The original article is more interesting and sheds light on the hidden assumptions that go into the calculation. It also considers under which sets of assumptions the answer is not 1/2.
> Now, we should mention at this point that we are discussing mathematical children who have no existence outside the world of probabilistic brainteasers. So we’re not worrying about the fact that a disproportionate number of children are born on Mondays and Tuesdays, since C-sections aren’t usually scheduled for the weekends. We’re also not worrying about twins or triplets. Or any other sort of “real world” consideration that might occur to you.
The one that occurred to me is that the probability of getting a boy or girl is not 50%, and the probability of getting two same-gender children is not 50% either. Here are some nice real stats: http://www.in-gender.com/XYU/Odds/Gender_Odds.aspx
I also thought the original article was better, and clearer. As you say—it's the hidden assumptions which are interesting.
In maths exams at school the problems are always presented in the form of: "a random family was chosen from ...". (Similar to "a smooth cube is resting on a smooth surface".
In real life, it's more difficult. The original article looks at real life, whereas the 300-word version has reduced it to the exam version. Of course it's "easier" to understand the exam version, but this wasn't the point.
What I dislike about this question is the flavor text. It seems designed to mislead the reader by invoking their “realism node,” and then requiring them to suspend it.
From archgoon’s link:
> "This puzzle confounds people legitimately, however, because most of the ways in which you are likely to find out that X has at least one boy contain an implicit bias which changes the answer.”
Exactly right. The phrasing in the problem asks the reader to consider a particular scenario, not isolated facts. Consider what happens if you take the question at its word:
> If a person on the street says to you, "I have two children," what is the probability that they're both boys?
Hell if I know. Maybe one is a boy and the other is a grown man? Or maybe he has five children, BBGGG. “I have two children, both of which are boys” is technically correct if there’s two boys and three girls, right? Or maybe he’s just outright lying! How do I assign a probability to any of those?
Consider this phrasing instead:
“If you flip a coin twice, and the first and/or the second is heads, what’s the probability that both are heads?” I’d expect more people to get this right than the boy/girl question.
Also, as an aside, punctuation is important:
“I have two children. At least, one of them is a boy.”
"I met a man on the street, he said he had two children. I asked him, is at least one of them a boy? He said yes. What's the probability both of them are boys?"
This removes the whole ambiguity of why he would assert "at least one of them is a boy," but still makes the question a bit tricky/counter-intuitive, which is useful for teaching the quirks of probability.
This is incorrect. There's no reason to assume that the stranger will say, "at least one of them is a girl" only in the case of GG.
In the case of BB, the stranger will always say, "At least one is a boy." Given GB or BG, they have even odds of saying "at least one is a boy" vs "at least one is a girl." Given GG, the stranger will always say, "At least one is a girl." The correct diagram would be this: http://abughrai.be/pics/boy_probability.png
For the probabilities in that post to apply, the stranger would say, "I have two children." Then you ask, "Is at least one of them a boy?" and they answer, "Yes."
Edit: Of course, the answer to, "Given there are two siblings and one is a boy, what are the chances both are boys?" is 1/3. But the phrasing of the puzzle does not reflect that question.
Fine, then I pose to you the same problem description, with the added fact that the stranger is proud of having a boy, will only draw attention to a boy and will thus say "at least one is a boy", independent of whether the other is a girl, while he would never say "at least one is a boy".
You can't solve a logic puzzle by arguing the psychology of a persona depicted in the puzzle.
You can’t solve a logic puzzle that way, but you can explain the pragmatics of why the “logical” answer is contrary to our expectations based on interaction with real people.
The blog post[1] that the author based his short explanation on goes into this. Peter Winkler, a math professor at Dartmouth, is quoted in it[2]:
Suppose the puzzle is phrased this way: X says “I have two children and at least one is a boy.” What is the probability that the other is a boy?
Put this way, the puzzle is highly ambiguous. Computer scientists, cryptologists and others who must deal carefully with message-passing know that what counts is not what a person says (even if she is known never to lie) but under what circumstances would she have said it.
The way the puzzle is phrased, the only way to get 1/3 as an answer is to assume that parents with one girl will never say, "At least one of them is a girl."
Thanks for the clarification, though I think my point stands.
> The way the puzzle is phrased, the only way to get 1/3 as an answer is to assume that parents with one girl will never say, "At least one of them is a girl."
Right. But that is assumed, because the puzzle states that the parent says, "At least one of them is a boy."
In other words, "Put this way, the puzzle is highly ambiguous." No, not it is not. It is absolutely not ambiguous.
I made this point in a different comment for a different blog post with the same example, but: both things you link to contain a fallacy, as follows.
