The hand-wavy way I've always thought about the Monty Hall problem is as follows:
When the contestant picked a box he was making a random choice between three boxes and had a probability of 1/3 of picking the car. The probability that one of the other two boxes was the car is thus 2/3. When Monty Hall opens a box it's still the case that the probability of the car being in the other two is 2/3 but the contestant can now eliminate the open box and knows that the probability that the car is in the other box is 2/3. Thus it's worth switching.
The other phrasing of the problem that I've seen help people is this (I think was also originally from Marilyn Vos Savant): Suppose there are a thousand doors, and only one is a car. Monty opens 998 of them to reveal goats, leaving door #628 still hidden. Do you switch?
That helps people understand that their original probability of picking the car the first time was very small, and didn't change because of what Monty did.
Most paradoxes vanish, become obvious this way (when you increase the numbers). The brothers paradox mentioned: 'I have two children. At least one is a boy.'/'I have two children. At least one is a boy born on Tuesday.'; 'What is the probability I have two boys?' drastically changes when you change to 'I have two children. At least one is a boy born at 7h:35m:42s.' -- it becomes clear that the succeeding information exactly specifies the boy as it becomes specific, in the limit turning into 'I have two children. The first is a boy.' (so the probability goes from 1/3 to 1/2).
You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one.
Whether Monty opens the door or not doesn't mean much.
That is only equivalent if Monty (as intended but usually not stated clearly) always picks a door without a car using knowledge of where the car is. If, instead, Monty picks randomly from the other two doors and just happens not to have revealed the car this time, odds are the same whether you stick with your door or switch.
Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.
I like to explain it with a wager: pick a star from the night's sky. If you pick the star I'm thinking of, I'll give you a dollar. After you've made your selection, I'll give you the opportunity to switch to one other star, guaranteeing that either your star or this one alternative will be the star I had in mind. Do you want to switch?
That was always the confusing unstated assumption for me, that makes me think the problem is only confusing because of how the wording usually de-emphasizes that distinction.
If he chooses a door, you can benefit from the secret information he reveals sometimes by doing so.
If the door was chosen at random, he is not adding any information, so you can't act on it?
At the point when I read it, I didn't realize Monty was always choosing the door with pre-knowledge of which one the car wasn't behind.
It wasn't until somebody wrote in with a computer program example that showed the benefits of switching, in which I inspected the source code, that I understand what Monty was doing.
personally, I think this has to do with ambiguous way word problems are stated, and how people model the word problem mentally.
Agreed, but the Monty Hall problem wouldn't make sense (at least from the game show perspective) if Monty's selection were random. If his selection were random, and he picked the car, it would just be "OK, I randomly decided to pick a door, and look, it's the car. Congratulations, game over."
Yeah, there really isn't any way to implement that version of the Monty Hall problem in the real world. You could say that Monty Hall randomly picks a door, and if he picks the winning door, that round of the game is aborted and you (the contestant) are given a drug that makes you forget that round ever happened. The rounds continue until Monty Hall happens to reveal a losing door, and that's the only one you will ever remember.
This is so detached from normal events in life that the paradoxical nature is much less impactful, so it's nowhere near as enticing of a thought experiment. It is, however, similar (equivalent, methinks) to the "God's Coin Toss" problem, which is also popular and which also has gotten some attention on Hacker News: http://www.scottaaronson.com/democritus/lec17.html
Eh, the math works out the same whatever is done in the case of him revealing a car. Reasonable options seem to include:
1) contestant wins (notion being the contestant retains the option to switch to any door, and now knows where the car is)
2) contestant loses (notion being Monty picked right and "won" in place of the contestant)
3) Round is aborted, things are repositioned, and the round is replayed (doesn't require any drugging, aborted rounds may or may not be aired but player learned nothing relevant to future rounds).
That's the whole reason why the problem is tough. It's carefully worded to be ambiguous about that point so that most people assume it was random (most of the erroneous math people pull out is valid were it random) but the problem actually makes little sense if Monty picked randomly since it completely ignores the potential possibility that he picked the car door.
The problem isn't probabilistically difficult and most people's intuitions would be on point if it weren't set up to deceive.
The wording in the "Ask Marilyn" column appears to have been:
'Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?'
