This is basically Monty Hall right? The other child is a girl with 2/3 probability, because the first one being a boy eliminates the case where both children are girls, leaving three total cases, in two of which the other child is a girl (BG, GB, BB).
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.
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.
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.
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.
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.
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.
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.
On the one hand, you might be right. This could be akin to flipping two coins and saying that at least one is heads. You’ve only eliminated GG, so BG, GB, and BB are all possible, so there’s only a 1/3 chance that both children are boys.
On the other hand, you could say this is akin to flipping two coins and saying that the one on the left (or the one who opened the door) is heads. In that case, you haven’t just ruled out GG, you’ve ruled out GB. Conditional probability is witchcraft
I don’t think the problem is conditional probability, it’s translating word problems to maths problems.
If you make the assumptions I made, the maths is unambiguous. Namely, I assumed that a child has a 50/50 probability of being born a boy or a girl. I assumed the child who opens the door is random. I don’t think I made any other assumptions that could have been made any other way. With those assumptions, I’m pretty confident my answer is the only correct one, though I’d love to see an argument otherwise.
If the probability of a child being a girl is different, say, 52%, that will affect the result.
More interestingly, if the probability of which child opens the door is different, that will affect the result. If there’s a 100% chance the elder child opens the door, it goes to 50/50 of the gender of the second child. This makes it like the “coin on the left” example you gave.
If we said the elder child is going to open the door 75% of the time…well, the maths becomes more complicated than I can be bothered with right now. But it’s an interesting scenario!
Assuming the chance of either sex is equal, this problem can be broken down into multiple cases. The first is that there are two unseen kids in the house. What’s the probability they are both boys? 1/4. Now the door opens and you see two boys. The probability both are boys is 1/1. But if you only see one boy, the problem simplifies into the probability of a child being a boy. One of the probabilistic events postulated in the original problem is fixed at 1. So the answer is 1/2.
Think of it as the two coin flip, except one coin has two heads. That simplifies to a one coin flip.
No, because knowing the first child is a boy doesn’t tell you any information about the second child.
Three doors, Girl Girl Boy
You select a door and Monty opens a door to show a Girl. You had higher odds of picking a girl door to start (2/3). So switching gives you better odds at changing to the door with the Boy because you probably picked a Girl door.
Here the child being a boy doesn’t matter and the other child can be either.
Well not really, right? BG and GB are the same scenario here, so it’s a 50/50 chance.
Even if, say, the eldest child always opened the door, it’d still be a 50/50 chance, as the eldest child being a boy eliminates the possibility of GB, leaving either BG or BB.
Ironically you’ve got the right answer, but (as you can see in some of the other conversation here) not necessarily for the right reason. It’s not necessarily that BG and GB are the same, but that BB and BB are two different scenarios worthy of being counted separately.
Why is it that BB and BB are being counted separately? I thought that order didn’t matter: you could have two girls, a boy and a girl (or vice versa, same thing), or a two boys. (And then by eliminating two girls you’d have a 50/50 chance).
Because describing it as I did in that comment is a (very) shorthand way of getting at how it was explained in full by @Hacksaw@lemmy.ca in this comment.
This is basically Monty Hall right? The other child is a girl with 2/3 probability, because the first one being a boy eliminates the case where both children are girls, leaving three total cases, in two of which the other child is a girl (BG, GB, BB).
Actually, it’s a Monty Hall problem because a door is opened.
Unironically, yes. A door is opened, and the opening of that door reveals information about the problem and eliminates some possible world-states.
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.
So does this problem. There was the GG possibility.
True, this problem doesn’t have that element.
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.
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.
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.
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.
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.
I’m guessing you haven’t read the rest of the thread? My first comment was incorrect and the correction has been made.
It’s somewhat ambiguous!
On the one hand, you might be right. This could be akin to flipping two coins and saying that at least one is heads. You’ve only eliminated GG, so BG, GB, and BB are all possible, so there’s only a 1/3 chance that both children are boys.
On the other hand, you could say this is akin to flipping two coins and saying that the one on the left (or the one who opened the door) is heads. In that case, you haven’t just ruled out GG, you’ve ruled out GB. Conditional probability is witchcraft
I don’t think the problem is conditional probability, it’s translating word problems to maths problems.
If you make the assumptions I made, the maths is unambiguous. Namely, I assumed that a child has a 50/50 probability of being born a boy or a girl. I assumed the child who opens the door is random. I don’t think I made any other assumptions that could have been made any other way. With those assumptions, I’m pretty confident my answer is the only correct one, though I’d love to see an argument otherwise.
If the probability of a child being a girl is different, say, 52%, that will affect the result.
More interestingly, if the probability of which child opens the door is different, that will affect the result. If there’s a 100% chance the elder child opens the door, it goes to 50/50 of the gender of the second child. This makes it like the “coin on the left” example you gave.
If we said the elder child is going to open the door 75% of the time…well, the maths becomes more complicated than I can be bothered with right now. But it’s an interesting scenario!
Assuming the chance of either sex is equal, this problem can be broken down into multiple cases. The first is that there are two unseen kids in the house. What’s the probability they are both boys? 1/4. Now the door opens and you see two boys. The probability both are boys is 1/1. But if you only see one boy, the problem simplifies into the probability of a child being a boy. One of the probabilistic events postulated in the original problem is fixed at 1. So the answer is 1/2.
Think of it as the two coin flip, except one coin has two heads. That simplifies to a one coin flip.
No, because knowing the first child is a boy doesn’t tell you any information about the second child.
Three doors, Girl Girl Boy
You select a door and Monty opens a door to show a Girl. You had higher odds of picking a girl door to start (2/3). So switching gives you better odds at changing to the door with the Boy because you probably picked a Girl door.
Here the child being a boy doesn’t matter and the other child can be either.
It’s 50/50 assuming genders are 50/50.
Well not really, right? BG and GB are the same scenario here, so it’s a 50/50 chance.
Even if, say, the eldest child always opened the door, it’d still be a 50/50 chance, as the eldest child being a boy eliminates the possibility of GB, leaving either BG or BB.
Ironically you’ve got the right answer, but (as you can see in some of the other conversation here) not necessarily for the right reason. It’s not necessarily that BG and GB are the same, but that BB and BB are two different scenarios worthy of being counted separately.
Why is it that BB and BB are being counted separately? I thought that order didn’t matter: you could have two girls, a boy and a girl (or vice versa, same thing), or a two boys. (And then by eliminating two girls you’d have a 50/50 chance).
Because describing it as I did in that comment is a (very) shorthand way of getting at how it was explained in full by @Hacksaw@lemmy.ca in this comment.
Just read it: makes sense now. Thanks!
That is an excellent, succinct explanation
It’s also wrong, I have since learnt. See the conversation below for why.