"Suppose we know that a certain man has two children and we also know that the older one is a boy" IMPLIES "Now suppose we know simply that a man has two children and that one of them is a son. This time we would reason that there is no possibility that the person has two girls. It follows that the sexes of his two children, ordered from oldest to youngest, are either BB, BG or GB...."
You can't conclude, as these authors do, from the first case, that the other child's gender is 1/2, and from the second, it is 1/3. It's not that the two cases are different (in any important way); they are not; one implies the other.
It's that in the 1/2 case, the person is naive, and in the 1/3 case, the person takes into account the known fact that there is a 1/2 chance per child of getting a certain gender, thus allowing the BB/BG/GB/GG analysis.
Afaict tell the initial range of possibilities is:
GG GB BB
The relative ages of the children play no part, they could be twins for all I care. So the initial probabilty of being BB is 1/3. Knowing one of them is a boy then removes GG as an option, and the probability for BB unsurprisingly becomes 1/2.
You only misunderstand because the author is presenting it stupidly (as you state, incorporating order where order is irrelevant), leading to his own stupid conclusion. Read my post elsewhere in the thread for the full explanation, and if you find this kind of thing interesting, look up the Monty Hall problem (the point of which is that changing the information changes the odds).
What occurred to me is that as more conditions are added to the question the answer seems to tend toward the naive assumption of 1/2 (in his explanation 0.25, 0.33, 0.42, 0.48). This is like a strange cross-over between word problems, probability and unbelievably calculus. Essentially:
lim[generic -> specific] f(P("is one child of two a boy")) = 1/2
Obviously I have no proof for it but it is an interesting observation.
If we're going to bring reality into this, I think we have to take into account the fact that only very weird people tell you that "at least one child is a boy born on a Tuesday" and we have to consider the possibility that they are lying in order to test your knowledge of probability.
And if we're really going to bring reality into this, then we'll need to consider the ages of the children, because of the at-birth sex ratio (105 newborn boys for every 100 newborn girls[0]) and differing mortality rates for each sex.
I think it's probably good to be precise in this case. It should state that "at least one child is a boy born on a Tuesday, but one is older than the other by at least a day".
This explanation ignores the two perspectives of probability - assertions on states of the world (the "frequentist" view) vs. states of the mind (the "bayesian" view).
As far as I can tell, that blog post contains a "bait and switch." Here it is:
> But suppose that instead you had asked, "Is your eldest child a boy?" and the mathematician had answered "Yes." Then the probability of the mathematician having two boys would be 1/2. Since the eldest child is a boy, and the younger child can be anything it pleases.
No, it cannot be "anything it pleases." That is the whole point of the blog post that we are all commenting about here on HN, which is re-told on the blog post you are linking to.
Since the post you are linking to is misrepresenting one side of the "argument," I strongly suggest that there is no actual apparentl contradiction here that needs to be resolved.
Yes, you can talk about the actual probability of something vs. making the best possible prediction you have. A good example raised in this post is the point about a biased coin; you know it is biased but not which way, so you assign odds of .5 for heads and .5 for tails. Fine. This is not some massive philosophical conundrum.
The fact is that actual knowledge is neither purely "out there" (e.g. "in the cards") nor purely "in here" (in the mind). This is the classic rationalism vs. empiricism debate, which was settled by Ayn Rand's theory of objective knowledge, where she answers that knowledge is in man's mind, but it is knowledge of external reality. So, "both."
Not understanding this, and other basic epistemological errors, seem to plague lesswrong.com. It is embarassing to see them supposedly championing "reason" under a non-objective epistemology. They come to bizarre conclusions, and that is, at least partially, why.
Well, there are only five reasonable ways to disagree.
(1) Do you hold that concepts and principles are purely "in" the referents?
(2) Do you hold that they are purely "in" man's mind?
(3) Do you hold that they are "in" man's mind, but refer to properties of the referents? (Rand's conclusion)
(4) Do you hold that Rand's argument and conclusion is correct, but somebody else made it earlier?
(5) Do you hold that there is insufficient evidence to answer the question (which then raises questions about why you would even engage in an intellectual discussion)? For example, you could "argue" for a subjectivist view of knowledge.
An actual answer along the lines of 1-5 would be useful, but stating "I don't think so" does not provide any reasoning, and thus is utterly devoid of intellectual content.
You seem to be dismissing a theory without considering and responding to its evidence, which is fraudulent science. Nobody wants to see that kind of "argument," even people who nonetheless do agree with the conclusion.
I do think you have the intellectual right to merely state that you disagree, but you have to say that you don't intend to defend it, or let yourself be subject to having it pointed out by someone, as I am.
That is because too many people today hide behind the assumption that agreement or disagreement without evidence presented is enough to establish or disestablish truth. As I said before, that is fraudulent science.