Given "the host, who knows what's behind the doors", I think the original phrasing was less ambiguous than many rephrasings I encounter. It certainly could have been still clearer.
Also, her answer - which was presumably read by those writing in to asset her answer was wrong - plainly states that the host always picks a losing door.
Ah. In my generation, who actually watched the program, there is no ambiguity. No intent to deceive was detected by me. Its not supposed to be a trick question. Its supposed to illustrate how far off our intuition is about statistical behavior.
Monty himself mentions that he would occasionally try hard to argue contestants into swapping. They would always refuse. He would even explain that 'his door' had the better chance - nobody would listen. But I suppose that's confounded by the suspicion that he was trying to trick them.
This may be very true and I think that if you're more familiar with the program and recognize that the host must know the answer then it becomes more a question of statistical intuition.
I'm confident that there are lots of people who still fail to make the connection (though we're getting to the point where I think people are failing less due to a lack of statistical intuition and more due to a symbolic/physical model mismatch issue), but I think this problem wouldn't be as renowned as it is if it weren't for all of the even expert statisticians who are getting fooled.
In their case it's definitely a matter of them misinterpreting the situation and the way that it's worded, for someone unfamiliar with the show, is at least a little "tricky".
I agree it would help to see the problem in action.
Again, the part I missed was "it would be silly for him to reveal a car".
I get that now, but that was not something that I intuited, and so the omission of that information I will continue to argue is at least a little bit deceptive.
Even if only to those like myself who have not seen the program.
I know what you mean! I sort-of understand why people don't make this clear when they're stating the problem, but I really don't understand the vast majority of people who don't explain it clearly when they're stating the answer to the problem. They'll go on for paragraphs, when they could have just said, "Monty only chooses doors without cars, so by choosing a door he gave you more information."
It's not just that he knows, but that he uses that knowledge to pick a door instead of choosing a door at random.
But yeah, that the problem is moot if it were random is exactly my point.
I was deceived by the wording and it's frustrating to hear "answers" to this problem that ignore that deception because I got the wrong answer and so I feel "dumb" for being decieved so I am trying to defend how i'm not "dumb" for not seeing the probability, just dumb for not thinking through the fact that it would be unlikely for Monty to reveal a car at that point.
I guess a lot of people go "oh, he won't reveal a car".
That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a door arbitrarily, and it just so happens to be empty, it doesn't matter if you switch.
To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.
The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door.
Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching.
Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.
"Something must make him always choose a losing door."
The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play.
Consider:
"You have $X, your opponents have $Y and $Z. You select Potent Potables for $400 and it's a daily double. How much should you wager?"
I think you'll agree that is going to be read near-universally as a statement of a particular situation that could arise in Jeopardy, not a statement of how Jeopardy games always go.
> The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play.
I'm not sure how that matters. The question is whether, in this scenario, you choose to switch after Monty Hall reveals the door. The terms of the thought experiment dictate that Monty Hall will reveal a losing door. Regardless of what force is actually causing the revealed door to be a losing door, you should switch, because you're essentially being given the option to take your original one door, or to take both of the other doors (at least one of which is a losing door).
The terms of the thought experiment are intended to dictate that Monty will always reveal a losing door. However, they are often worded such that a reading that Monty merely has this time revealed a losing door is a valid interpretation. In that case, the problem is under-specified, as we don't know how Monty picks (randomly or always a losing door) and that changes the answer.
That does sound underspecified, although I would think that it still makes sense to switch. If Monty reveals a winning door, it obviously doesn't matter whether you switch (both other doors are losers), but if he reveals a losing door you're better off switching because you're still essentially getting 2 doors instead of the 1 door you originally selected.
If he might have opened a losing door, your odds are no longer 2:1 switching. This surprised me as well (in fact, I wrote a simulation for myself intending to reassure me that it was not the case, and learned better).
To waffle on this point: He has a higher (double) chance of picking a goat if you've picked the car. So by unplannedly revealing a goat, he gives you more information about whether you're in a world where you've picked the car, or one where you've picked a goat.
If you pick randomly between a fair coin and a double-headed coin, there's a 50/50 chance of picking either. If Monty then flips the coin and it comes up heads, that suggests you're more likely to be in a world where you've picked the biased coin. If, on the other hand, Monty deliberately takes your coin and places it so that it's heads (and he was gonna do that whatever coin you picked) then you have no new information.
