Interesting. I hadn’t heard of the Child Born on Tuesday Problem. I think it’s actually quite relevant to Sleeping Beauty, but I won’t go into that here...
Both problems (your 1 and 2) aren’t well-defined, however. The problem is that in real life we do not magically acquire knowledge that the world is in some subset of states, with the single exception of the state of our direct sense perceptions. One could decide to assume a uniform distribution over possible ways in which the information we are supposedly given actually arrives by way of sense perceptions, but uniform distributions are rather arbitrary (and will often depend on arbitrary aspects of how the problem is formulated).
Here’s a boys/girls puzzle I came up with to illustrate the issue:
A couple you’ve just met invite you over to dinner, saying “come by around 5pm, and we can talk for a while before our three kids come home from school at 6pm”.
You arrive at the appointed time, and are invited into the house. Walking down the hall, your host points to three closed doors and says, “those are the kids’ bedrooms”. You stumble a bit when passing one of these doors, and accidentally push the door open. There you see a dresser with a jewelry box, and a bed on which a dress has been laid out. “Ah”, you think to yourself, “I see that at least one of their three kids is a girl”.
Your hosts sit you down in the kitchen, and leave you there while they go off to get goodies from the stores in the basement. While they’re away, you notice a letter from the principal of the local school tacked up on the refrigerator. “Dear Parent”, it begins, “Each year at this time, I write to all parents, such as yourself, who have a boy or boys in the school, asking you to volunteer your time to help the boys’ hockey team...” “Umm”, you think, “I see that they have at least one boy as well”.
That, of course, leaves only two possibilities: Either they have two boys and one girl, or two girls and one boy. What are the probabilities of these two possibilities?
The symmetrical summaries of what is learned are intentionally misleading (it’s supposed to be a puzzle, after all). The way in which you learned they have at least one girl is not the same as the way you learned that they have at least one boy. And that matters.
Your problem is both more, and less, well-posed than you think.
The defining feature of the “older child” version of the Two Child Problem has nothing to do with age. It is that you have ordered the children independently of gender, and identified the gender of a child in a position within that order. Age works well here, since it is easy to show why BB, BG, GB, and GG must be equiprobable by examining the event of the second birth.
But any gender-independent ordering works. It could be alphabetizing names, their seats around the dinner table (clockwise from Mother), or which bedroom each child has. You picked a specific child in an order by looking in a specific room, so the genders of the other two are independent of it and each other. So gBB, gBG, gGB, and gGG are equiprobable at that point in your acquisition of knowledge.
But your second acquisition depends on whether similar help is needed for other sports, and how many gender-specific sports there are. And why there isn’t one for girls’ sports, since we know there is a girl.
My problems are well-posed for what I intended. You didn’t “stumble upon” the information, a source with absolute knowledge told it to you, with no hint of any discrimination between genders. There is an established solution in such cases; it’s called Bertrand’s Box Paradox. That name did not, originally, refer to a problem; it referred to the following solution. It is best illustrated using a different probability than what I asked for:
I know Mr. Abbot’s two children. At least one is a boy.
I know Mrs. Baker’s two children. At least one is a girl.
I know the Curry’s two children. At least one has the gender that I have written inside this sealed envelope.
In each case, what is the probability that the family has a boy and a girl?
Clearly, the answers A1 and A2 must be the same. This is not using uniform distributions, although that is a valid justification. Probability is not about what is true in a specific instance of this disclosure of information—that’s a naive mistake. It is about what we can deduce from the information alone. It is a property of our knowledge of a world where it happens, not the world itself. Since our information is equivalent in Q1 and Q2, that means A1=A2.
But you have no significant information about genders in Q3, so A3 must be 1⁄2. And that can be used to get A1 and A2. Bertrand argued simply that if the envelope were opened, A3 had to equal A1 and A2 regardless of what it said, so you didn’t need to open it. Any change would be a paradox. But there is a more rigorous solution.
If W represents what is written in the envelope, the Law of Total Probability says:
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A2
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A1
A3 = [Pr(W=”Boy”) + Pr(W=”Girl”)]*A1
A3 = A1 = A2 (which all equal 1⁄2).