I responded—note that this was completely spontaneous—"What on Earth do you mean? You can't avoid assigning a probability to the mathematician making one statement or another. You're just assuming the probability is 1, and that's unjustified."
While the equations of probability theory are agreed upon, there are multiple interpretations[1]. In some ways, it's like QM.[2]
In oversimplifying your table for potential idiot readers by introducing the silent variable of older vs younger child, you have introduced order into a statistics problem wherein order is irrelevant and masked the fact that what matters is which of the children you already know to be a boy. You grant this difference for the BG pairing, thus producing two options, but not for the BB pairing. There is not ONE way of knowing this for BB, as you present in the table, but two: you know the older is the boy, or you know the younger is the boy. 2/4 = 1/2. QED.
What you should have done in the first place, however (unless your aim is to produce a blog post which, apparently, can convince otherwise intelligent people that irrelevant information can magically become relevant) was simply remove the one boy from the equation and reformulate the question. What is the probability that this other child is a boy (and thus that both are boys)? 1/2. Fuck the table. Had you not used a table, this would never have happened. But it feels authoritative, right? Thank you for the psychology lesson.
.
.
.
Downvotes but not refutations, because there aren't any. I'll assume it's my tone. My opinion of the competence of the average reader on this site has plummeted reading the other comments, however, so hey, fuck you guys too. :)
Your reasoning works if the person says "The older is a boy." You claim that this is correct.
Consider ...
Assuming the sex of a child is 50:50, then consider all couples with exactly two children. Of those, consider only those who can truthfully say "at least one of my children is a boy." Of those couples, 1/3 will have two boys.
Interpreting the initial problem in this manner, the answer of 1/3 is right. What I'd like to know is why you think the initial problem should be interpreted in any different way.
> ... in the table, he DOES tell you that in the BG case.
> Otherwise there aren't two spots; there's only one. He
> introduces the older/younger thing, then fails to apply
> the new information across all cases.
I'm having real trouble understanding you here. The explanation lists all the possibilities, and to do so it's necessary to distinguish between the children. The most obvious way to do that is to talk about the older and younger.
we know that in families with two children about half the time you have one of each sex. You only get that if you distinguish between the children in some sense so that there are four overall possibilities: BB, BG, GB, GG. If you don't distinguish between the children then there are only three possibilities: Both boys, both girls, one of each. Doing real world trials clearly shows that model to be flawed. We must distinguish between the children when enumerating cases.
Forgive me if this is all obvious to you, but I honestly can't see your argument, so it's necessary to lay down much more detail to try to find where your reasoning varies from mine.
So now consider the situation I laid out. Take all families with two children. There are four equally likely possibilities. Eliminate those who cannot truthfully say "At least one child is a boy." You are left with three equally likely possibilities. In only one of those do we have two boys. Thus one out of three possibilities has both children boys.
The probability of both children being boys is 1/3.
Can you explain where that reasoning is faulty?
Also, computer simulations clearly show that under this model of what's going on, the chance of two boys is 1/3. If you explain the faulty reasoning, you'll also need to explain why the computer simulation gives the same answer.
Yeah, I actually deleted it before seeing your response because I thought it was unclear. My bad.
>The most obvious way to do that is to talk about the older and younger.
Exactly. The father doesn't talk about older and younger, but the table does. What it should look like, because you're correct that age is irrelevant: B (given) * [b|g] -> [Bb|Bg], or if you account for age: [B is first|B is second] * [b|g] -> [Bb|bB|Bg|gB]. But you have to pick one or the other! Either way, it works, but you have to account on both ends, not just one. He's calculating for BG as if age were relevant, but it's not.
The computer "simulation" is wrong because it is grounded in the logic that one child's sex wasn't a given all along, when in fact it was. NB: if you're throwing results out, that's a bad sign.
Yea/nay?
Or, again, you can just step back and say: X is a boy. What is the probability that Y is a boy? Why on earth would it matter what X was? Do you meet 2/3 women because you are a man?
After enumerating all the options and their respective probabilities, yes, you discard all those that are inconsistent with the information to hand.
You bring up the Monty Hall problem. The options there are:
[ c ] [ g ] [ g ]
[ g ] [ c ] [ g ]
[ g ] [ g ] [ c ]
These are all equally likely.
Then you choose a door. Let's assume you choose door 1. To keep things equally likely, let's suppose Monty flips a coin. Heads he opens the left-most unchosen door that doesn't reveal a car, Tails he opens the right-most. Now we have six options:
Heads:
[ c ] [ g*] [ g ]
[ g ] [ c ] [ g*]
[ g ] [ g*] [ c ]
Tails:
[ c ] [ g ] [ g*]
[ g ] [ c ] [ g*]
[ g ] [ g*] [ c ]
As you can see, there are now 6 equi-probable choices, and in 4 of them it's better to switch.
