It doesn't, though. The Monty Hall problem utilizes the fact that there were more possibilities before one was eliminated AND that it cannot eliminate the "best" outcome. No such qualities are at play here.

The question being asked here is "what is the gender of the second child?" The gender of the first child is completely irrelevent. Observed or unobserved, door open or closed, it doesn't impact the outcome of the second child.

I suspect it's not the question OP intended to ask, but it's the question they asked nonetheless.

[–] Zagorath@aussie.zone 2 points 11 months ago (2 children)

The Monty Hall problem utilizes the fact that there were more possibilities before one was eliminated

So does this problem. There was the GG possibility.

AND that it cannot eliminate the "best" outcome.

True, this problem doesn't have that element.

The question being asked here is "what is the gender of the second child?" The gender of the first child is completely irrelevent. Observed or unobserved, door open or closed, it doesn't impact the outcome of the second child.

I don't agree. First, I'd say your use of the term "second child" is ambiguous, because normally that would mean "the younger of two children", which obviously isn't what's meant here. What you mean to say here is "the child that we have not already seen". It's in that rephrasing that it becomes obvious that having observed the first child matters, because there cannot be a second until there has been a first. And it's in that observation that the outcome is altered.

If we haven't seen the first child and are asked "what will be the gender of the second child to walk through the door?" we would have to answer 50/50. But having seen one child, we eliminate one of four possibilities of gender pairs (BB, BG, GB, GG). This we are left with 3 equally possible cases, 2 of which will be the opposite of the gender of the child we saw first.

Of course, we could easily simulate this experiment to arrive at an empirical answer. Randomly generate 2 genders, randomly select one of those. If they're a girl, end the experiment and move to another iteration (because they didn't fit the parameters). If a boy, record the gender of the other child. Repeat a few dozen times and see how many times the second child was a girl.

[–] Glide@lemmy.ca 16 points 11 months ago* (last edited 11 months ago)

Well, I guess OPs point is demonstrated. People will in fact argue about it.

What you're trying to present has multiple holes, but only one matters: you're not paying attention to the question that's being asked. You can say first, second, alpha, beta, Leslie, whatever you want to assign the child in question as, but the question only asks you the gender of a singular child. The door opening child doesn't matter, because it isn't part of the question. No one asked what gender that child is. No one asked what the odds they have a female child is. It just isn't a part of the question.

Yes, I referred to it as the second child because the question that was asked happens to have a child in it and ask you about another. Because we're communicating in a hilariously precise language, we have to say "the other child". But that doesn't make the door opening child a part of the equation. The question could be "there is a child in a box. What are the odds the child is female? Oh, it has a brother by the way." Cool, who cares, the sibling wasn't a part of the question.

The Monty Hall problem spreads multiple outcomes across multiple choices and then eliminates one. The outcomes and options have a relation. This question just asks you about a singular variable with two possible outcomes and throws around an unrelated red herring.

[–] calcopiritus@lemmy.world 1 points 11 months ago (1 children)

You can't just end the experiment if the randomly chosen child doesn't "fit the parameters", by doing that you aren't accounting for half the girls in the whole event pool. Half of the girls have siblings that are girls.

Being 2 girls was a possible event at the start, you can't just remove it. This time it happened to be a boy who opened the door, but it could've been as likely for a girl to open it.

If it was phrased like "there are 2 siblings, only boys can open doors. Of all the houses that opened their doors, how many have a girl in them?", then it will be 2/3. In this example, there is an initial pool of events, then I narrowed down to a smaller one (with less probability). If you "just" eliminate the GG scenario, then the set of events got smaller without reducing the set's probability.

[–] Zagorath@aussie.zone 1 points 11 months ago (1 children)

Being 2 girls was a possible event at the start

It depends on what you classify as the "start". If all households with two children everywhere? Sure. But the story explicitly starts with us knowing a boy came to the door. By the parameters of the story, we know that's what happened.

We don't just eliminate GG, we also eliminate any BG or GB where it just so happened that the girl came to the door. Because that's what we already know is true, and we're asking for the conditional probability given that this has already happened.

[–] calcopiritus@lemmy.world 1 points 11 months ago (1 children)

You didn't eliminate BG and GB where a girl opens the door though. If you do that, then the answer is 50%. Because you remove half the probability from BG and GB and you remove none from BB.

I know you didn't eliminate those cases because you said "That leaves us with 3 possibilities with equal probabilities". That would be false, BB is twice as likely.

[–] Zagorath@aussie.zone 1 points 11 months ago

I know you didn’t eliminate those cases because you said “That leaves us with 3 possibilities with equal probabilities”. That would be false, BB is twice as likely.

I'm guessing you haven't read the rest of the thread? My first comment was incorrect and the correction has been made.