I was asked the question “a man states he has two children, and at least one of them are boys, what are the odds that the man has two boys?” I’ve been told the answer is 1/3, but I can’t wrap my head around it. Additionally, there is another version of the problem that states he has at least one boy born on a Tuesday. How does that change the odds? Why?
Edited to add (so people don’t have to sort through replies): the answer is 1/3 because “at least one boy” is accounting for B/G & G/B. The girl can be the first or second child. You can move the odds to 50/50 by rewording the question to “my first of two children is a boy, what are the odds the other child is a boy”
In: Mathematics
I think the biggest mistake that people are making in visualizing the problem as a coinflip is considering the 2 flips to be of the “same” coin. This produces a false equivalence between the first and second child. If you call Heads girls, and Tails boys, and the first born is a Quarter, and the second born is a Dime. This becomes more obvious:
Heads Quarter, Heads Dime => GG
Heads Quarter, Tails Dime => GB
Tails Quarter, Heads Dime => BG
Tails Quarter, Tails Dime => BB
If you eliminate the HH combination, that leaves 3 possible outcomes for the father in the question. GB and BG cannot be conflated.
For those arguing 2/3, does this mean that of those that have 2 kids, one of which is a boy, that 2/3 of them have a girl.
It’s 50:50, which can be manipulated to 2/3 if you set some funky conditionals and phrasings, or you lock the background population (like the Monty hall problem). Birth gender is always a fun one for stats because the 50:50 is the only metric that impacts it. A good example of this is “does the 1 child policy (stop after 1 boy or repeat until you have one, impact gender ratios – no it doesn’t, always 50:50, reduced eventually cos women live longer unless there’s female infanticide, and that shows up on the stats very sharply.
This question has many variants but to solve this version.
If you have a family with 2 children and at least 1 is a boy but gave no more information then the only possible outcomes for the family can be BB GB BG. That is 1 in 3.
If you cant wrap your head around it,
Imagine walking into a room with 2 silhouettes and you were told one of these is a boy what are the chances both are boys? As you don’t know which one is the boy but one has to be a boy then your options can only be BB GB BG. 1 in 3.
A fun way to think of this is a lottery, where each qualifying person can draw only once, and each person who draws has an equal chance for winning a prize for the family. There is always exactly one winner.
The families competing are two person families. The representatives are the two children, and their father, in this case.
The question is, once the qualifiers are given, what are the odds that a two son family wins?
If each father draws, it’s a 25%, obviously. One chance for each family to win.
If each eldest son draws, it’s 50%. Same for each youngest son. Half of the two-son’s competitors are disqualified, so the chances double.
If each daughter draws, it’s 0%. The two son families are disqualified, so how can they win?
If each son draws, it’s 50% again. While there are twice as many son-daughter pairs, and the two-daughter families are disqualified, the two-son families get two chances each, and even the odds. But what if they didn’t get that second ticket?
This is the two children problem. The lottery asks for a father to draw, but in order to do so, they must approach with exactly one son of their family. This is the father stating that they have two kids, and at least one is a boy. Even if they have two sons, their family only gets one ticket. And, remember, the two-daughter families are unable to draw, and the son-daughter families are still twice as numerous and still get their one ticket.
With their advantage lost, a two-son family will win 1/3 of the time, and a son-daughter family will win 2/3 of the time.
TL;DR at the bottom.
So.
I was confident in my answer of 1/3, because this is what is the typical mathematical answer. It lines up with the questions and answers you would get in this field. But I admit, that it is not the sole correct answer in the sense, that the question itself might be ambiguous, if we disregard some reasonable assumptions.
I recommend reading though the [wiki page](https://en.m.wikipedia.org/wiki/Boy_or_girl_paradox). If you don’t get something, try to keep on reading to see if you understand it after getting some more details. I also recommend reading at least some of the sources (and I mean published papers) to get some insight, it helped to clear up some things for me.
The answer depends on assumptions that some make, and some might not.
We assume:
* the father knows the sex of both kids (reasonable)
* the father doesn’t lie about the sex of his kids
* the options are only boy or girl, with a 50-50% probability for each child independently
* if the father has at least a boy, he will say he has at least a boy
Each of these assumptions are required for the answer to stay 1/3.
