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Probability does not bite; describing partial information in English bites. It's not actually true that the probability is 1/3, nor that the probability is 1/2

Probability does not bite; describing partial information in English bites. It's not actually true that the probability is 1/3, nor that the probability is 1/2. (Same with 13/27 vs 1/2). The problem is underspecified. Here's two different more specified versions for which the answer is clear: 1. Sample from all two-child families with at least one boy. What portion of these families have two boys? (answer, rot13: n guveq) 2. Choose a random two-child family, then knock on their door. A boy answers. What are the odds the other child is a boy? (rot13: bar unys) These are both consistent with the description "at least one child is a boy"! The day-of-week versions: 3. Sample from all two-child families with at least one boy born on a Tuesday. The odds both are boys? (nyzbfg unys) 4. Knock on the door of a random two-child family. A boy born on Tuesday answers. Odds both are boys? (n unys)

kgwgk 3y ago
> It's not actually true that the probability is 1/3, nor that the probability is 1/2. You’re right. Those who are satisfied with the 1/3 answer may want to consider 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 the pair is single-sex. 1/3 > 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 the pair is single-sex. Also 1/3 > I tell you I have two children, and ask you what you think is the probability that the pair is single-sex. 1/2 So if I tell you that I have two children you think that the probability that they are of the same sex is 1/2. And when I tell you the gender of one of them, whatever it is, you will think that the probability goes down to 1/3?
- HWR_14 3y ago
  Your math is accurate. Once you are told the gender of one child with no other information, the odds of being all the same gender go down. Probability is tricky.
  - kgwgk 3y ago
    > I have two children… Oh, you have two children? The probability that they are of the same sex is 1/2. > and the sex of at least one of them is… Say no more! If at least one of them is of some sex the odds that they are both of the same sex go down to 1/3. I said 1/2 before but that was before knowing that at least one of them is either a boy or a girl. That changes everything! (Probability is tricky.)
    - kqr 3y ago
      It is not sufficient to know one of them is of some sex. For the probability to be 1/3, you need to be asked what the probability is that one of them is a specific sex, not just any sex.
      - kuboble 3y ago
        I think the trickiest part is that the other party willingly shared some information and their motives affect probabilities way more than any math. I find it easier to think about this problem stated like this: let's say you go around asking people " do you have exactly 2 children and at least one of them is a boy?". What are the odds of them having 2 boys if they answered yes.
        HWR_14 3y ago
        All probability questions suffer from the same bias. The Monty Hall problem doesn't work if the person offering the choice has some agency and motives.
    - phalf 3y ago
      Nice! What's really fun about this problem is that you can have very convincing arguments for 1/2 being the correct answer, and very convincing arguments for 1/3 being the correct answer. And for either you can make subtle reformulations that supposedly illustrate how ridiculous this answer is. And there is no way to know. There is no gold standard for designing an experiment that would show whether 1/2 or 1/3 is correct. You could set up something that generates millions of pairs of (virtual) kids and then count the pairs that fit. But each of these experiments will have built-in the assumption on which the response is ultimately already predicated on. The only thing really convincing would be if everybody, all "sides", could agree on an experiment with an outcome that they would feel bound to. Then one could settle this once and for all, whether it's 1/2 or 1/3 or 13/27 or 729/1459 or whatnot. But people will never agree on such an experimental setup. Which tells me that this is not a mathematical problem. This problem is either underspecified or it's contradictory. If it was uniquely specified then we could just use probability theory with its axioms and inference rules to derive at the correct answer. But we obviously can't, since nobody can agree on how to formally note this down.
- mofeien 3y ago
  The statement "at least one of them is a boy" (<=> "I don't have two daughters") is a little more subtle than "I tell you the gender of one of them" since the former excludes one out of four possibilities (FF, thus letting us update our belief on the single-sex question to a third) while the latter implies fixing the gender of a specific one of the children (without specifying which one, and in either case the probability of the other being M is still a half, thus not giving us information towards the single sex question). So if you tell me the gender of a specific one of them, say the youngest, then I haven't learned anything that makes my subjective probability go down that the other is the same gender. I think in real life you will come across the second kind of statement (e. g. "my oldest is a girl") than the first kind (e. g. "I do not have two boys") But it does not feel too weird to me that "at least one of them is a girl" will reduce the probability of the pair being single-sex to a third. In fact if you further tell me that both "an least one of them is a girl" and "at least one of them is a boy", the probability of the pair being single-sex will go to zero and this seems perfectly reasonable
  - kgwgk 3y ago
    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 the pair is single-sex. 1/3 > 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 the pair is single-sex. 1/3 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 the pair is single-sex. Will your answer change after you have a chance to check your messages?
