One has to consider the data-generating process behind puzzles such as these. I am going to assume throughout that it is what seems to me the simplest consistent with the statement of the problems.
Every child is independently equally likely to be a boy or a girl.
Every birth day of the week is equally and independently likely.
The puzzle narrator is chosen equally at random from all members of the population for which the statements that the narrator makes are true.
One can imagine all manner of other data-generating processes, which in general will give different answers. Most quibbles with such problems come down to imagining different processes, especially for statement 3. Some examples are in the OP and the comments. However, the above assumptions seem the simplest, and the ones intended by people posing such problems. If this question came up on an exam, I would be sure to begin my answer with the above preamble.
Puzzles 1 and 2 can be unified by considering a common generalisation to there being N days in a week, one of which is called Tuesday. Puzzle 1 has N=1. Puzzle 2 has N=7.
When two children are born, there are 4 N^2 possibilities for their sexes and birthdays. Which of these are left after being given the information in the puzzle?
The sexes must be either boy-boy, boy-girl, or girl-boy.
For boy-boy, there are 2N-1 ways that at least one is born on a Tuesday.
For boy-girl, there are N ways the boy could be born on a Tuesday.
For girl-boy, by symmetry also N.
So the probability of boy-boy is (2N-1)/(2N-1 + N + N) = (2N-1)/(4N-1).
For N=1 this is 1⁄3, which is the generally accepted answer to Puzzle 1.
For N=7, it is 13⁄27. In general, the larger N is, the closer the probability of the other being a boy is to 1⁄2.
For Puzzle 3, the obvious answer is that before learning the day, it’s as Puzzle 1, 1⁄3, and after, it’s as Puzzle 2, 13⁄27. Is obvious answer correct answer? I have (before reading the OP’s solution) not found a reason not to think so. The OP says, by a different intuitive argument, that the obvious answer is 1⁄3, and that obvious answer is correct answer, but WilliamKiely’s comment raises a doubt, finding ambiguity in the statement of the data-generating process. This leaves me as yet undecided.
I started thinking about the ways that extra information (beyond “I have two children, at least one of whom is a boy”) affects the probability of two boys, and came up with these:
Puzzle 4. I have two children, of whom at least one is a boy with black hair. In this fictional puzzle, everyone knows that all the children in a family have the same hair colour. What is the probability that the other is also a boy?
Puzzle 5. I have two children, of whom at least one is a boy born on a Tuesday. In this fictional puzzle, everyone knows that two consecutive children never have the same birth day. What is the probability that the other is also a boy?
Puzzle 6. I have two children. The elder one is a boy. What is the probability that the younger is also a boy?
Puzzle 7. I have two children, called Alex and Sam. (In this fictional puzzle, these names communicate no information about gender.) Alex is a boy. What is the probability that Sam is a boy?
Puzzle 8. I have two children, of whom at least one is a boy. What is the probability that the elder child is a boy?
I believe the answers to these are 4: 1⁄3. 5: 6⁄13 (slightly smaller than the 13⁄27 of Puzzle 2). 6: 1⁄2. 7: 1⁄2. 8: 2⁄3.
All of these can be found by the same method of considering all the possibilities that are consistent with all the information given, and counting the proportion where both children are boys.
But there is also a general idea underlying them. When you are given extra information about the boy that is said to exist, that will push the probability of the other being a boy towards 1⁄2, to the extent that the extra information breaks the symmetry between them.
In Puzzle 1, no extra information is given, and the answer is 1⁄3. In Puzzle 4, the extra information does not break the symmetry, so the probability remains 1⁄3. In Puzzles 2 and 5, it partly breaks the symmetry, and the answers lie between 1⁄3 and 1⁄2. In 6 and 7, it completely breaks the symmetry, and the answer is 1⁄2.
Puzzle 8 is Puzzle 1 with a different question, equivalent to: what is the expected proportion of boys among my children?
Puzzle 7a. I have two children, of whom at least one is a boy. Their names are Alex and Sam. (In this fictional puzzle, these names communicate no information about gender.) Alex is a boy. What is the probability that Sam is a boy?
Puzzle 1a. I have two children, of whom at least one is a boy. Now I shall toss a coin and accordingly choose one of the two children. (Does so.) The child I chose is a boy. What is the probability that the other is a boy?