A: So now let's take another experiment. I tell you I'm going to flip two coins until at least one of them is heads. I do that. Now I ask you to bet on whether or not there is a tail. What do you think are fair odds?
B: Now I hunt among couples with two children until I find one that doesn't have two girls, and I ask: what are the odds they have two boys?
C: Finally, I hunt among couples with two children until I find one that has at least one boy, and I ask: what are the odds they have two boys?
This last is how the question is usually interpreted.
There are three options.
1: Your answers to A, B, and C are not all the same;
2: They are all the same, and not 1/3;
3: They are all the same, and all 1/3.
If your answers to A, B and C are not all the same, I'd like to know why. If they are not all 1/3, I'd like to play game A with you. If your answer to C is 1/3, then you've agreed with the usually interpretation and contradicted yourself.
> To keep things equally likely, let's suppose Monty flips a coin.
Aha! But he can't! Do you see? He can't open a door with a car, so he can't flip a coin in the event that the car is behind a door other than yours. He is bound by the rules not to do so.
1/3 your door is good; 2/3 your door is bad. If bad, he MUST leave you with the good door to switch to. So in 2/3 of scenarios, you have a 100% likelihood of winning if you switch.
The answer to all 3 is 1/2. GG + "at least one is a boy" is not a possible world, but you are calculating the probability of the others, initially, as if it were, and then asking another question on top of it. You can't do that. BB is only 1/2 as likely as BG/GB if GG is a possible world for the question to begin with.
>> To keep things equally likely, let's suppose
>> Monty flips a coin.
> Aha! But he can't! Do you see? He can't open a
> door with a car, so he can't flip a coin in the
> event that the car is behind a door other than
> yours. He is bound by the rules not to do so.
Read again ... I said:
> To keep things equally likely, let's suppose
> Monty flips a coin. Heads he opens the left-most
> unchosen door that doesn't reveal a car, Tails he
> opens the right-most.
So I allow for that. I have him always flip a coin to ensure that every branch of the game tree is equally likely, even though in 2 out of 3 cases it doesn't actually affect his choice of door. It's a standard trick that is proven to work in the real world. More, it's supported by the appropriate measure theory versions of probability.
Please, stop assuming everyone except you is an idiot.
I did an experiment: I kept flipping two coins [1]. If it came up tails, tails I ignored it. If it came up heads, heads I noted it down in one column. If it came up heads, tails I noted it down in another column. After many flips the ratio was 1:3.
As far as I can tell this is an accurate simulation of the problem. We have two things, coins or children, which have a 50:50 chance of being one value or another. We are told one combination of values isn't the case, so we discount them from the simulation. We then use the frequency of the target combination to approximate its probability. We haven't added any artificial ordering into the problem.
I know it's not well argued logical reasoning, but it supports the articles approach. Can you point out where I tripped up in the simulation?
[1] Of course I didn't actually sit there flipping coins. I wrote a program to do it.
(let [sample-size 10000000] (/ (get (frequencies (take sample-size (filter #(not= % [:girl :girl]) (repeatedly #(vector (rand-nth [:boy :girl]) (rand-nth [:boy :girl])))))) [:boy :boy]) sample-size))
Instead of actually programming, I flipped 2 coins this morning and left them on my desk. They are still sitting there. Either HT, TH, HH or TT
I just remembered one of them is tails, so is there is a 1/2 chance that they are both tails now? Nope it's 1/3 chance they are both tails. Either HT, TH, or TT.
Of course that's a much simpler way to simulate the problem, but then you get people trying to argue that HT and TH should be collapsed into a single state.
By using a simulation with repetition it's easier to pose the states as 'TT' and 'not TT' and use frequencies to show 'not TT' occurs more frequently than 'TT'. I find this starts to help convince people that the states shouldn't be collapsed, even if it doesn't explicitly explain why.
>We haven't added any artificial ordering into the problem.
Yes you have! To run the experiment, you have to treat TH and HT as different but you "pretend" not to do so by adding them into the same pile. You're pretending to know that the father tells you which child is the "known boy", but by the statement of the problem, you don't know that. Why is this an error? Because the father must follow a rule - he must give you a boy, and there are two ways for him to do that in BB but you only recognize one. If you go the "table-minus" route, which both the author and your experiment do. But you can't do that without affecting the odds! By telling you they aren't both girls, he lowers the chance that either of them might be. Yes, I'm serious.
Correct formulation of table:
a) BG; b) GB; c) BB and he gives you the first; d) BB and he gives you the second.
Isn't the probability of c and d half that of a and b? If BG is the case, there's 100% chance he's talking about the first child. If BB is the case, there's a 50% chance he's talking about the first one, and a 50% chance he's talking about the second one.