Your argument isn't really wrong as far as it goes, but it doesn't explain why it matters whether Monty deliberately opens a door without a car or picks randomly and just happened to open a car-less door this time.
Love the George Carlin quote referenced in the section defining The Reasonable Person Principle:
"Have you ever noticed when you're driving that anybody driving slower than you is an idiot, and anyone going faster than you is a maniac?"
I'll include the canonical quip by Laplace:
"The theory of probabilities is at bottom nothing but common sense reduced to calculus; it enables us to appreciate with exactness that which accurate minds feel with a sort of instinct for which ofttimes they are unable to account."
-Introduction to Théorie Analytique des Probabilitiés
Hm. That guy writing the intro was extremely optimistic about human nature. Most folks find statistics very unintuitive. Folks routinely overestimate extremely unlikely events (lotto) and underestimate near-certainties (auto accident). Its the idea that we might get away with something, or maybe it won't happen to me, that guides most human activities.
Its this warping of the probability curve that is responsible for major facets of human society: war, witchcraft, gambling, marriage, on and on.
I think those problems are a little different. For example, no matter how you approach it, from a rational point of view joining a war for your country makes no sense. You are highly likely to die and highly unlikely to change to outcome of the war.
But if you think in an evolutionary sense, if each individual in a tribe acted rationally, we'd get a sort of Tragedy of the Commons: the enemy tribe wins the war, and slaughters your men. So it may happen that if each individual acts rationally the group, each individual included, suffers; so it makes sense individual sense to develop mechanisms to go to war, which by symmetry must include yourself.
Even less intuitively even if the probability of death is lower by not going to war (no tragedy of the commons in traditional sense), which means no individual would rationally choose war or develop mechanisms to force going to war, it would be rational for the tribe to choose war, in an evolutionary sense, provided many more of your men are killed than the enemy tribe (so they do better in evolutionary terms).
I know I've heard this before, and I've not taken an evolutionary biology class before so I do give a lot of weight to the idea that there is little influence at the group level, but one of the first paragraphs states: "As of yet, there is no clear consensus among biologists regarding the importance of group selection."
In particular, when selecting for war, there's a clear individual level selection pressure: if you don't cooperate with your group, you likely die. This is a just so story, I have no idea how irrationality actually developed, but say you have small groups of hunter gatherers and one individual develops a preference for cooperating in raids against their best interest. If this means that on average you now have two versus one whenever this individual participates, there's a clear competitive advantage which could allow that gene to propagate. The key here in this hypothetical is that the mutation occurs at the individual level in a lone individual, who then cooperates with someone else who wouldn't necessarily follow them, but is happy to have the help. This results in a disproportionate gain (100%) in effectiveness vs lone opponents. As time evolves, the gene becomes more widespread by its early disproportionate effectiveness and groups that fail to cooperate are killed on average, thus eliminating individual competitor genes.
Again, I have no idea how war actually evolved, but it seems easy to believe that when a trait influences whether one group kills another, that it would cause the killers to have a reproductive advantage. I'd love someone more educated on this topic to send me up though.
Between the skinner box gaming industry and web, and the circus that is politics and war, the fact that we are both unabashedly exploiting this and unabashedly being exploited by this, makes me kind of pessimistic.
> Studies have shown that reward uncertainty rather than reward per se, will magnify mesolimbic DA, both in monkeys (Fiorillo et al., 2003; de Lafuente and Romo, 2011) and healthy human participants (Preuschoff et al., 2006). In PG, accumbens DA is maximal during a gambling task when the probability of winning and losing money is identical—a 50% chance for a two-outcome event representing maximal uncertainty (Linnet et al., 2012). Although non-dopaminergic neurons might also be involved in the coding of reward uncertainty (Monosov and Hikosaka, 2013), these results based on electrophysiological and neuroimaging techniques indicate that DA is crucial for the coding of reward uncertainty. This suggestion is corroborated by a large number of behavioral studies, showing that mammals and birds respond more vigorously to conditioned cues predicting uncertain rewards (Collins et al., 1983; Anselme et al., 2013; Robinson et al., under review) and tend to prefer an uncertain food option over a certain food option in dual-choice tasks (Kacelnik and Bateson, 1996; Adriani and Laviola, 2006), sometimes despite a lower reward rate (Forkman, 1991; Gipson et al., 2009). According to Greg Costikyan, an award-winning game designer, games cannot hold our interest in the absence of uncertainty—which can take many forms, occurring in the outcome, the game's path, analytical complexity, perception, and so on (Costikyan, 2013).