    - kqr 3y ago
      In this scenario, you are subtly changing the meaning of "single-sex". In the first two cases, "single-sex" means "the same specific sex as the child you know the sex of" whereas in the last case it means "the same sex as a child that can still have two possible sexes". If you would say, > 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 the pair are both girls? and then follow up with another question, > what you think is the probability that the pair are both boys? and then add the two probabilities up equally weighted, you might see why 1/2 is the reasonable answer in that case. (And why opening up the email in question would reduce the probability of one of the questions to 0, and the other to 1/3.)
- anon4242 3y ago
  There are four cases to consider, MM (both kids are Male), FF (both kids are Female), MF and FM. So there's a 50% chance of same gender kids and a 25% chance for both kids to be female. So if you know the gender, say female, of one kid but not if they are the older or younger, you have these possibilities FF, FM or MF. And FF is 1/3 of that.
  - kgwgk 3y ago
    If I understand correctly what you said: If I tell you that one kid is male, you think that the probability that there is one male and one female is 2/3. If I tell you that one kid is female, you think that the probability that there is one male and one female is 2/3. (Right?) If I don't tell you anything - beyond the fact that I have two kids - what's the probability that there is one male and one female?
    - dragonwriter 3y ago
      There are four equally likely combinations (under the [both false!] assumptions of equal and independent sexes for children in the same family): MM, FM, MF, and FF; if you know that there is at least one male (or at least one female) you eliminate one of those possibilities, leaving the relative probabilities of the other three still equal. So, knowing no additional information, the chance of one male and one female is two-fourths, or one-half. Knowing that there is at least one male (eliminating FF), or at least one female (eliminating MM), the probability of one male and one female is 2/3. If you know the sex and birth order of one, you eliminate two possibilities, retaining the relative probabilities of the remaining ones as equal, so if you know the first is male, eliminating FM and FF, then the probability of one male and one female is 1/2 (and similarly, mutatis mutandis, with other sex and birth order combinations, which produce the same result eliminating different pairs of possibilities.)
      - kgwgk 3y ago
        > Knowing that there is at least one male (eliminating FF), or at least one female (eliminating MM), the probability of one male and one female is 2/3. Don't you always know that there is at least one male or one female? I mean, if A="there is at least one male" and B="there is at least one female" you're telling me that if you know that A holds the probability is 2/3 and if you know that B holds the probability is 2/3. But, knowing no additional information, you KNOW that A and/or B holds! 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.
movpasd 3y ago
#2 is not actually equivalent to "at least one child is a boy". It is rather equivalent to "the first child is a boy". The difference may seem trivial, but one implies the other without the converse being true. This changes the probabilities — it's not an issue with underspecification. I think your example #1 makes it much clearer why the 1/3 arises, at least in a frequentist analysis. I would like to offer a similar interpretation but from a Bayesian lens. The 1/3 as rises due to the artificiality of the knowledge condition. Given real-world constraints, we expect any information collected to cleave neatly between the two children in our imagined information gathering scenario. So we implicitly translate "at least one child is a boy" to "we've checked one child, it's a boy". Consider the following related problem: I have two faucets next to each other, each has a 50% chance of dripping overnight. I leave one shared bucket under both of them. The next day, the bucket is wet. What's the odds that _both_ faucets dripped? This setup makes the correlative nature of the information much clearer, and I think most people would be less likely to jump to 1/2 as an answer.
- nstbayless 3y ago
  The bucket formulation is very elegant. I still feel the problem arises from English, not probability. It's clear that "we've checked one child, it's a boy" implies "at least one child is a boy." But furthermore, If someone tells me "at least one the two kids is a boy," I do not know how they arrived at that information. It could either have been through the bucket method or the knock-at-door method. From a Bayesian perspective, we should consider both as possible with priors P and 1-P (i.e. the answer is somewhere between 1/3 and 1/2). On the other hand, from the perspective of someone taking a math test, I'd rather like the professor to tell me their own prior -- which, given they felt confident enough to put this on a test, they must believe it's basically 0 or basically 1. Ultimately, both scenarios are describable by the same English phrase, and it feels proscriptivist to just consider one of them, even if it happens to have the least entropy in this case. There should always be the followup question asked: "_how_ did you know this?" and if it's kicked back to " because someone told me," either we need to ask how that person learned it or else bust out some priors.
  - movpasd 3y ago
kqr 3y ago
Another classic example of "sampling method matters" is that the average arrival time between trains is longer for passengers than the train operator. (Because a randomly selected passenger is more likely to be one of the many waiting for a delayed train, than one who happened to get on an earlier train.)