BG, giving the first - 1/3
GB, giving the second - 1/3
BB, giving the first - 1/6
BB, giving the second - 1/6
[Edit] You're right. Tables can be a bad way of thinking about it, because you have to make sure each option has an equal probability for them to work. This is okay when you have a single variable (like in the original post), but gets tricky when you introduce a new variable (as you did: is the father thinking about the first or second). Once you get past one variable, trees are probably a better but still simple representation.
The problem is the interpretation of the question in precise terms. We are given some background information and then told the observation of a particular random variable. In the given article, the random variable is interpreted as "yes/no: at least one child is a boy born on a Tuesday". Rambling about "order" does not change the fact that the article is correct, given the definition of the random variable that is observed.
The problem is that this interpretation is only correct if you imagine that you are always given the observation of the same random variable no matter what the genders or birthdays of the children actually are. For example, if I had two girls born on Wednesday, I would then tell you "no child is a boy born on a Tuesday". That's an unlikely scenario which is what makes it legitimately surprising.
It's like the Monty Hall problem. What if the host picks the door in advance? What if the host only opens a door with a goat under the condition that you picked the door with the prize? What if the host only opens a door with a goat under the condition that you picked the wrong door?
The article has problems but "wrong" is not one of them.
> so hey, fuck you guys too
Hacker News is a terrible place to visit if you want to take the voting system personally. Don't make this about you. It's about the post.
As I see it, there are three options - either it's a boy-boy pair, a girl-girl pair or a boy-girl pair. Now, since OP introduced the age variable, let's give the kids names to make things easier to understand. Let's say that the boys are named John and Jack, the girls Jill and Judy and if it's a boy-girl pair then they're named Jack and Jill.
Now if it's a boy-girl pair, we have exactly two options:
either Jack is older than Jill or Jill is older than Jack. This is shown in the first picture.
However, we also have two options for boy-boy and girl-girl pairs: Either Jack is older than John or John is older than Jack. Either Jill is older than Judy or Judy is older than Jill. This is not shown.
Since we know that one of them is a boy, we can dismiss both girl-girl pairs. Now we're left with four options. In 50% of these options the other child is a boy.
The reason age was introduced was simply was a way of differentiating the two children (giving them identity). You introduced names to differentiate the two children, which is another perfectly good way to do it.
But then you _also_ introduced age. Because you now have two different ways of labelling the children, you've introduced a new variable, and that changes the probability.
A better way to look at it is with two unisex names: Drew and Sam. This completely replaces the age thing as a way of identifying them, so forget all about age.
Our possibilities are:
Drew: boy, Sam: boy
Drew: boy, Sam: girl
Drew: girl, Sam: boy
Drew: girl, Sam: girl
We know Drew and Sam can't both be girls, because at least one is a boy, so that leaves:
Drew: boy, Sam: boy
Drew: boy, Sam: girl
Drew: girl, Sam: boy
There's a 1/3 chance that both Drew and Sam are boys.
> As I see it, there are three options - either it's a boy-boy pair, a girl-girl pair or a boy-girl pair.
Yes, but those three options are not equally likely.
Let's say the parents decide that their firstborn will be named Jack if male and Jill if female, and the second child will be John if male and Judy if female. Now you see that there are four possibilities: Jack+John, Jack+Judy, Jill+John, Jill+Judy. The last one is eliminated when you know that there is one boy, which gives 1/3 chance that it's the first.
You can't give a correct answer to an incomplete question. Sit down and define the problem and if you come up with a different answer it's because you made a different assumption about the definition of the observed random variable.
Let's suppose there are four universes.
* Two boys (25% chance)
* Two girls (25% chance)
* Boy and girl (50% chance)
If we assume that the observed random variable has a preimage that contains all possibilities where the statement is true, then we come to the 1/3 conclusion, which is counterintuitive but true. This is the position of the article. I have come to the same conclusion without using order, but I have used the same definition of the observed random variable.
Tell me what you think the random variable is.
We could assume that the random variable has a preimage which only contains the case where both children are boys. Bam, probability 100%, not 1/2 or 1/3. Or we can assome that the random variable has a preimage which only contains the case where one child is of each gender. Bam. Probability 0%. So any answer between 0% and 100% is defensible.
God, I knew this was coming. Since you won't accept my strategy of building up to the complete question (because it is in fact your formulation which is incomplete, and thus I, like you, will not respond to it), let me try this instead:
You misunderstand Monty Hall. I would know; I learned about it by being asked the question and getting it right, not by reading someone else's explanation (which leaves room for misinterpretation). The key to Monty Hall is none of what you stated (or rather, rhetorically asked); it's that the host must pick a door that is both a) bad and b) not yours. If your door is good, then he can leave either, but if your door is bad, then he must leave the good one. So, two possibilities: the door he left was either bad or good, BUT the rule driving his choice wasn't 50/50, so the odds of it being bad or good aren't 50/50, so neither are the odds of yours, even though there are only two left. Take a minute, think about it.
