of consensus, or even on the general grounds that counterintuitive results are the best kind of results.

Perhaps this is why many will accept the Monty Hall problem or the Two-Child problem and so on, before fully understanding the math. Before understanding, for example, why it is that if a random mother tells you “one of my two children is a girl born on a Tuesday,” the probability of her having two daughters is 13/27, while if you ask a random mother of two if “one of them a girl born a Tuesday,” and she says “yes,” the chance of both being daughters is 1/2.10

Maybe this is in part an attempt to avoid finding themselves in the camp of those who scolded Marylin vos Savant for her correct answer to the Monty Hall problem. Though I imagine something different goes on with the layperson and the professional (the former can still be seen all over YouTube rejecting the 2/3 Monty Hall solution, often emphatically offering their own “obvious” solution; while the latter, as well as ambitious amateurs such as myself, take 2/3 as obvious).

I’ve seen articles in reputable popular publications attempting to explain the above-noted 13/27 “born on a Tuesday” result, ultimately getting it wrong. Those do sometimes carry a tinge of modesty, but I prefer the modestly skeptical approach taken by George Johnson in his review of Leonard Mladinow’s 2008 book, The Drunkard’s Walk. In that book, Mladinow further complicates the Two-Child problem: What if you learn one of the children is a girl named Florida? Johnson writes in response: “Even weirder, and I’m still not sure I believe this, [Mladinow] demonstrates that the odds change again if we’re told that one of the girls is named Florida.”

At any rate, I think such results should be displayed with care. I fear that treating them as obvious threatens to trivialize a field that needs to be taken more, rather than less, seriously by a general public that might think the intuitions of experts have been so beaten into submission and bent out of shape by formal training that the experts will accept absurd results over obviously common sense ones—or in simpler terms: the expert will miss what a layperson sees readily.

Sometimes this worry seems legitimate. It would be interesting to make a list of strange things believed by experts, both in terms of ideas with many adherents, and those with few. I can think of several recent philosophers and scientists who’d make both categories (one need only read a few writings on the nature of consciousness to get started).

As Augustín Rayo once put it on the Elucidations podcast, in a droll response to a question about why anyone would think there could exist no dinosaurs while the number of dinosaurs was not zero: “You would only think that after years of training as a philosopher.” Interestingly, later in that fascinating conversation, which was about the construction of logical space, Rayo himself makes an argument essentially claiming that numbers have objective, independent existence of some sort. This belief, generally referred to as “mathematical platonism,” strikes me as strange as a belief in ghosts, but many mathematicians, philosophers, and physicists seem to view it as obvious.

I myself take for granted that Edmund Gettier got it right in his 1963 paper11 in which he shows that justified true belief is not always knowledge, as it had for centuries been believed to be. For reasons I won’t get into here, I doubt his paper would have convinced pre-20th-century experts (or perhaps even pre-WWII experts). To me and seemingly most contemporary epistemologists, though, there’s nothing strange about it.

More aptly, this worry can be found in the context of statistics as well. For example, when Taleb, in The Black Swan (2010, second edition), writes things like: “[Chris] Anderson is lucky enough not to be a professional statistician (people who have had the misfortune of going through conventional statistical training think we live in Mediocristan)” (page 223); or when he urges “the intelligent reader” without formal training to skip a section “meant to prove a point to those who studied too much to be able to see things with clarity” (page 352).

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Footnotes:

1. More thorough explanations can be found at my aforementioned post.
2. van Fraassen, B. (1989). Laws and Symmetry. Oxford: Oxford University Press.
3. Keynes, J. M. (1921). A Treatise on Probability. London: Macmillan.
4. van Fraassen, B. (1989). Laws and Symmetry. Oxford: Oxford University Press.
5. Jaynes, E. T. (1973). The Well Posed Problem. Foundations of Physics, 4(3):477– 492.
6. Marinoff, L. (1994). A Resolution of Bertrand’s Paradox. Philosophy of Science, 61(1):1–24.
7. Mikkelson, J. M. (2004). Dissolving the Wine/Water Paradox. British Journal for the Philosophy of Science, 55:137–145.
8. Shackel, N. (2007). Bertrand’s Paradox and the Principle of Indifference. Philosophy of Science, 74:150–175.
9. Admittedly, though I’ll forego attempting one, I likely owe some explanation here of what I mean by “obvious,” as many of these things seem like they couldn’t be anything but true to an intuition that’s been developed in a certain way. I hope context is enough to understand what I mean here by “obvious.”
10. I won’t explore this here, but there’s a political dimension to this as well: people often seem to accept or reject certain scientific claims according to—or sometimes as a corollary to—their political affiliation rather than due to understanding those claims.
11. Is Justified True Belief Knowledge?“, Edmund L. Gettier, Analysis, Vol. 23, No. 6. (Jun., 1963), pp. 121-123.

6 Replies to “Three Strange Results in Probability: Cognitive States and the Principle of Indifference (Monty Hall, Flipping Coins, and Factory Boxes)”

1. JeffJo says:

Where to begin? I just skimmed this, but here are some random thoughts. The names here might get confusing, so be careful.

