[kgwgk]

The odds don’t “sound equal”. According to you, they are equal (2/3).

Saying

“before opening the email I think the probability that you have one boy and one girl is 1/2 but one of two things will happen, I either find that you have at least one boy and I will conclude that the probability that you have one boy and one girl is 2/3, or I will find that you have at least one girl and I will reach the same conclusion”

is like saying

“under this cup there is either a dime or a quarter, it’s a dime the probability of heads is 1/2 and if it’s a quarter the probability of heads is also 1/2”

and claiming that the probability of heads before I tell you whether it’s a dime or a quarter is something other than 1/2 and changes always to 1/2 when I let you know what it is.

I understand the Monty Hall problem. I also understand this one.

I wrote a detailed solution here https://news.ycombinator.com/item?id=37206445 making clear the additional assumptions needed to make the solution of original problem 1/3.

With those assumptions the probability that there are a boy and a girl are 2/3 if I tell you that there is at least a boy and 0 if I tell you that there is at least a girl. The probability that the email says that I have at least a boy are 3/4 (I would only say that I have a girl if I didn’t have any boys). You can calculate the probability that I have one boy and one girl before opening the email as 3/4 * 2/3 + 1/4 * 0 and it equals 1/2 as it should.

[HWR_14]

You need to look at the odds for all events. You cannot just look at the odds for a just specific event for deciding that the specified gender in the email is irrelevant. The fact that the rest of the odds are different means that it's 1/2 when the email is sent.

Your coin question is totally different. Whether the coin is heads or tails is independent from which coin it is. Whether you mention you have at least one boy is not independent of the gender of the children.

Your last paragraph has correct math. But the math works equally well with "specify a girl if you have one" or "flip a coin and use a random kids gender"

[kgwgk]

> Your last paragraph has correct math. But the math works equally well with "specify a girl if you have one" or "flip a coin and use a random kids gender"

That’s the point.

The math works well with "specify a boy if you have one" and then the answer to A [I tell you I have two children and that (at least) one of them is a boy, and ask you what you think is the probability that I have one boy and one girl.] is 2/3 and the answer to B [I tell you I have two children and that (at least) one of them is a girl, and ask you what you think is the probability that I have one boy and one girl.] is 0.

The math works well with "specify a girl if you have one" and then the answer to A is 0 and the answer to B is 2/3.

The math works well with "flip a coin and use a random kids gender" and then the answer to A is 1/2 and the answer to B is 1/2.

If every parent with two kids says either “at least one is a boy” or “at least one is a girl” there is no way to make the math work so the answer to A is 2/3 and the answer to B is 2/3.

——-

As I explain in another comment for that the two following conditions need to be met:

P(you tell me that you have at least one boy | you have two boys) = P(you tell me that you have at least one boy | you have one boy and one girl)

P(you tell me that you have at least one girl | you have two girls) = P(you tell me that you have at least one girl | you have one boy and one girl)

There are ways to make the math “work”. For example: if you have two boys or two girls flip a coin and if you get heads talk about the weather, if you get tails say [I have two kids and at least one is a boy/girl], if you have one boy and one girl say [I have two kids and at least one is a …] using a coin flip to decide if you say “girl” or “boy”.

However, they seem quite unnatural and hardly a justification to claim that “any arguments for 1/2 are just wrong.”

[HWR_14]

You misunderstand me.

No matter how you choose the statement "I have at least one (girl/boy)", (prefer one, flip a coins, etc) once you tell me it's always 2/3 boy-girl. Until you tell me it's 1/2. Any algorithm to choose which to say works as long as it's true and you don't convey more information about the children like "my older child is male".

Your counter arguments are wrong, but you don't seem to even acknowledge that I am saying that. I'm willing to try to explain why, but not if you don't want to learn and just want to insist you are correct. Ask yourself how long you would spend trying to explain Monty Hall to someone who kept insisting it was 1/2 to change.

[kgwgk]

> Your counter arguments are wrong, but you don't seem to even acknowledge that I am saying that.

I do acknowledge that you're saying that I'm wrong. That's why we're still exchanging arguments! What I don't know exactly is what do you think that it's wrong with my arguments so I try to find where do we agree - and where we don't.

Do you think that there is something wrong with the answer I wrote down here https://news.ycombinator.com/item?id=37206445 ?

It seems that you don't agree that the answer depends on the (relative) value of P(you tell me that you have at least one boy|you have two boys) and P(you tell me that you have at least one boy|you have one boy and one girl).

> I'm willing to try to explain why, but not if you don't want to learn and just want to insist you are correct.

Well, I could also say that just want to insist that my arguments are wrong but I sincerely hope that you want to learn as much as I do.

> Ask yourself how long you would spend trying to explain Monty Hall to someone who kept insisting it was 1/2 to change.