Now ask yourself which "door" (read child) the father told you was a boy. There's the gotcha. Under your/the author's formulation of the question, he doesn't remove a door in the BB universe, but he does in the BG universe. There are two BG's because the table acknowledges that he can tell you which of the two is definitely a boy, and thus he has two options for doing so BUT HE DOESN'T, HE HAS ONE BUT THERE ARE TWO BG UNIVERSES, SO IT LOOKS LIKE TWO.
As mentioned in other replies, it's clear now that you are interpreting the question in a manner other than the usual interpretation, and then being incredibly rude about anyone who disagrees with you.
I have in my hand a device that lets me call up on demand most of the knowledge of the world, and I use it to look at pictures of cats, and to argue with strangers.
> Since you won't accept my strategy of building up to the complete question (because it is in fact your formulation which is incomplete, and thus I, like you, will not respond to it),
No, his formulation is complete and inevitably leads to the correct answer 1/3. You are constantly avoiding a complete formulation of the problem by diverting to a muddled discussion of an entirely different problem. This is NOT about Monty Hall.
Or you're trolling. In which case you are doing it masterfully.
Hahahaha oh my god what a response. The fact that you think you're pointing anything out to anyone with that second paragraph is just priceless.
In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second.
Your brain will of course refuse to process the above due to your earlier remark and the sweet, sweet motherfucking irony it would mean.
Right, now it's clear that you are answering a different question, and in that question, the answer is 1/2. Many people insist that to be the right question, because the answer matches their intuition, but it is a different question.
Not recognizing the difference is what lets some poker players make money.
Consider this question. A chap flips a coin in the morning and thereby chooses one of his two children. Today he will only talk about that child. Later I meet him and he talks about his son. I thereby deduce that he has at least one boy. What are the chances he has two sons? (corrected in edit - thank you)
Answer: 1/2.
Another version: I meet a man and his son, and in conversation it emerges that he has two children. I can see he has at least one son, what are the chances he has two sons?
Answer: 1/2.
Another version: I meet a man and it emerges in conversation that he has two children. I observe that he is carrying a shopping bag with a child's dress in it, from which I conclude that he has at least one girl. What are the chances he has two girls?
Answer: 1/3. (I had to change gender because I couldn't think of anything that would be specific to a boy.)
Your dismissal of the idea that there are other interpretations than your own is disappointing. The language in which you couch your dismissals even more so. Are you like this in person?
"Consider this question. A chap flips a coin in the morning and thereby chooses one of his two children. Today he will only talk about that child. Later I meet him and he talks about his son. I thereby deduce that he has at least one boy. What are the chances he has two children?
Answer: 1/2."
Do you mean "what are the chances he has two sons, not children?
By the way, thanks for your input on this thread, I'm sure I speak for many in saying that I really appreciate it!
No. In person I can see what the other party is thinking.
The two are independent variables, man. I don't know how to explain it any more clearly than that. What's tripping people up is the idea that removing GG from the possibilities doesn't affect the local probability of either particular child being girl. It does. You must reformulate. THAT's the counterintuitive part. It's just hella hard to explain, because there are too many 2's and 1/2's floating around the problem, so when you say one cancels another, people think you're not incorporating the first to begin with, so you must be one of the dolts the author is trying to learn some knowledge.
The following, you can take or leave: probability is my thing. I got Monty Hall right when I was in 6th grade. I was an AIME invitee, if you know what that is. I got 800 on the math SAT; 36 on the math ACT in freshman year. None of these things mean anything, logically speaking, I know. I say them only to encourage you (and only you, because you're putting in effort) to consider what I'm saying as fully as you're considering what the crowd is saying.
> No. In person I can see what the other party is thinking.
So when you have less knowledge about the other person you are more rude. I feel that's logically inconsistent.
> The two are independent variables, man. I don't know
> how to explain it any more clearly than that.
"The two" ?? What "two"? Your "explanations" are very unclear.
> What's tripping people up is the idea that removing GG
> from the possibilities doesn't affect the local probability
> of either particular child being girl. It does. You must
> reformulate.
Again, it's clear that you are talking about a different problem. Yes, the problem you are talking about has an answer of 1/2. No, it's not the problem we're talking about.
> It's just hella hard to explain, ...
You will find many things difficult if you don't learn how to slow down, take time, be precise, and explain clearly. Be more like Feynman - take time to explain things clearly and precisely.
> ... you must be one of the dolts ...
Ah.
> The following, you can take or leave: probability is my
> thing. I got Monty Hall right when I was in 6th grade.
> I was an AIME invitee, if you know what that is. I got
> 800 on the math SAT; 36 on the math ACT in freshman year.
