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

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.

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.

This is really similar to the Monty Hall Problem.

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).

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.