kgwgk 3y ago

> It's not actually true that the probability is 1/3, nor that the probability is 1/2. You’re right. Those who are satisfied with the 1/3 answer may want to consider 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 the pair is single-sex. 1/3 > 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 the pair is single-sex. Also 1/3 > I tell you I have two children, and ask you what you think is the probability that the pair is single-sex. 1/2 So if I tell you that I have two children you think that the probability that they are of the same sex is 1/2. And when I tell you the gender of one of them, whatever it is, you will think that the probability goes down to 1/3?

HWR_14 3y ago
Your math is accurate. Once you are told the gender of one child with no other information, the odds of being all the same gender go down. Probability is tricky.
- kgwgk 3y ago
  > I have two children… Oh, you have two children? The probability that they are of the same sex is 1/2. > and the sex of at least one of them is… Say no more! If at least one of them is of some sex the odds that they are both of the same sex go down to 1/3. I said 1/2 before but that was before knowing that at least one of them is either a boy or a girl. That changes everything! (Probability is tricky.)
  - kqr 3y ago
    It is not sufficient to know one of them is of some sex. For the probability to be 1/3, you need to be asked what the probability is that one of them is a specific sex, not just any sex.
    - kuboble 3y ago
      I think the trickiest part is that the other party willingly shared some information and their motives affect probabilities way more than any math. I find it easier to think about this problem stated like this: let's say you go around asking people " do you have exactly 2 children and at least one of them is a boy?". What are the odds of them having 2 boys if they answered yes.
      - HWR_14 3y ago
        All probability questions suffer from the same bias. The Monty Hall problem doesn't work if the person offering the choice has some agency and motives.
  - phalf 3y ago
    Nice! What's really fun about this problem is that you can have very convincing arguments for 1/2 being the correct answer, and very convincing arguments for 1/3 being the correct answer. And for either you can make subtle reformulations that supposedly illustrate how ridiculous this answer is. And there is no way to know. There is no gold standard for designing an experiment that would show whether 1/2 or 1/3 is correct. You could set up something that generates millions of pairs of (virtual) kids and then count the pairs that fit. But each of these experiments will have built-in the assumption on which the response is ultimately already predicated on. The only thing really convincing would be if everybody, all "sides", could agree on an experiment with an outcome that they would feel bound to. Then one could settle this once and for all, whether it's 1/2 or 1/3 or 13/27 or 729/1459 or whatnot. But people will never agree on such an experimental setup. Which tells me that this is not a mathematical problem. This problem is either underspecified or it's contradictory. If it was uniquely specified then we could just use probability theory with its axioms and inference rules to derive at the correct answer. But we obviously can't, since nobody can agree on how to formally note this down.
mofeien 3y ago
The statement "at least one of them is a boy" (<=> "I don't have two daughters") is a little more subtle than "I tell you the gender of one of them" since the former excludes one out of four possibilities (FF, thus letting us update our belief on the single-sex question to a third) while the latter implies fixing the gender of a specific one of the children (without specifying which one, and in either case the probability of the other being M is still a half, thus not giving us information towards the single sex question). So if you tell me the gender of a specific one of them, say the youngest, then I haven't learned anything that makes my subjective probability go down that the other is the same gender. I think in real life you will come across the second kind of statement (e. g. "my oldest is a girl") than the first kind (e. g. "I do not have two boys") But it does not feel too weird to me that "at least one of them is a girl" will reduce the probability of the pair being single-sex to a third. In fact if you further tell me that both "an least one of them is a girl" and "at least one of them is a boy", the probability of the pair being single-sex will go to zero and this seems perfectly reasonable
- kgwgk 3y ago
  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 the pair is single-sex. 1/3 > 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 the pair is single-sex. 1/3 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 the pair is single-sex. Will your answer change after you have a chance to check your messages?
  - kqr 3y ago
    In this scenario, you are subtly changing the meaning of "single-sex". In the first two cases, "single-sex" means "the same specific sex as the child you know the sex of" whereas in the last case it means "the same sex as a child that can still have two possible sexes". If you would say, > 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 the pair are both girls? and then follow up with another question, > what you think is the probability that the pair are both boys? and then add the two probabilities up equally weighted, you might see why 1/2 is the reasonable answer in that case. (And why opening up the email in question would reduce the probability of one of the questions to 0, and the other to 1/3.)
anon4242 3y ago
There are four cases to consider, MM (both kids are Male), FF (both kids are Female), MF and FM. So there's a 50% chance of same gender kids and a 25% chance for both kids to be female. So if you know the gender, say female, of one kid but not if they are the older or younger, you have these possibilities FF, FM or MF. And FF is 1/3 of that.
- kgwgk 3y ago
  If I understand correctly what you said: If I tell you that one kid is male, you think that the probability that there is one male and one female is 2/3. If I tell you that one kid is female, you think that the probability that there is one male and one female is 2/3. (Right?) If I don't tell you anything - beyond the fact that I have two kids - what's the probability that there is one male and one female?
  - dragonwriter 3y ago
    There are four equally likely combinations (under the [both false!] assumptions of equal and independent sexes for children in the same family): MM, FM, MF, and FF; if you know that there is at least one male (or at least one female) you eliminate one of those possibilities, leaving the relative probabilities of the other three still equal. So, knowing no additional information, the chance of one male and one female is two-fourths, or one-half. Knowing that there is at least one male (eliminating FF), or at least one female (eliminating MM), the probability of one male and one female is 2/3. If you know the sex and birth order of one, you eliminate two possibilities, retaining the relative probabilities of the remaining ones as equal, so if you know the first is male, eliminating FM and FF, then the probability of one male and one female is 1/2 (and similarly, mutatis mutandis, with other sex and birth order combinations, which produce the same result eliminating different pairs of possibilities.)
    - kgwgk 3y ago
      > Knowing that there is at least one male (eliminating FF), or at least one female (eliminating MM), the probability of one male and one female is 2/3. Don't you always know that there is at least one male or one female? I mean, if A="there is at least one male" and B="there is at least one female" you're telling me that if you know that A holds the probability is 2/3 and if you know that B holds the probability is 2/3. But, knowing no additional information, you KNOW that A and/or B holds! 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.