Good for you. I speak for and contribute to the training of IOM teams, I've been an invited speaker at the Museum of Mathematics in New York, the Dutch National Mathematics days, Matematikbiennalen in Sweden, and more math masterclasses than I can count. This year I'm speaking in London, Switzerland, Belgium, and giving around 150 talks, masterclasses, and workshops. I have a Bachelors, Masters, and PhD.
> None of these things mean anything, logically speaking,
Right.
> I know. I say them only to encourage you (and only you,
> because you're putting in effort) to consider what I'm
> saying as fully as you're considering what the crowd is
> saying.
Hmm. Well, thanks for the encouragement. I was only saying these things to point out that I have considered these things, and have some form in the area. I forgot to mention that I was at the G4G when the Tuesday Boy problem was first mentioned, and spent much of the evening with Tanya Khovanova, Gary Foshee, John Conway, Richard Guy, and others, talking about the various interpretations. I didn't talk about it with Martin Gardner when I spent an afternoon with him, we were talking about my work.
I mention these things only to point out that you might consider that you are wrong.
Well don't I feel like an idiot for saying any of that. Why were you running a "computer simulation" of the problem? Were you, even?
>"The two" ?? What "two"? Your "explanations" are very unclear.
The only two there are: the two children. The sex of one is independent from the sex of the other.
>You will find many things difficult if you don't learn how to slow down, take time, be precise, and explain clearly. Be more like Feynman - take time to explain things clearly and precisely.
Okay, I'll try, good buddy. Monty Hall is different because it precludes upfront the possibility of there being any ratio other than 1:2. If we were, however, to say that any door could have either a goat or a car, then Monty's elimination of some other door would not inform you on the odds of your door YOUR door. Yea or nay?
The only way for Monty to give you any information at all about whether to switch is if the rule binds him to tell you something about the odds of the unchosen doors. In the original problem, this happens, because the ratio is bound 1:2. In our revised problem, this does not happen - unless he tells you he can't remove a door because neither is a goat. But! Using this as an analogue to our children problem, we've already precluded that possibility upfront by eliminating GG, thus Monty can't provide us any information because his B-G choice is unbound by any rule.
> Why were you running a "computer simulation" of the
> problem? Were you, even?
I was reconstructing what happened when Erdős was first told of the Monty Hall problem. He answered 1/2, and was very frustrated for a time. In the end he said something like "But you're not telling me the why!"
Moving on ...
>> "The two" ?? What "two"? Your "explanations" are
>> very unclear.
> The only two there are: the two children. The sex
> of one is independent from the sex of the other.
So we are assuming that this chap has exactly two children, and that their sexes are independent with probability 1/2.
> Monty Hall is different because it precludes upfront
> the possibility of there being any ratio other than
> 1:2. If we were, however, to say that any door could
> have either a goat or a car, then it wouldn't matter
> whether you stayed or switched. Monty's elimination
> of some other door would not inform you on the odds
> of your door YOUR door. Yea or nay?
Yes, agreed.
> The only way for Monty to give you any information at
> all about whether to switch is if the rule binds him
> to tell you something about the odds of the unchosen
> doors.
Yes.
> In the original problem, this happens, because the
> ratio is bound 1:2.
OK.
> In our revised problem, this does not happen - unless
> he tells you he can't remove a door because neither
> is a goat.
Yes.
> Using this as an analogue to our children problem,
> we've already precluded that possibility upfront
> by eliminating GG, thus Monty can't provide us any
> information because his B-G choice is unbound by
> any rule.
I fail to see how that is in any way relevant. There are four equally likely possibilities for the children. If we label them older first then we have BB, BG, GB, GG. Only by labelling them in some consistent order to we get equally likely possibilities. The father says that at least one child is a boy. That eliminates the option of GG. Are you somehow suggesting that the probability distribution of the other options is changed by this?
If so, how? Just as Monty's choice gives no information about the door we have selected (and which therefore remains as chance 1/3 of being the winning door), telling us that it's not the case that both children are girls gives us no further information about the other three equally likely possibilities.
A: So now let's take another experiment. I tell
you I'm going to flip two coins until at least
one of them is heads. I do that. Now I ask you
to bet on whether or not there is a tail. What
do you think are fair odds?
B: ...
C: ...
There are three options.
1: Your answers to A, B, and C are not all the same;
2: They are all the same, and not 1/3;
3: They are all the same, and all 1/3.
That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it.
>That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it.
I misread A. I agree that it's 1/3 if you flip pairs until at least one is heads. I disagree that living in a world where pairs of tails don't exist and living in a world where pairs of tails do exist but are ignored are the same thing.
>Are you somehow suggesting that the probability distribution of the other options is changed by this?