French Mathematician Joseph Bertrand has three “paradoxes” named for him. One is called Bertrand’s Box Paradox. Today, most people apply it to the problem he used to illustrate it, since conflicting solutions are claimed. But he meant it for a commentary on the incorrect solution. Since the Monty Hall problem is equivalent to his, I’ll use it.

Say you pick Door #1. Some claim that if Monty Hall opens Door #2 to show a goat, that Door #1’s probability changes from 1/3 to 1/2. If that were true, then the same change occurs if he opens Door #3 to show a goat. So, what if instead of opening a door, Monty says “My assistant, Lovely Linda, is standing behind a door that you didn’t pick, and that has a goat. She will open that door in a moment; but first, you can trade your door for the one she isn’t going to open.”

If the 1/2 argument were correct, since it applies regardless of where Lovely Linda is, THE CHANGE HAS ALREADY TAKEN PLACE. But since it is always possible for her to find a qualifying door, no change can have occurred. So the 1/2 argument can’t be correct.

The only answer that can be correct, based on this paradox, is 1/3 for staying and 2/3 for switching. But notice that I didn’t say that this is a correct solution. (Answer=value, solution=method that produces an answer.) It just shows that any other answer isn’t right. But it is more of a solution than anything Marilyn vos Savant has ever published. She has only ever asserted an answer.

The same paradox says the Tuesday Girl answer can’t be 13/27, and learning a name can’t affect anything about gender.

Bertrand’s other two paradoxes are just called “Bertrand’s Paradox.” But one applies to Economics, and the other to Probability, so they don’t get confused. But the one that applies to Probability addresses the same issue as your Box Problem. “Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle?”

Bertrand discussed three ways to pick a random chord: between two random points on the circle, the perpendicular at a random point on a random radius, and the line through a random point such that the point insects the chord. The answers for the three methods were 1/3, 1/2, and 1/4, respectively.

I’ll add another problem that better illustrates the issue: which is more likely: it will rain tomorrow, or the neighbor’s dog will raid my trash can (again)? You don’t have any reason to say one is more, or less, likely than the other. But that doesn’t mean the Principle of Indifference applies, since you also don’t have any reason to compare the two events. It’s apples vs. oranges.

In both your Box problem, and Bertrand’s Chord problem, there is no justification for directly comparing the ranges, or using the Principle of Indifference. They may be more similar than rain and dogs, but it’s still apples vs. oranges. (Well, a former Professor of mine, Edwin Jaynes, does claim a way exists for one method. http://bayes.wustl.edu/etj/articles/well.pdf. But it wasn’t probability I took from him, it was Thermodynamics.)

1. Dan says:

Hi JeffJo,

Thanks for another instructive response. In the comment in which I asked you to have a look at this post, I almost mentioned the Bertrand circle paradox, mainly to point out that, for whatever reason, it doesn’t strike me as thought-provoking as the Factory Box situation. My thought was that this might be because the circle “paradox” has mostly to do with an insufficiently specified probability model that can be resolved (which is to say I presume some model will tell us in fact what the probability in question is), while the Factory Box problem seems to me to lack such an a priori solution. (I don’t see a problem with applying the principle of indifference there in principle, once a range has been chosen [ie, for edge or face].)

Thanks for sharing the Jaynes article. I look forward to reading it and thinking more about the above problems. I JUST bought his The Probability Theory: Logic of Science, which I’m eager to get to (after a couple other books I’m working through). Cool that you studied with him.

2. Dan says:

PS: Regarding the Linda example, couldn’t a one-half-er just say, “It comes down to a 1/2 chance between the first door I picked and the door Linda doesn’t open”? It’s not correct, but I don’t see how it’s more obviously incorrect than the one-half position given the standard version of the problem.

Also, in response to letters protesting her initial answer to the Monty Hall Problem, vos Savant posted what seems to me to count as a solution (from http://marilynvossavant.com/game-show-problem/, where you can also read some of the letters):

Good heavens! With so much learned opposition, I’ll bet this one is going to keep math classes all over the country busy on Monday.

My original answer is correct. But first, let me explain why your answer is wrong. The winning odds of 1/3 on the first choice can’t go up to 1/2 just because the host opens a losing door. To illustrate this, let’s say we play a shell game. You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you’ve chosen, we’ve learned nothing to allow us to revise the odds on the shell under your finger.

The benefits of switching are readily proven by playing through the six games that exhaust all the possibilities. For the first three games, you choose #1 and “switch” each time, for the second three games, you choose #1 and “stay” each time, and the host always opens a loser. Here are the results.

(Table appears here showing possible results.)

When you switch, you win 2/3 of the time and lose 1/3, but when you don’t switch, you only win 1/3 of the time and lose 2/3. You can try it yourself and see.

Alternatively, you can actually play the game with another person acting as the host with three playing cards—two jokers for the goat and an ace for the prize. However, doing this a few hundred times to get statistically valid results can get a little tedious, so perhaps you can assign it as extra credit—or for punishment! (That’ll get their goats!)