movpasd 3y ago

#2 is not actually equivalent to "at least one child is a boy". It is rather equivalent to "the first child is a boy". The difference may seem trivial, but one implies the other without the converse being true. This changes the probabilities — it's not an issue with underspecification. I think your example #1 makes it much clearer why the 1/3 arises, at least in a frequentist analysis. I would like to offer a similar interpretation but from a Bayesian lens. The 1/3 as rises due to the artificiality of the knowledge condition. Given real-world constraints, we expect any information collected to cleave neatly between the two children in our imagined information gathering scenario. So we implicitly translate "at least one child is a boy" to "we've checked one child, it's a boy". Consider the following related problem: I have two faucets next to each other, each has a 50% chance of dripping overnight. I leave one shared bucket under both of them. The next day, the bucket is wet. What's the odds that _both_ faucets dripped? This setup makes the correlative nature of the information much clearer, and I think most people would be less likely to jump to 1/2 as an answer.

nstbayless 3y ago
The bucket formulation is very elegant. I still feel the problem arises from English, not probability. It's clear that "we've checked one child, it's a boy" implies "at least one child is a boy." But furthermore, If someone tells me "at least one the two kids is a boy," I do not know how they arrived at that information. It could either have been through the bucket method or the knock-at-door method. From a Bayesian perspective, we should consider both as possible with priors P and 1-P (i.e. the answer is somewhere between 1/3 and 1/2). On the other hand, from the perspective of someone taking a math test, I'd rather like the professor to tell me their own prior -- which, given they felt confident enough to put this on a test, they must believe it's basically 0 or basically 1. Ultimately, both scenarios are describable by the same English phrase, and it feels proscriptivist to just consider one of them, even if it happens to have the least entropy in this case. There should always be the followup question asked: "_how_ did you know this?" and if it's kicked back to " because someone told me," either we need to ask how that person learned it or else bust out some priors.
- movpasd 3y ago

kqr 3y ago

Another classic example of "sampling method matters" is that the average arrival time between trains is longer for passengers than the train operator. (Because a randomly selected passenger is more likely to be one of the many waiting for a delayed train, than one who happened to get on an earlier train.)