As long as needed. Souls are saved one at a time. Here we go.

-----

> No matter how you choose the statement "I have at least one (girl/boy)", (prefer one, flip a coins, etc) once you tell me it's always 2/3 boy-girl. Until you tell me it's 1/2. Any algorithm to choose which to say works as long as it's true and you don't convey more information about the children like "my older child is male".

That's wrong and I'm going to try to show you that with an example (the mathematical proof is in the link above). Hopefully I'm not misrepresenting your position - please tell me if I do.

You are in an auditorium with 600 people. Each of them has two kids. (Let's assume there is no strange thing going on like "meeting of parents with twins" and the sex of the kids is equally probable and independent.)

Q: What's the probability that a given person has one boy and one girl?

A: 1/2

Q: How many of them do you estimate that have one boy and one girl?

A: 300

Each of them write into a paper their name and "I have at least one (girl/boy)" (they never lie and if there is a choice the choose however they want: prefer one, flip a coin, etc.).

You have the 600 papers in front of you, but have not read them yet.

Q: What's the probability that a given person has one boy and one girl?

A: Still 1/2

Q: How many of them do you estimate that have one boy and one girl?

A: Still 300

You can win $100 if you guess correctly whether there are more than 350 or less than 350 people with one boy and one girl.

Q: What's your guess?

A: Less than 350, because my estimate is 300.

Q: What will be the probability that a given person has one boy and one girl after you've read the papers?

A: 2/3 because once they tell me it's always 2/3 boy-girl.

Q: How many of them will you estimate that have one boy and one girl after you've read the papers?

A: 400

Q: Do you want you want to change your guess to "more than 350"?

A: No, until I read the papers the probability is 1/2 and my estimate is that 300 people have one boy and one girl.

Q: So you keep your "less than 350" guess even though you know with certainty that in a few minutes you will estimate that the right answer is around 400 and you will wish you had answered "more than 350"?

A: Yes, I'm happy with that. I think I could get the $100 if I answered "more than 350" now but I refuse to do it until I read the papers.

You read the papers.

Q: What's the probability that a given person has one boy and one girl?

A: 2/3

Q: How many of them do you estimate that have one boy and one girl?

A: 400

Q: Do you want to change your guess for the $100 prize?

A: Yes, now I’d like to answer "More than 350". Thanks for letting me change my guess!

Unfortunately you lose, because in a group of 600 pairs of kids we expected around 300 pairs of boy and girl. Writing things on a paper leaves the children unaffected.

[HWR_14]

I think now I understand. "I will make you make exactly one of two statements and each results in a 2/3 chance of BG" doesn't make sense. "You freely made one of two statements and each results in a 2/3 chance of BG" does. I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works.

So it comes down to if we decide on the question (however we do that) before we look at the kids or after.

[kgwgk]

> I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works.

The point is that original problem says "I tell you I have two children and that (at least) one of them is a boy". It doesn't say "I tell you I have two children and [when you ask me to confirm whether (at least) one of them is a boy] that (at least) one of them is a boy".

Reasoning from the cases "BB", "BG", "GB", "GG" - and discarding the last one to get p(BB)=1/3 - is implicitly using the cases "BB and I tell you that at least one of them is a boy", "BG and I tell you that at least one of them is a boy", "GB and I tell you that at least one of them is a boy", "GG and I tell you something else like at least one of them is a girl".

That breaks the symmetry between the "I tell you that at least one of them is a boy" and the "I tell you that at least one of them is a girl" problems. Using the cases in the previous paragraph the answer for the former is p(BB)=1/3 but the answer to the latter is p(GG)=1.

If you want to have the same solution when you switch girl <-> boy ["one of them is a boy, what is the probability that I have two boys" <-> "one of them is a girl, what is the probability that I have two girls"] you should treat them equally.

A quite natural way to do so would be to consider the eight equiprobable cases (some of them repeated)

    BB and I tell you that at least one of them is a boy
    BB and I tell you that at least one of them is a boy
    BG and I tell you that at least one of them is a boy
    BG and I tell you that at least one of them is a girl
    GB and I tell you that at least one of them is a boy
    GB and I tell you that at least one of them is a girl
    GG and I tell you that at least one of them is a girl
    GG and I tell you that at least one of them is a girl

but then the answer to both problems is 1/2.

You can make the answer to both problems 1/3 but the eight cases that you would need to consider for that are quite unnatural:

    BB and [... discard this line somehow ...]
    BB and I tell you that at least one of them is a boy
    BG and I tell you that at least one of them is a boy
    BG and I tell you that at least one of them is a girl
    GB and I tell you that at least one of them is a boy
    GB and I tell you that at least one of them is a girl
    GG and I tell you that at least one of them is a girl
    GG and [... discard this line somehow ...]