These problems are actually variations of one of the oldest “probability paradoxes” ever. And I put that in quotes, because in 1889 when Joseph Bertrand published it, “Bertrand’s Box Paradox” meant how he proved that one proposed answer could not be right, because it produced a contradiction.
Here is that paradox, applied to Problem #1. With a slight modification that changes nothing except using a complimentary probability. In each question, what is the probability that I have a boy and a girl?
I have two children, at least one of whom is a boy.
I have two children, at least one of whom is a girl.
I have two children, and have written the gender of at least one in this sealed envelope.
The answers to #1 and #2 have to be the same. It is tempting to say that answer is 2⁄3.
BUT, in #3, there are only two possibilities for what is written in the envelope. Once opened, it reduces to either #1, or #2. So the Law of Total Probability tells us that #3, even before the envelope is opened, also has that same answer.
But with no information given, the answer to #3 is clearly 1⁄2. So it can’t be 2⁄3. This is the paradox in Bertrand’s Box Problem. If you add a fourth box to that one, with a gold and a silver coin, it is identical to this Boy or Girl problem. I’m not saying that this directly proves that 1⁄2 is right. Just that no other answer can be right, and 1⁄2 can. Much like how proof by contradiction works.
Martin Gardner popularized this problem in the May, 1959 issue of ScientificAmerican. But his wording was “Mr. Smith has two children. At least one is a boy.” At first he said the answer was 2⁄3, but he withdrew that answer the next October, in a column titled “Probability and Ambiguity.” His problem statement was ambiguous because it did not specify how the information came to be known. As others have pointed out, some methods lead to the answer 2⁄3, and others to the answer 1⁄2.
This wording is also ambiguous, but for a different reason. Whoever “I” is, knows the genders of the two children, and picked one of them. What we don’t know, is how. But a Bayesian approach says that if he has a boy and a girl, we should assume he picked at random. This makes the answer 1⁄2.
To demonstrate this, Gardner published the Three Prisoners Problem in that October 1959 column. It is identical to the more modern puzzle, the Monty Hall Problem. With one exception: Gardner specified that a coin flip was used to choose what information is revealed, when both kinds of information are possible. That is usually not made explicit in Monty Hall, yet it is a necessary assumption to get that switching doors has a 2⁄3 probability of winning.
These problems are actually variations of one of the oldest “probability paradoxes” ever. And I put that in quotes, because in 1889 when Joseph Bertrand published it, “Bertrand’s Box Paradox” meant how he proved that one proposed answer could not be right, because it produced a contradiction.
Here is that paradox, applied to Problem #1. With a slight modification that changes nothing except using a complimentary probability. In each question, what is the probability that I have a boy and a girl?
I have two children, at least one of whom is a boy.
I have two children, at least one of whom is a girl.
I have two children, and have written the gender of at least one in this sealed envelope.
The answers to #1 and #2 have to be the same. It is tempting to say that answer is 2⁄3.
BUT, in #3, there are only two possibilities for what is written in the envelope. Once opened, it reduces to either #1, or #2. So the Law of Total Probability tells us that #3, even before the envelope is opened, also has that same answer.
But with no information given, the answer to #3 is clearly 1⁄2. So it can’t be 2⁄3. This is the paradox in Bertrand’s Box Problem. If you add a fourth box to that one, with a gold and a silver coin, it is identical to this Boy or Girl problem. I’m not saying that this directly proves that 1⁄2 is right. Just that no other answer can be right, and 1⁄2 can. Much like how proof by contradiction works.
Martin Gardner popularized this problem in the May, 1959 issue of Scientific American. But his wording was “Mr. Smith has two children. At least one is a boy.” At first he said the answer was 2⁄3, but he withdrew that answer the next October, in a column titled “Probability and Ambiguity.” His problem statement was ambiguous because it did not specify how the information came to be known. As others have pointed out, some methods lead to the answer 2⁄3, and others to the answer 1⁄2.
This wording is also ambiguous, but for a different reason. Whoever “I” is, knows the genders of the two children, and picked one of them. What we don’t know, is how. But a Bayesian approach says that if he has a boy and a girl, we should assume he picked at random. This makes the answer 1⁄2.
To demonstrate this, Gardner published the Three Prisoners Problem in that October 1959 column. It is identical to the more modern puzzle, the Monty Hall Problem. With one exception: Gardner specified that a coin flip was used to choose what information is revealed, when both kinds of information are possible. That is usually not made explicit in Monty Hall, yet it is a necessary assumption to get that switching doors has a 2⁄3 probability of winning.