1) The person that gives the statement only knows the sex of one kid.
In the case of a wording with a father, it mostly makes sense that they know both of their kids’ sex. But if a bunch of two-kid families are randomly selected, then one of their kids are randomly checked if they are boy or girl, the probabilities get tilted back to 1/2. (For details see the wiki page’s Analysis of the Ambiguity section.)
2) The father lies about his kids
This obviously invalidates the “trick” in the question, since if we can’t exclude one of the four options from the BB/BG/GB/GG set, the probability of both kids being boys will depend on the probability of the dad lying about his kids.
3) We don’t live in a “perfect” world
In reality, the probability of a male or female baby being born is not equal, and we can’t necessarily disregard eg. intersex babies either. These all would influence the answer:
eg. 105 boys (50.7%), 100 girls (48.3%), 2 intersex (1.0%) babies would result in a somewhat different probability
4) The father is *not* biased to the extreme
We assume that the dad of GB/BG kids will *always* say that he has at least a boy. If we don’t make this assumption, then we have to factor in, that eg. only half the time does a father say that he has at least a boy, while at other occasions he says he has at least a girl.
Here this probability will again heavily influence what exactly the result is. For example in a 50-50 scenario, we can adjust the probabilities to the following:
BB – 1
BG – 0.5
GB – 0.5
GG – 0
The eliminate the GG option, and we reach 1/(1+0.5+0.5) = 1/2 as the answer.
The above list might not be exhaustive and complete, but it has some examples of how we can divert the answer from the intended 1/3. “Diverting” here is the key, since getting to answers other than 1/3 requires you to go out of your way to “misunderstand” the task. (And that is not what most of the comments saying 1/2 are doing.)
For the Tuesday question, it is also counterintuitive, and can be ambiguous in how the data was acquired.
For the generally correct answers (that a university course’s prof would almost definitely accept), I recommend reading The Guardian’s [article](https://www.theguardian.com/science/2019/nov/18/did-you-solve-it-the-two-child-problem) on it. (For anybody faced with a paywall, here is an [archived version](https://web.archive.org/web/20230821020730/https://www.theguardian.com/science/2019/nov/18/did-you-solve-it-the-two-child-problem) of it. Haven’t read this one, but skimming through the text, it seemed like the contents are the same as on the live page currently.)
TL;DR
Good answers are 1/3 (og question), and 13/27 (if we are given the day of birth of “the” boy), but due to the ambiguity of the wording, we can argue that you could turn it into basically any percentage depending on assumed context.
Sources: [wiki](https://en.m.wikipedia.org/wiki/Boy_or_girl_paradox) (and its sources), [The Guardian](https://web.archive.org/web/20230821020730/https://www.theguardian.com/science/2019/nov/18/did-you-solve-it-the-two-child-problem)
Humans are born male or female with probability 50% (for the purposes of simple probability problems).
So in a scenario where you’re trying to guess a human’s sex with no additional information, the probability would be 50%. For example: “This is Jimmy. He has 1 sibling. Do you think he has a brother or a sister?” Here, his sibling is a human, therefore there is 50% probability they are either sex.
This scenario is like if you toss your favourite pet coins, Athena and Barbara, and look at Athena but not Barbara. Barbara, like all coins, is 50/50.
The interpretation that results in 1/3 is if you read the words “I have two children. At least one of them is a boy. What is the probability I have two sons?” and say okay, I don’t have any information about any human. There is only a whole circumstance including two humans: at least one is a boy.
It’s like if you’re on holiday, and you ask your friend to toss your coins and tell you whether there are any heads. If they tell you yes, what do you think both of them are? Note I can’t say “the other one” because there is *no* “*the other one*” — *you don’t know* which one is heads. The three different equally likely possibilities in the circumstances as described are HT TH HH (where I’m writing Athena on the left and Barbara on the right).
Whether this interpretation is a reasonable way to read these words is something that people disagree about.
I hope this helps!
One fourth of all parents with two children have two girls, one fourth have two boys and a half has a girl and a boy.
So, if you consider 400 pairs, 100 of them have two boys and 300 have one or two boys. 100 is a third of 300.
Maybe sometimes it helps to convert a question about probablility into a question about distributions.
Latest Answers