[prolways]

Is there a more rigorous explanation of why they count the probability space how they do? Watching that video I feel like the ordering of the kids and the striking of one of the "b2b2" entries seems wrong to me. If we care which kid was first... which doesn't seem to matter... then the first b2b2 and the second b2b2 seem like they're different and shouldn't "cancel."

Then again... it took me multiple explanations to understand the Monty Hall problem.

[kevinwang]

Yeah, it definitely seems wrong to me. B2B2 has twice the probability of any other event listed in that table.

edit: I guess it's more nuanced than that. The explanation and interpretation sections on this blog post [1] and on wikipedia talk about the controversy.

[1]: https://jakubmarian.com/the-day-of-the-week-boy-or-girl-para...

[2]: https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

[Houshalter]

Even if that's the case it still doesn't make sense. That changes the probability to 50% girl instead of 52%. But it should be 2/3rds.

[ralfd]

This B2/B2 elimination also immediately jumped out to me as being wrong.

But a Youtube comment wrote: > "But the thing is, I ran a few tests through a big randomized sample set, and... he's right. It's super weir … the second boy-girl problem had ~51.9% change of containing a girl. Keep in mind this was about 100 million randomized samples too."

I can't wrap my head around it.

[chillee]

So I think intuitively, something that might make sense to you is that the reason that BB is more common with the restriction that a boy must be on Tuesday is that having 2 boys increases the probability that you will have a boy that was born on Tuesday.

Here's another way to think about it. It's twice as likely to have one girl and one boy (either BG or GB) compared to having 2 boys. Thus, the version of the paradox where you're given that the parent has a boy results in a 2/3 chance of the other child being a girl.

However, the more unlikely it is that any given boy satisfies the condition (in this case the condition would be being born on a Tuesday), the more likely the BB case becomes compared to BG or GB.

More concretely, if only 1/n of the boys satisfies some condition, you would be left with only 1/n of BG or GB, while you would be left with 2/n - 1/(n^2) of BB. In this case, let the population be all parents with at least one boy (this consists 1/3 of BB, 1/3 of BG, and 1/3 of GB). Letting n = 7, (BG union GB) represents (1/7)(2/3)=2/21 of the population, while BB represents (2/7 - 1/49)(1/3) = 13/147 of the population.

Now, (2/21)/(2/21+13/147) = 14/27, our desired result.

I hope that's a more intuitive way of thinking about it. The important part is realizing the differences between knowing that the parent has a boy vs knowing that the parent has a boy born on Tuesday.

[Houshalter]

>having 2 boys increases the probability that you will have a boy that was born on Tuesday.

I guess. But the fact that it happened to be Tuesday isn't really important. The person could just as easily have had a kid on Wednesday. And all the logic would be the same. And the kid has to be born on some day of the week. How does finding out what day it was give us any additional information about the other child? It's completely independent!

[chillee]

So that's what confused me to begin with as well.

I think the main thing here depends on your interpretation of how the parent is chosen.

Let's say the question is the same, but we relax the requirement that the boy is born on Tuesday. Do you think the probability that the other child is a girl is 2/3 or 1/2?

[Arnavion]

>while you would be left with 2/n - 1/(n^2) of BB

For those like me who didn't immediately understand where this came from:

    P(atleast one of two boys satisfies the condition)
    = 1 - P(neither of two boys satisfies the condition)
    = 1 - [P(a boy doesn't satisfy the condition)]^2
    = 1 - [1 - P(a boy satisfies the condition)]^2
    = 1 - [1 - 1/n]^2
    = 2/n - 1/n^2

[kevinwang]

dammit, you're right.