You are talking about the probability of a single point event, which is also always 0 on that same sigma algebra.

The OP point is not completely trivial because the event contains an infinite (but countably) number of elements. It is fairly easy to understand though since by its very definition, the P[{X in Q}] = sum P[{x}] taken over every rational number (since Q is countable), and each P[{x}] is 0.

A deeper statement is that there exists uncountable sets of probability 0.

But this "You know what Mr. Jones has told you, but you don't know under what circumstances he would have told you this" objection is objecting to some other problem than the one posed; the problem posed had nothing to do with Mr. Jones saying anything.

I understand the reason for worrying about this, because many probability riddles ARE poorly worded or presented in such a way as that this becomes an issue, but it wasn't the case here. (Note: I haven't watched rest of the video and have no comment on it; I'm just considering the wording of this individual question within it)

There was never any claim that Mr. Jones said anything, and no one was called to infer anything from any actions taken by Mr. Jones. He could be a lifelong mute. Rather, the fact that Mr. Jones has two children was presented, by an omniscient narrator, and then a counting question was asked.

(Indeed, Mr. Jones himself is completely irrelevant to the problem asked, except as a way of framing the counting question to be about two-children families. The question asked might as well have been "What proportion of two-children families with a boy born on Tuesday have girls?". It was very slightly differently worded, but not in such a way as makes "We don't know what Mr. Jones was asked!" a relevant objection.)

Once you accept that you have an entropy source in the physical world, you can easily be injecting random real numbers (from some range) and in fact, usually are, which are then being binned into integers by ADCs.

problem 1) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy born on Tuesday. They say yes. What is the probability that they have a girl?

problem 2) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy. They say yes. You then ask them which day of the week a boy they have was born on. They say Tuesday. What is the probability that they have a girl?

The original problem that was posed is equivalent to problem 1, but not equivalent to problem 2. This could be what is confusing you, because in problem 2 the extra information plays no role in the selection process, while it does play a role in problem 1. In problem 2, the answer is the standard 2/3. Why are the probabilities different between problem 1 and 2? Here's why:

Think about the set of people who could answer yes to the question in problem 2. The ratio of these groups is important. A parent with BB (two boys) is equally likely to answer yes to problem 2 (100% likely to be exact) as a parent with BG and GB (also 100% likely to answer yes), which leads to the correct solution of 2/3. However, in problem 1 a parent with BB is NOT EQUALLY LIKELY to answer yes as a parent with BG. This is because we added an extra qualifier (must be born on Tuesday). The parent with BB has two chances to meet this qualifier because they have two boys, so the parent with BB is actually more likely to answer yes to the question than the parent with BG. As the qualifier becomes more and more rare (day of lunar cycle), the probability of the BB parent answer yes P(yes|BB) approaches twice the value of P(yes|BG). So now you're left with some subset of parents with BB, BG, and GB, but in this scenario you've sampled from BB approximately twice as much as you've sampled from each of the BG and GB groups, leaving you with approximately the same number of people from group BB as the combined amount from groups BG and GB. This is why the probability approaches 50%

I spend a while writing this, so hopefully it helps!

If K1 is a girl and K2 is a boy born on Tuesday, this still counts as the family (Mr. Jones, if you like) having a boy born on Tuesday. There are 27 cases that count as the family having a boy born on Tuesday, all equiprobable. And out of those, 14 also count as the family having a girl.

As for your noting that we can split these cases even more finely, so that there's no distinguished end-all, be-all partitioning of cases, sure, you can do that. What I'm really saying is this:

1/2 of two-child families have their elder child being a boy. 1/2 of two-child families have their elder child being a girl. [On conventional idealizations for these problems. You surely do not dispute this, yes? You may not care about this number, but you don't dispute it, right?]

In each of those subgroups, 1/7 of families have their elder child born on Sunday, 1/7 have their elder child born on Monday, etc. [Do you dispute this?]

In each of THOSE subgroups, 1/2 of families have their younger child a boy, and 1/2 have their younger child a girl. [Any dispute?]

And in each of THOSE subgroups, 1/7 of families have their younger child born on Sunday, 1/7 of families have their younger child born on Monday, etc. [Any dispute?]

And some amount of those have low birth weight, some have high birthweight, some have 5 freckles, etc., but we needn't figure out those numbers.

So now I've carved the world up into 2 * 7 * 2 * 7 groups, based on gender and birth-date-of-week for older and younger child. We can carve the world up into groups in different ways also, more finely or more coarsely or just differently. But making the four conventional assumptions we just made, the 2 * 7 * 2 * 7 grouping based on gender and birth-date-of-week for older and younger child is such that each particular such group takes up 1/2 * 1/7 * 1/2 * 1/7 of all families; these are all equifrequent groups.

And that having been done, we find that in 27 of these groups, there is at least one boy born on a Tuesday. In 13, the elder child is a boy born on Tuesday but not the younger child; in 13, the younger child is a boy born on Tuesday but not the elder child; in 1, both children are boys born on Tuesday.

But the question was not intended to be about a specific boy. The question was intended to be "Out of families that have a boy born on Tuesday (meaning at least one boy born on Tuesday), what proportion have a girl?". Any family with at least one boy born on Tuesday counts as having "a boy born on Tuesday", and even families with two boys born on Tuesday count, with no particular of their two boys given any distinguished status.

Perhaps you read the question differently; that, then, is a problem with the phrasing of the question for communicating to you its intent. But when it was asked "What is the probability Mr. Jones has a girl, given that he has a boy born on Tuesday?", what the author indeed intended this to mean, and would be generally taken in the conventional language of probability to mean, was "Out of families that have at least one boy born on Tuesday, what proportion have a girl?".

And we find that, out of the 27 equally sized groups of families that have at least one boy born on Tuesday, 14 of them have a girl, so that the answer to this question becomes 14/27.

This is a testable prediction. I predict that making the source of your knowledge explicit eliminates the paradox.

To me, it feels strange that "the probability that Mr. Jones has a girl given that he has a boy born on Tuesday" is ~1/2. However, it feels normal that "You ask Mr. Jones weather he has a boy born on Tuesday, and he says yes. What is the probability that he has a girl?" is ~1/2.

Do other people agree?