2. JeffJo says:

Consider two statements of the PoI: “You have no basis to claim the cases behave differently” and “You have a basis to claim they work the same”. These could be called weak PoI and strong PoI. Only the strong is valid.

The rain/dog example is a blatant attempt to use the weak version, but it can hard to see why the other examples are. The chord problem is less blatant, but it isn’t hard to see how it is using the weak. The paradox in your box example arises because you apply the weak version in two ways that clearly violate the strong.

In the Lovely Linda problem, the three doors clearly arrived at the current state by different paths: your door couldn’t be opened, one conceals LL, and one could have concealed her but doesn’t. So the strong PoI can’t be applied.

And that supposed solution by MvS, while it gets the right answer, is wrong. (And note that it is an attempt at a solution even though she claims it to be explanation for why 1/2 is wrong.) Incomplete might be a better description, but what it lacks is the reason why the strong PoI doesn’t apply. So I call it wrong.

The winning odds for the first door can go up, but only if Monty Hall doesn’t choose randomly when he has the choice of two doors. If you picked a goat, it was a 100% chance to open the door he did. If you picked the car, he had a choice. Let’s call the probability he would choose as he did Q. This makes the odds you picked a goat 1:Q, or a probability of 1/(1+Q). If Q=1, this is 1/2. If Q=0, this is 100%. But the (strong) PoI says we must use Q=1/2, so it is 2/3.

1. Dan says:

Hi JeffJo,

I still don’t see why if only told, say, that a machine randomly produces a piece of string between greater than 0” and no more than 12″, you shouldn’t assume (in accordance with commonly accepted notions about probability) a PoI in terms of range lengths the machine will produce.

Also, isn’t your strong PoI just a way of saying the things in question are known to be equiprobable? Under that constraint, we wouldn’t be able to say a given six-sided die is fair unless we have a basis to claim each side has the same likelihood (or propensity or what have you) of coming up. I doubt many dice actually are fair, but we assume they are until given reason to think otherwise (e.g., after rolling it many many times to uncover its bias). Or is apparent symmetry basis enough?

I guess I don’t get what the Lovely Linda example is meant to demonstrate that the standard version of the MHP problem—in which it’s understood that MH always, 100% without fail, reveals a goat—doesn’t already demonstrate. In the alternative case that he chooses a door at random, I indeed understand that we apply PoI in assuming he could choose either door (and/or, I suppose, that the people who place the goats do a good job of randomizing where they put them, in case, e.g., MH, when guessing, tends to pick the door farthest to the left); I point that out—or things that imply that—at various points in stuff I’ve recently posted (in my original post about MHP, I also consider some examples in which MH flips a coin to choose which door to pick, which I think suggest some interesting implications for our intuitions about the problem insomuch as it involves MH’s attitudes and beliefs).

For some reason the table from MvS’s explanation didn’t show up in my comment. But she does give a table with possible outcomes, demonstrating that you win 2/3 of the time by switching. Seems to me to be a sufficient explanation of why 2/3 is the answer for switching, or at least of why one should accept the 2/3 answer. I also took her comment “As I can (and will) do this regardless of what you’ve chosen” to imply that the game-master always knows what’s under the shells. I agree with you, however, that this doesn’t explain why 1/2 is wrong; it would have been better had she done that and explicitly explained the differences between the 1/2 and 2/3 results.

1. Dan says:

At any rate, though I have many lingering questions, I might be Principle-of-Indifference’d out for the moment. What I’ve found in my readings is that some experts view it as leading to dangerous, unsurmountable contradictions (despite there being clever solutions, such as Jaynes’, to stuff like Bertrand’s cube and chord problems–e.g., as noted by Bas van Fraassen, whom I cite in my post, but see also and especially his paper “Indifference: Symmetries of Probability”: https://fitelson.org/probability/vanfraassen_pi.pdf); some cautiously claim PoI can be refined, or at least coherently applied in more cases than generally thought (Jaynes, e.g., in the article you shared); and others say there are no contradictions at all so long as PoI is properly applied (e.g., this guy: http://www.ccsenet.org/journal/index.php/ijsp/article/viewFile/59617/32596).

It’s not clear to me that those saying it must be properly applied, or who recommend only a strong PoI condition, aren’t just really rejecting PoI: aren’t the sorts of gaps in definition and knowledge that attract the weaker version precisely the sorts of things the PoI is (or at least originally was) meant to fill in? The thought of rejecting the weak version doesn’t bother me (that’s kinda my underlying point in all this), but I’m not sure of what that means for the strong version. Also, the general idea seems to be that one should reject PoI according to whether it results in contradictions, rather than due to comparing apples and oranges (even van Fraassen seems to accept that PoI survives Bertrand’s challenges thanks to solutions from Poincaré and Jaynes… but this doesn’t mean PoI doesn’t lead to contradictions elsewhere) .

But, as I said, a bit PoI’d out today. What I’m more interested in (and bothered and perplexed by) is the relationship between a single trial and the set(s) of trials of which that single trial is considered to be a member, particularly when the sets are composed of hypothetical frequencies. More on that later…

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