This solution is also used for the famous Monty Hall Problem, even if those using it do not realize it. The most common solution uses the assertion that “your original probability of 1⁄3 can’t change.” So, since the open door is revealed to not have the car, the closed door that you didn’t pick must now have a 2⁄3 probability.
The assertion is equivalent to my sealed envelope. You see the door that gets opened, which equivalent to naming one gender in Q1 and Q2. Since your answer must be the same regardless of which door that is, it is the same as when you ignore which door is opened.
Interesting. I hadn’t heard of the Child Born on Tuesday Problem. I think it’s actually quite relevant to Sleeping Beauty, but I won’t go into that here...
Both problems (your 1 and 2) aren’t well-defined, however. The problem is that in real life we do not magically acquire knowledge that the world is in some subset of states, with the single exception of the state of our direct sense perceptions. One could decide to assume a uniform distribution over possible ways in which the information we are supposedly given actually arrives by way of sense perceptions, but uniform distributions are rather arbitrary (and will often depend on arbitrary aspects of how the problem is formulated).
Here’s a boys/girls puzzle I came up with to illustrate the issue:
The symmetrical summaries of what is learned are intentionally misleading (it’s supposed to be a puzzle, after all). The way in which you learned they have at least one girl is not the same as the way you learned that they have at least one boy. And that matters.
Your problem is both more, and less, well-posed than you think.
The defining feature of the “older child” version of the Two Child Problem has nothing to do with age. It is that you have ordered the children independently of gender, and identified the gender of a child in a position within that order. Age works well here, since it is easy to show why BB, BG, GB, and GG must be equiprobable by examining the event of the second birth.
But any gender-independent ordering works. It could be alphabetizing names, their seats around the dinner table (clockwise from Mother), or which bedroom each child has. You picked a specific child in an order by looking in a specific room, so the genders of the other two are independent of it and each other. So gBB, gBG, gGB, and gGG are equiprobable at that point in your acquisition of knowledge.
But your second acquisition depends on whether similar help is needed for other sports, and how many gender-specific sports there are. And why there isn’t one for girls’ sports, since we know there is a girl.
My problems are well-posed for what I intended. You didn’t “stumble upon” the information, a source with absolute knowledge told it to you, with no hint of any discrimination between genders. There is an established solution in such cases; it’s called Bertrand’s Box Paradox. That name did not, originally, refer to a problem; it referred to the following solution. It is best illustrated using a different probability than what I asked for:
I know Mr. Abbot’s two children. At least one is a boy.
I know Mrs. Baker’s two children. At least one is a girl.
I know the Curry’s two children. At least one has the gender that I have written inside this sealed envelope.
In each case, what is the probability that the family has a boy and a girl?
Clearly, the answers A1 and A2 must be the same. This is not using uniform distributions, although that is a valid justification. Probability is not about what is true in a specific instance of this disclosure of information—that’s a naive mistake. It is about what we can deduce from the information alone. It is a property of our knowledge of a world where it happens, not the world itself. Since our information is equivalent in Q1 and Q2, that means A1=A2.
But you have no significant information about genders in Q3, so A3 must be 1⁄2. And that can be used to get A1 and A2. Bertrand argued simply that if the envelope were opened, A3 had to equal A1 and A2 regardless of what it said, so you didn’t need to open it. Any change would be a paradox. But there is a more rigorous solution.
If W represents what is written in the envelope, the Law of Total Probability says:
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A2
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A1
A3 = [Pr(W=”Boy”) + Pr(W=”Girl”)]*A1
A3 = A1 = A2 (which all equal 1⁄2).
This solution is also used for the famous Monty Hall Problem, even if those using it do not realize it. The most common solution uses the assertion that “your original probability of 1⁄3 can’t change.” So, since the open door is revealed to not have the car, the closed door that you didn’t pick must now have a 2⁄3 probability.
The assertion is equivalent to my sealed envelope. You see the door that gets opened, which equivalent to naming one gender in Q1 and Q2. Since your answer must be the same regardless of which door that is, it is the same as when you ignore which door is opened.