You know that a couple has two children. You go to the couple’s house and one of their children, a young boy, opens the door. What is the probability that the couple’s other child is a girl?
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).
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 @[email protected] in this comment.
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
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.
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!
It is 50-50, though. The remaining possible states are BG and BB. Both are equally likely. Any further inference is narrative… not statistics.
The classic example of this is flipping 100 coins. If you get heads 99 times in a row… the last coin is still 50-50. Yes, it is obscenely unlikely to get heads 100 times in a row. But it’s already obscenely unlikely to get heads 99 times in a row. And it is obscenely unlikely to alternate perfectly between heads and tails. And it is obscenely unlikely to get a binary pattern spelling out the alphabet. And it is obscenely unlikely to get… literally any pattern.
Every pattern is equally unlikely, with a fair coin. We see 99 heads in a row versus 1 tails at the end, and think it narrowly averted the least-probable outcome. But only because we lump together all sequences with exactly one tails. That’s one hundred different patterns. 1-99 is not the same as 99-1. We just treat them the same because we fixate on uniformity.
Compare a non-binary choice: a ten-sided die. Thirty 1s in a row is about as unlikely as 100 heads in a row. But 1 1 1… 2 is the same as 1 1 1… 3. Getting the first 29 is pretty damn unlikely. One chance in a hundred million trillion. But the final die can land on any number 1-10. Nine of them upset the pattern our ape brains want. Wanting it doesn’t make it any more likely. Or any less likely.
It would be identically unlikely for a 10-sided die to count from 1 to 10, three times in a row. All the faces appear equally. But swap any two events and suddenly it doesn’t count. No pun intended.
If this couple had eight children, for some god-forsaken reason, and you saw seven boys, the eighth kid being another boy is not less likely for it. The possibility space has already been reduced to two possibilities out of… well nine, I suppose, if order doesn’t matter. They could have 0-8 boys. They have at least 7. The only field that says the last kid’s not a coin toss is genetics, and they say this guy’s chromosome game is strong.
You’re right, but it’s not a subversion of the Gambler’s Fallacy, it’s a subversion of conditional probability. A classic example is that I have two kids, and at least one of them is a boy. What is the probability that I have two boys?
The intuitive answer is 50%, because one kid’s sex doesn’t affect the other. But when I told you that I have two kids, there were four possibilities: GG, GB, BG, or BB. When I told you that at least one of them is a boy, all I did was take away the GG option. That means there’s only a 1 in 3 chance that I have two boys.
But by having one child answer the door, I change it yet again–now we know the sex of a particular child. We know that the child who opened the door is a boy. This is now akin to saying “I have two children, and the eldest is a boy. What is the possibility that I have two boys?” It’s a sneaky nerd snipe, because it targets specifically people who know enough about statistics to know what conditional probability is. It’s also a dangerous nerd snipe, because it’s entirely possible that my reasoning is wrong!
Two more for funsies! I flipped two coins. At least one of them landed on heads. What is the probability that both landed on heads? (Note: this is what my comment originally said before I edited it)
I have two children. At least one of them is a boy born on a Tuesday. What is the probability that I have two boys?
Oops, I changed it to a more unintuitive one right after you replied! In my original comment, I said “you flip two coins, and you only know that at least one of them landed on heads. What is the probability that both landed on heads?”
And… No! Conditional probability strikes again! When you flipped those coins, the four possible outcomes were TT, TH, HT, HH
When you found out that at least one coin landed on heads, all you did was rule out TT. Now the possibilities are HT, TH, and HH. There’s actually only a 1/3 chance that both are heads! If I had specified that one particular coin landed on heads, then it would be 50%
No. It’s still 50-50. Observing doesn’t change probabilities (except maybe in quantum lol). This isn’t like the Monty Hall where you make a choice.
The problem is that you stopped your probably tree too early. There is the chance that the first kid is a boy, the chance the second kid is a boy, AND the chance that the first kid answered the door. Here is the full tree, the gender of the first kid, the gender of the second and which child opened the door, last we see if your observation (boy at the door) excludes that scenario.
1 2 D E
B B 1 N
B G 1 N
G B 1 Y
G G 1 Y
B B 2 N
B G 2 Y
G B 2 N
G G 2 Y
You can see that of the scenarios that are not excluded there are two where the other child is a boy and two there the other child is a girl. 50-50. Observing doesn’t affect probabilities of events because your have to include the odds that you observe what you observed.
If you really wanna see a bloodbath, watch this:
You know that a couple has two children. You go to the couple’s house and one of their children, a young boy, opens the door. What is the probability that the couple’s other child is a girl?
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).
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 @[email protected] in this comment.