Yes. I'm saying it's an unwitting double Monty Hall. The fact that he can't give you 3 cars changes the odds. Going (invisibly) from 4 possible worlds (including GG) to 3 changes the prob dist of the remaining 3, by which I mean 2. If Monty must guarantee you at least one boy, then there aren't really 3 remaining possibilities. The unguaranteed can be a boy, or it can be a girl. What is the third?
>> The father says that at least one child is a boy.
>> That eliminates the option of GG. Are you somehow
>> suggesting that the probability distribution of
>> the other options is changed by this?
> Yes. ... Going (invisibly) from 4 possible worlds
> (including GG) to 3 changes the prob dist of the
> remaining 3, by which I mean 2. If Monty must
> guarantee you at least one boy, then there aren't
> really 3 remaining possibilities. The unguaranteed
> can be a boy, or it can be a girl. What is the
> third?
The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after.
In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases.
Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable?
> A: So now let's take another experiment.
> I tell you I'm going to flip two coins
> until at least one of them is heads. I
> do that. Now I ask you to bet on whether
> or not there is a tail. What do you
> think are fair odds?
And you say:
> I misread A. I agree that it's 1/3 if you
> flip pairs until at least one is heads.
OK. Now B:
> B: Now I hunt among couples with two children
> until I find one that doesn't have two girls,
> and I ask: what are the odds they have two
> boys?
I think I'm ready to admit being wrong and move on with my life. Thanks for your ridiculous patience, Colin. I will be leaving my comments up as a monument to idiocy. Good god am I frustrated with myself right now.
It's hard to say this without sounding condescending - please take it at face value ...
I'm pleased we got to the heart of it. I remember many times being in a similar situation as yours and I owe much to those too numerous to mention here who were patient with me. All I would say now is that in the future when you find yourself in disagreement, ask if you can learn something, don't just assume you are right. I found that easy, because I had more determination than talent, and was wrong often enough to be reminded constantly that there was still much to learn.
>Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable?
Because in Monty Hall, the effect of changing probability is produced not just via elimination but via process of elimination - which is produced only because sufficient dependencies exist for it: a) the guaranteed 1:2 ratio, b) Monty must eliminate a goat, and c) the goat he eliminates can't be behind your door. These dependencies don't exist in the children problem. Unless Monty's choice occurs under the constraint of avoiding the door we chose, nothing changes. Monty: "You door has a goat behind it" -- so what? Now both of the other doors are 50/50. No similar conflict exists in the other problem. No conflict at all exists in the other problem. Elimination occurs, but it doesn't allow you to do anything further. "No GG" is just one rule, and if one elimination rule isn't sufficient for Monty Hall probabilities to change, why here?
And elimination of what? The elimination not of a possibility, but of a probability? Does that statement even have rational meaning? I'm being dead serious - we can give it mathematical meaning quite easily, but does it even have meaning to begin with?
Yes, I apparently misread that one too. But again, I disagree on the issue of whether the form of the proposed analogue matches the original problem.
>> OK. Now B:
>> > B: Now I hunt among couples with two children
>> > until I find one that doesn't have two girls,
>> > and I ask: what are the odds they have two
>> > boys?
>> Do you agree that it's 1/3?
> Yes, I apparently misread that one too.
OK, so you agree that this is 1/3. Now on to C:
>> > C: I hunt among couples with two children until
>> > I find one that has at least one boy, and I
>> > ask: what are the odds they have two boys?
Do you agree that this is 1/3?
Let's assume for fun that you do agree with that. The original problem was:
A: I met a man who said that he has two children,
at least one of which is a boy. What is the
probability that both children are boys?
Let's compare that with this one:
B: Among all the couples who have exactly two
children, and such that at least one of their
children is a boy, what is the probability
that both are boys?
Reading this:
> But again, I disagree on the issue of whether
> the form of the proposed analogue matches the
> original problem.
It seems like you're going to claim that these aren't the same problem.
I've made it very clear exactly what my model is for problem B. If you claim that problem A is different, then you need to provide me with a very clear model for your interpretation.
> The fact that you think you're pointing anything out to anyone with that second paragraph is just priceless.
Posturing does not make your arguments more convicing.
> In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second.
How is "ways of telling" you possibly relevant to the puzzle?
You're still apparently committing the same basic fallacy of ennumerating possibilities without assigning them probabilities and basing your argument on the wrong assumption that they must be equally probable.
> Your brain will of course refuse to process the above due to your earlier remark and the sweet, sweet motherfucking irony it would mean.
That describes exactly your behaviour in every single comment here.
> In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second.
With respect to only the sentence above, why is the following table incorrect?
http://scienceblogs.com/evolutionblog/2011/11/08/the-tuesday...
The attempt at shortening it down to under 300 words omits all the complications and subtlety involved in the problem.