Sorry, but the Carlin quite has always bothered me. No one actually feels the way that he's mocking. Drivers don't regard everyone slower as an idiot, only those who also unnecessarily hold up traffic. (eg camping the passing lane) And they don't find all faster drivers to be maniacs, only those who also make dangerous near misses.
(And yes, I'm sure your interpretation of the joke was always consistent with the above, but some people consider it an actual insight in addition to being funny.)
To be sure, the general point is sound, but that is a really bad, false example. A better version might be "Ever notice how when you're angry, it's because of a legitimate injustice, but when someone else is angry, it's because they're just in a bad mood?"
Peter Norvig often covers really interesting stuff. A long time ago here on HN I read his article on the probability of there being no set in the card game SET [1]. I had never heard of SET then, but picked it up based on the article. It's a simple game, but has interesting properties. Analytical answers are hard to find for a lot of the cases, so simulation to the rescue. I ended up running more simulations than Peter did, with some interesting results [2].
Also, the interview with Peter Norvig in the book "Coders at Work" [3] is great - one of my favorites in the book (actually, the whole book is great).
Another interesting paradox in probability is Bertrand’s Paradox[1]. Given an equilateral triangle inscribed in a circle, what is the probability that chord of the circle chosen at random is longer than the side of the triangle? (Hint: The answer depends on your method of selecting a random chord.)
For me the most messy part is being allowed to rotate the triangle to match it to one of the described cases.
Normally, if you assume triangle does not move, and then draw a random chord, then depending on relative position of the chord and triangle you use one of three methods described by Bertrand to judge whether it is longer or shorter than triangle side.
Then you need a way to count probability of each of those cases to happen and then count the weighted average of all three "paradoxical" cases.
EDIT: in fact I would just bet on Method 3 as a correct solution, because it is the most general: it does not require rotating the triange.
If you are rotating the triangle then you are in fact changing problem definition during solving it, which leads to some hidden assumptions (i.e. Method 2 is only good for chords which are parallel to one side of triangle, and Method 1 is only good for chords that do not cross with triangle sides)
Actually, as the wikipedia entry mentions, one can argue that Method 2 is the correct one: it is the only one respecting reasonable translation and scale invariance constraints.
Man, that's just beautiful - I love reading his notebooks. Just skimming them yields unexpected little gems like "the value of money is roughly logarithmic", which I had never heard before and gets me thinking about progressive taxation.
The final step of the St. Petersburg paradox, which he took far further than I had ever seen before, might be implementing the Kelly Criterion, where the amount you bet is related to the size of your own personal bankroll, but I don't know offhand how to relate that to a probability distribution since the Kelly Criterion is normally calculated with one odds value and one payoff value. At the least, I'd like to see a calculation of how many times you'd have to place the bet before you could be reasonably sure of a positive payout - you could do that easily enough with a monte carlo simulation I guess.
Good point, tunesmith, I should define a version of `util` that lets you start at your current personal bankroll, rather than at 0, and show how the amount you are willing to pay to play depends on how much you already have.
My TLDR summary: all probabilities are conditional. If you have two probabilities for the same event and they aren't the same, you're just assuming different priors in the statement of the probability. (and if you can't find any different priors, maybe you've discovered something new)
If you feel comfortable that the answer to the Monty Hall problem is "switch", you might want to test your understanding to be sure you reach the correct answer here as well. As Wikipedia states, Three Prisoners "is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively". But mathematically equivalent may not make it intuitively equivalent, and the answer may feel like it contradicts the correct Monty Hall answer.