Just read it: makes sense now. Thanks!
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
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.
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!
That is an excellent, succinct explanation
It’s also wrong, I have since learnt. See the conversation below for why.
Cheeky bastard.
It is 50-50, though. The remaining possible states are BG and BB. Both are equally likely. Any further inference is narrative… not statistics.
The classic example of this is flipping 100 coins. If you get heads 99 times in a row… the last coin is still 50-50. Yes, it is obscenely unlikely to get heads 100 times in a row. But it’s already obscenely unlikely to get heads 99 times in a row. And it is obscenely unlikely to alternate perfectly between heads and tails. And it is obscenely unlikely to get a binary pattern spelling out the alphabet. And it is obscenely unlikely to get… literally any pattern.
Every pattern is equally unlikely, with a fair coin. We see 99 heads in a row versus 1 tails at the end, and think it narrowly averted the least-probable outcome. But only because we lump together all sequences with exactly one tails. That’s one hundred different patterns. 1-99 is not the same as 99-1. We just treat them the same because we fixate on uniformity.
Compare a non-binary choice: a ten-sided die. Thirty 1s in a row is about as unlikely as 100 heads in a row. But 1 1 1… 2 is the same as 1 1 1… 3. Getting the first 29 is pretty damn unlikely. One chance in a hundred million trillion. But the final die can land on any number 1-10. Nine of them upset the pattern our ape brains want. Wanting it doesn’t make it any more likely. Or any less likely.
It would be identically unlikely for a 10-sided die to count from 1 to 10, three times in a row. All the faces appear equally. But swap any two events and suddenly it doesn’t count. No pun intended.
If this couple had eight children, for some god-forsaken reason, and you saw seven boys, the eighth kid being another boy is not less likely for it. The possibility space has already been reduced to two possibilities out of… well nine, I suppose, if order doesn’t matter. They could have 0-8 boys. They have at least 7. The only field that says the last kid’s not a coin toss is genetics, and they say this guy’s chromosome game is strong.
You’re right, but it’s not a subversion of the Gambler’s Fallacy, it’s a subversion of conditional probability. A classic example is that I have two kids, and at least one of them is a boy. What is the probability that I have two boys?
The intuitive answer is 50%, because one kid’s sex doesn’t affect the other. But when I told you that I have two kids, there were four possibilities: GG, GB, BG, or BB. When I told you that at least one of them is a boy, all I did was take away the GG option. That means there’s only a 1 in 3 chance that I have two boys.
But by having one child answer the door, I change it yet again–now we know the sex of a particular child. We know that the child who opened the door is a boy. This is now akin to saying “I have two children, and the eldest is a boy. What is the possibility that I have two boys?” It’s a sneaky nerd snipe, because it targets specifically people who know enough about statistics to know what conditional probability is. It’s also a dangerous nerd snipe, because it’s entirely possible that my reasoning is wrong!
i hate it when ppl do nb erasure for their stupid math text problems. use anything else pls
What’s the probability of the other kid being non-binary?
And don’t forget that there’s always a slim chance that no matter the gender, the other child is GOAT.
Two more for funsies! I flipped two coins. At least one of them landed on heads. What is the probability that both landed on heads? (Note: this is what my comment originally said before I edited it)
I have two children. At least one of them is a boy born on a Tuesday. What is the probability that I have two boys?
50%, since the coins are independent, right?
Oops, I changed it to a more unintuitive one right after you replied! In my original comment, I said “you flip two coins, and you only know that at least one of them landed on heads. What is the probability that both landed on heads?”
And… No! Conditional probability strikes again! When you flipped those coins, the four possible outcomes were TT, TH, HT, HH
When you found out that at least one coin landed on heads, all you did was rule out TT. Now the possibilities are HT, TH, and HH. There’s actually only a 1/3 chance that both are heads! If I had specified that one particular coin landed on heads, then it would be 50%
No. It’s still 50-50. Observing doesn’t change probabilities (except maybe in quantum lol). This isn’t like the Monty Hall where you make a choice.
The problem is that you stopped your probably tree too early. There is the chance that the first kid is a boy, the chance the second kid is a boy, AND the chance that the first kid answered the door. Here is the full tree, the gender of the first kid, the gender of the second and which child opened the door, last we see if your observation (boy at the door) excludes that scenario.
1 2 D E
B B 1 N
B G 1 N
G B 1 Y
G G 1 Y
B B 2 N
B G 2 Y
G B 2 N
G G 2 Y
You can see that of the scenarios that are not excluded there are two where the other child is a boy and two there the other child is a girl. 50-50. Observing doesn’t affect probabilities of events because your have to include the odds that you observe what you observed.