A couple has two children. At least one of them is a boy. Assuming the probability of having a boy or girl is 50%, what is the probability that both children are boys?
If you answered 1/2, you’re not without comrades, but the generally accepted answer by statisticians (though not without debate) is 1/3. This is because there are four possible combinations: boy-boy, boy-girl, girl-boy and girl-girl. Since we are told one of the children is a boy (but we don’t know if it’s the first or second child), we can rule out the girl-girl combination, leaving three remaining options. Only one out of 3 is boy-boy, so we get a 1/3 chance.
After the Wikipedia article, I think this is a good way to explain it if someone says it’s “1/2”:
Imagine “At least one is a boy” is equivalent to being shown one boy, and being asked what the other’s gender is.
G/G is ruled out. Now, some people might consider that “G/B” is not a valid possibility either, because we’ve been shown “B” — and so, that the only possibilities are B/G and B/B.
However, G/B is valid: Whether it were B/G or G/B, we’ve been shown the B part of it. So with B/B, B/G, and G/B as possibilities, the probability of each is 1/3.
The problem with your logic is such:
Not only is G/G not possible, but G/B is also not possible since we already know the gender of the first noted child (which is B).
The reason Ramon’s argument doesn’t work is that although the B could be considered either the first or the second child it DOES have to be one or the other (and in either case it still only yields two possible outcomes – not three).
take the 2 kids as 100% of a child, total being 200%, first child is 100% boy, 2nd one is 50% chance of being a boy, 100+50=150, 150/200= .75, .75×100=75%
Everyone, I have seen this question come up many times in the past and I can give you the real answer along with the problem in this solution.
The answer is 50%, sorry “statisticians.” The logical flaw in this explanation is that you list out: boy-boy, boy-girl, girl-boy. You are listing that girl-boy and boy-girl are different options but fail to realize boy-boy could be boy1-boy2 and also boy2-boy1. the REAL possible outcomes are boy1-boy2, boy2-boy1, boy1-girl2, and girl1-boy2 = 50% chance of having two boys. I hope this helps explain things.
Superwhiz Maths Person (also neutral as to not offend those who have a complex of themselves and others) says March 10, 2018 @ 20:01
I say…there is a missing factor here that complexifies the issue.
Firstly, there is the birth defect, where a person could be born a hermaphrodite, and only in the future will they potentially choose one of two well-known surgical procedures – namely a lopadicome or an adadictome. This shows that the percentage of knowing whether the other child is girl or boy. It is also why a person with two feet has more feet than the average person, due to more people having one or less feet than there are those with three or more feet (feet=tarsals and associated appendages).
Secondly, in this day of being able to say to a man “How’s your husband?” and not getting a punch in the face, it is even more difficult to determine, as because a person grows up and is groomed by the lobby groups for their own desires, a person who one sees as a boy may see themselves as a girl, and a girl may see themselves as a boy.
These areas make it a little more difficult in determining the statistics of the statisticians.
There are totally four different situations, which are girl/boy, boy/girl, girl/girl, boy/boy
Bayes’ theorem can solve this problem. Let’s assume Event A: at least one child is a boy, Event B: they have two boys.
By Bayes’s theorem, we can get P(B|A) = P(B)*P(A|B)/P(A)
Obviously, P(B|A) is the answer of this question.
From the above four different situations but with equal probabilities, we can get that P(B)=1/4, P(A)=3/4, P(A|B)=1, so in this way, P(B|A) = 1/3
11 Comments on "At Least One is a Boy"
Probability That Both Children Are Girls | Riddles and Brain Teasers says
November 5, 2013 @ 17:13
[…] Another variation of this can be found here. […]
Ramón says
April 17, 2015 @ 17:38
After the Wikipedia article, I think this is a good way to explain it if someone says it’s “1/2”:
Imagine “At least one is a boy” is equivalent to being shown one boy, and being asked what the other’s gender is.
G/G is ruled out. Now, some people might consider that “G/B” is not a valid possibility either, because we’ve been shown “B” — and so, that the only possibilities are B/G and B/B.
However, G/B is valid: Whether it were B/G or G/B, we’ve been shown the B part of it. So with B/B, B/G, and G/B as possibilities, the probability of each is 1/3.
Justin says
May 11, 2015 @ 15:13
The answer IS 1/2.
The problem with your logic is such:
Not only is G/G not possible, but G/B is also not possible since we already know the gender of the first noted child (which is B).
The reason Ramon’s argument doesn’t work is that although the B could be considered either the first or the second child it DOES have to be one or the other (and in either case it still only yields two possible outcomes – not three).
Audreybb18 says
February 4, 2016 @ 11:22
I think it’s 2/3.for boys.
This is really similar to the Monty Hall Problem.
logic says
April 6, 2016 @ 11:55
take the 2 kids as 100% of a child, total being 200%, first child is 100% boy, 2nd one is 50% chance of being a boy, 100+50=150, 150/200= .75, .75×100=75%
Jay says
June 7, 2016 @ 22:11
I’m like I’m not reading all of this
Jackie says
September 27, 2016 @ 22:17
Still 50%
Liberty says
November 10, 2016 @ 06:22
Everyone, I have seen this question come up many times in the past and I can give you the real answer along with the problem in this solution.
The answer is 50%, sorry “statisticians.” The logical flaw in this explanation is that you list out: boy-boy, boy-girl, girl-boy. You are listing that girl-boy and boy-girl are different options but fail to realize boy-boy could be boy1-boy2 and also boy2-boy1. the REAL possible outcomes are boy1-boy2, boy2-boy1, boy1-girl2, and girl1-boy2 = 50% chance of having two boys. I hope this helps explain things.
Superwhiz Maths Person (also neutral as to not offend those who have a complex of themselves and others) says
March 10, 2018 @ 20:01
I say…there is a missing factor here that complexifies the issue.
Firstly, there is the birth defect, where a person could be born a hermaphrodite, and only in the future will they potentially choose one of two well-known surgical procedures – namely a lopadicome or an adadictome. This shows that the percentage of knowing whether the other child is girl or boy. It is also why a person with two feet has more feet than the average person, due to more people having one or less feet than there are those with three or more feet (feet=tarsals and associated appendages).
Secondly, in this day of being able to say to a man “How’s your husband?” and not getting a punch in the face, it is even more difficult to determine, as because a person grows up and is groomed by the lobby groups for their own desires, a person who one sees as a boy may see themselves as a girl, and a girl may see themselves as a boy.
These areas make it a little more difficult in determining the statistics of the statisticians.
NOne ya buisness!!!! says
March 12, 2019 @ 21:30
i think 50%
Jingmin says
December 10, 2020 @ 06:29
There are totally four different situations, which are girl/boy, boy/girl, girl/girl, boy/boy
Bayes’ theorem can solve this problem. Let’s assume Event A: at least one child is a boy, Event B: they have two boys.
By Bayes’s theorem, we can get P(B|A) = P(B)*P(A|B)/P(A)
Obviously, P(B|A) is the answer of this question.
From the above four different situations but with equal probabilities, we can get that P(B)=1/4, P(A)=3/4, P(A|B)=1, so in this way, P(B|A) = 1/3
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