[kgwgk]

Do you agree with the following?

> 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.

2/3

> 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.

2/3

If you don’t, why not?

If you do, what’s your answer to the following question?

> I tell you I have two children and that I’ve just sent you an email with the sex of (at least) one of them, and ask you what you think is the probability that I have one boy and one girl.

[HWR_14]

I agree 2/3, I agree 2/3. The answer to the last question is 1/2.

[kgwgk]

Thanks for your answer.

What I don’t understand is that when you read the content of that email you will find yourself in either the first situation (I told you that I have two children and that (at least) one of them is a boy) or the second situation (I told you that I have two children and that (at least) one of them is a girl).

In both cases the probability will be 2/3 so why wouldn’t you conclude that the probability is 2/3 without waiting to find out the (irrelevant) details?

[HWR_14]

The odds only sound equal/the details irrelevant because you are only looking at one outcome from the set. In reality, the email will resolve the probabilities to: (BB: 1/3 BG:2/3 GG:0/3} or {BB:0/3 BG:2/3 GG:1/3}. Although the BG values are the same, the rest of the probabilities are not. Therefore, the details are relevant.

I don't have a great explanation as to why that's intuitively true, but it is. I can try again if things are still confusing. But if so it would help to know if you understand the Monty Hall problem.

[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.”