Wikipedia points out that it's originally from a 1950's Martin Gardner column, but I came across it in the textbook "Stastistical Inference" by Casella and Berger. Here's C&B's phrasing:
Three prisoners, A, B, and C, are on death row. The governor
decides to pardon one of the three and chooses at random the
prisoner to pardon. He informs the warden of his choice but
requests that the name be kept secret for a few days. The
next day, A tries to get the warden to tell him who had been
pardoned. The warden refuses. A then asks which of B or C
will be executed. The warden thinks for a while, then tells A
that B is to be executed.
Warden’s reasoning: Each prisoner has a 1 in 3
chance of being pardoned. Clearly, either B or C must be
executed, so I have given A no information about whether A
will be pardoned.
A’s reasoning: Given that B will be executed, then either A
or C will be pardoned. My chance of being pardoned has
risen to 1 in 2.
Who is right?
In what I found to be a parody of textbook tropes, C&B begin their explanation "It should be clear that the warden's reasoning is correct..."
While it's true that the warden's reasoning is correct, leading off with "it should be clear" seems cruel. Here's a slight variation of the Monty Hall problem, one of the most famous "paradoxes" of popular statistics, and you are going to start with a paraphrase of "it should be obvious to the reader" without the slightest sense of irony, even though this variation produces an answer superficially incompatible with the better known problem? Ah, the strange humor of textbook authors! The remainder of their answer (which is solid) can be found here on Section 1.3 page 22: http://people.unica.it/musio/files/2008/10/Casella-Berger.pd...
It's always great to see these notebooks from Peter Norvig. There's also some really interesting implementations (in several languages) of probabilistic algorithms on the website [0] of the book he co-wrote on artificial intelligence. Great book and great material.
Great advice on assuming other people are also reasonable, and try to understand the problem they are trying to solve.
Years ago, Peter Norvig was a tech editor for an AI book I wrote. I realized two things: he has a clearer way of analyzing things than I do, and also that I was very fortunate that he ended up spending several hours helping me with his review material.
Sorry, that was over ten years ago, and I can't find his review notes so I can't really answer your question except to say that I remember his comments to be very useful.
The Python code is mostly clear and beautiful. Just one small wish: please use collections.deque() instead of showing an O(n) insertion into a list using "deck[0:0] = [card]".
You're right! 2.4 was out when I wrote that, so I could/should have used dequeue. But I wanted to use things that I thought readers would be familiar with, and timing wasn't a big issue.
"Problem 3. One is a boy born on Tuesday. What's the probability both are boys?" - this is not a problem at all. The fact that one boy is born on Tuesday is completely irrelevant as the sex of the child does not depend on day of the week. That information is given only to mess with readers mind. So, if you ignore that Tuesday part then you can reduce Problem 3 to Problem 2.
Not really - like Norvig said, that's one interpretation, but there is another valid one.
Say you had a bunch of families with two children each. The children are evenly distributed in terms of gender and the the days of the week on which they were born. If you pick one parent from the crowd, the chance that they have at least one boy is 3 / 4, the chance that they have two is 1 / 4, and the chance that they have none is also 1 / 4:
| B| G|
--|--|--|
B |BB|BG|
--|--|--|
G |GB|GG|
Each of the four squares on the above table is equiprobable. However, if the person says they have at least one boy, they must be in either the left column or the top row, so in one of three squares. There is only one square in those three with both boys, so the chance of that parent having two boys is 1 / 3.
Now, for the day-of-the-week problem. If you ask a parent if they have a male child born on Tuesday, it is not equiprobable that they're in any of those three squares. In the BG group, they all have male children, so the chance that any parent chosen has a male child born on Tuesday is 1 / 7. Similarly in the GB group. However, in the BB group, either one of their children may be born on a Tuesday to satisfy the condition. The chance that either child is born on a Tuesday is the same as the inverse of neither child being born on a Tuesday, or
1 - (6/7 * 6/7) = 13/49
So the number of parents in the top left group (BB) who satisfy the condition is 13 / 49, whereas the number of parents in the top right (BG) is 1 / 7, and the bottom left (GB) is also 1 / 7. You're looking for the probability that a given parent in that satisfies the condition is in the top left group, which is
For me both problem 2 and problem 3 should also be 1/2 .
The variances are due to how people interpret the outcome space .
It is right that probability = favaroble out comes / total possible outcomes.