One has to consider the data-generating process behind puzzles such as these. I am going to assume throughout that it is what seems to me the simplest consistent with the statement of the problems.
Every child is independently equally likely to be a boy or a girl.
Every birth day of the week is equally and independently likely.
The puzzle narrator is chosen equally at random from all members of the population for which the statements that the narrator makes are true.
One can imagine all manner of other data-generating processes, which in general will give different answers. Most quibbles with such problems come down to imagining different processes, especially for statement 3. Some examples are in the OP and the comments. However, the above assumptions seem the simplest, and the ones intended by people posing such problems. If this question came up on an exam, I would be sure to begin my answer with the above preamble.
Puzzles 1 and 2 can be unified by considering a common generalisation to there being N days in a week, one of which is called Tuesday. Puzzle 1 has N=1. Puzzle 2 has N=7.
When two children are born, there are 4 N^2 possibilities for their sexes and birthdays. Which of these are left after being given the information in the puzzle?
The sexes must be either boy-boy, boy-girl, or girl-boy.
For boy-boy, there are 2N-1 ways that at least one is born on a Tuesday. For boy-girl, there are N ways the boy could be born on a Tuesday. For girl-boy, by symmetry also N.
So the probability of boy-boy is (2N-1)/(2N-1 + N + N) = (2N-1)/(4N-1).
For N=1 this is 1⁄3, which is the generally accepted answer to Puzzle 1.
For N=7, it is 13⁄27. In general, the larger N is, the closer the probability of the other being a boy is to 1⁄2.
For Puzzle 3, the obvious answer is that before learning the day, it’s as Puzzle 1, 1⁄3, and after, it’s as Puzzle 2, 13⁄27. Is obvious answer correct answer? I have (before reading the OP’s solution) not found a reason not to think so. The OP says, by a different intuitive argument, that the obvious answer is 1⁄3, and that obvious answer is correct answer, but WilliamKiely’s comment raises a doubt, finding ambiguity in the statement of the data-generating process. This leaves me as yet undecided.
I started thinking about the ways that extra information (beyond “I have two children, at least one of whom is a boy”) affects the probability of two boys, and came up with these:
Puzzle 4. I have two children, of whom at least one is a boy with black hair. In this fictional puzzle, everyone knows that all the children in a family have the same hair colour. What is the probability that the other is also a boy?
Puzzle 5. I have two children, of whom at least one is a boy born on a Tuesday. In this fictional puzzle, everyone knows that two consecutive children never have the same birth day. What is the probability that the other is also a boy?
Puzzle 6. I have two children. The elder one is a boy. What is the probability that the younger is also a boy?
Puzzle 7. I have two children, called Alex and Sam. (In this fictional puzzle, these names communicate no information about gender.) Alex is a boy. What is the probability that Sam is a boy?
Puzzle 8. I have two children, of whom at least one is a boy. What is the probability that the elder child is a boy?
I believe the answers to these are 4: 1⁄3. 5: 6⁄13 (slightly smaller than the 13⁄27 of Puzzle 2). 6: 1⁄2. 7: 1⁄2. 8: 2⁄3. All of these can be found by the same method of considering all the possibilities that are consistent with all the information given, and counting the proportion where both children are boys.
But there is also a general idea underlying them. When you are given extra information about the boy that is said to exist, that will push the probability of the other being a boy towards 1⁄2, to the extent that the extra information breaks the symmetry between them.
In Puzzle 1, no extra information is given, and the answer is 1⁄3. In Puzzle 4, the extra information does not break the symmetry, so the probability remains 1⁄3. In Puzzles 2 and 5, it partly breaks the symmetry, and the answers lie between 1⁄3 and 1⁄2. In 6 and 7, it completely breaks the symmetry, and the answer is 1⁄2.
Puzzle 8 is Puzzle 1 with a different question, equivalent to: what is the expected proportion of boys among my children?
Puzzle 7a. I have two children, of whom at least one is a boy. Their names are Alex and Sam. (In this fictional puzzle, these names communicate no information about gender.) Alex is a boy. What is the probability that Sam is a boy?
Puzzle 1a. I have two children, of whom at least one is a boy. Now I shall toss a coin and accordingly choose one of the two children. (Does so.) The child I chose is a boy. What is the probability that the other is a boy?