In problem 2, in my opinion the possible outcomes are not how it was suggested in the post but as below. When a family has 2 kids , the below are the only possible outcomes
1) Potential Outcome 1: Both are boys
2) Potential Outcome 2: One boy and one girl
3) Potential outcome 3: Both are girls
The 3rd one is not a legal potential outcome in our particular constraint of problem 2, since problem #2 statement already states 'at least one is a boy'
So Total possible legal outcomes = 2
Favorable out comes for our event (both boys) = 1
So probability for Problem 2 = 1/2
Similarly for problem #3, I think the post unnecessarily complicates the calculation of problem space. The fact that 'Tuesday' is mentioned is irrelevant in my opinon, if you state the problem #3 in a different way that is more clearly understood.
There are 14 baskets labelled as follows "Sunday Boy", "Sunday Girl", "Monday Boy", "Monday Girl",....."Saturday Boy", "Saturday Girl". A stork came and dropped 2 babies. One baby was dropped in "Tuesday Boy" basket. What is the probability that both are boys?
Now the total outcomes and favarable outcomes are :
Total possible outcomes = Number of ways second baby could have been dropped = 14 possible baskets = 14
Favorable outcomes = second baby dropped in 'boy' basket = 7 possible baskets = 7
If you understand problem 2, then you will get problem 3 as well, so let's focus on 2.
> It is right that probability = favaroble out comes / total possible outcomes.
No, not actually: It is only right if all outcomes are equally likely! (There's an old joke about the guy who has a 50% chance of winning the lottery, since either he will win it or he won't.)
In particular, you make that mistake here:
> 1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls
These three outcomes are not all equally likely. Outcome 1 has probability 1/4, outcome 2 has probability 1/2, and outcome 3 has probability 1/4. (This is if you assume that each child has a half chance each of being a boy/girl.)
Norvig gets rid of this problem by listing out all four possible outcomes, which are all equally likely.
Again, both Norvig and you are messing with ordering of events. If you list 4 events like this, assuming that ordering of boys and girls matters, then you should stick with that. So, if ordering mattered when children were born it should also matter when you are doing "checks". So you shouldn't formulate problem as "one of the children is boy", you should formulate problem as "first child is boy" (with ordering in place), which eliminates possibility 3. Otherwise you are "solving" problem by listing sample space of completely different problem.
The event "at least one child is a boy" is well-defined on the 4-state sample space. It is the set of events (first boy/second boy, first boy/second girl, first girl / second boy). It has probability 3/4.
When you say "1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls" you are right that those are three potential outcomes, but they are not equiprobable outcomes. One boy and one girl is twice as probable each of the others.
By that reasoning, in Problem 2 the fact that one child is a boy cannot affect the sex of his sibling so you can reduce Problem 2 to Problem 1. The point that Norvig is making is that there are other reasonable interpretations corresponding to different sample spaces (and therefore to other probabilities).
But if your sample space is just Cartesian of {B,G} and {1,2,3,4,5,6,7} then it is in fact just a noise. If you had some additional information, that, say, on Tuesdays there are two times as much boys born as girls, only then expanding sample space would make sense.
Also, for interpretation of problem 2a there is one fundamental flaw that I can see: The sample space listed as:'BB', 'BG', 'GB' is wrong, because 'BG' and 'GB' are in fact the same sample. Or, to put it in a different way: if you decide that ordering does not matter then you should list either 'BG' or 'GB', not both. And if order does matter then you should list 'BB' twice, for all the possible orderings. Both of which will make probability equal to 1/2.
EDIT: Moral of the story is: "If it looks like paradox you are doing it wrong". Which I think was the Norvig's point from the start.
"The sample space listed as:'BB', 'BG', 'GB' is wrong, because 'BG' and 'GB' are in fact the same sample."
Flip two coins. Exclude the case where both are tails. What is the probability the two coins are different? If 'BG' and 'GB' are "the same sample", is it also the case that 'HT' and 'TH' are the same sample?
When the contestant picked a box he was making a random choice between three boxes and had a probability of 1/3 of picking the car. The probability that one of the other two boxes was the car is thus 2/3. When Monty Hall opens a box it's still the case that the probability of the car being in the other two is 2/3 but the contestant can now eliminate the open box and knows that the probability that the car is in the other box is 2/3. Thus it's worth switching.