It might seem as though the answer is 1/2 again. However, this time there are 3 possibilities:
1) Older child is a boy and the younger a girl.
2) Older child is a boy and the younger a boy.
2) Older child is a girl and the younger a boy.
Each possibility seems to be equally likely. Therefore it seems as though the probability of both children being boys is 1/3.
OR IS IT?
Things are not always as they seem at first. How was Prof Ravenscroft's family chosen? Was his family chosen at random from all families with two children; or from all families with two children, at least one of whom is a boy?
Let us assume that the former is the case, and that the family were chosen at random from all families with two children. Now, families with two boys might be said to have "at least one boy" and families with two girls might be said to have "at least one girl", but families with one boy and one girl might be said to have either "at least one boy" or "at least one girl", with the phrase being chosen at random.
In this case, the probability that Prof Ravenscroft has two boys is 1/2, not 1/3.
For two children families, there are four equally probable possibilities:
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl
In only half of these scenarios are there two children of the same sex.
But one does not know how the families were chosen and nominated. If the family was chosen at random from all families with two children, at least one of whom is a boy, the probability would indeed be 1/3 as was calculated earlier.
It could even be that the family were chosen at randon from all two children families, but that all families with a boy were nominated as having "at least one boy". In this case the probability would again be 1/3, as all the boy/girl combinations would now be nominated as having "at least one boy" rather than only half of them.
SO WHAT IS THE ANSWER?
The answer is that the problem is not answerable without more information.
Could one conclude that there are two equally probable scenarios, and that the probability of Prof. Ravenscroft having 2 boys is therefore half the sum of the two possible answers, 1/2 and 1/3, ie 5/12? (For more understanding about this, research the Principle of Indifference.)
Another consideration might be whether it is safe to assume that the probability of someone's child being a boy is indeed 1/2. In China, for instance, the official birth rates are 113 to 100 in favour of boys, which the State Family Planning Commission attributes to the widespread use of selective abortions, the abandonment of baby girls and infanticide, brought about, in part, by the government's "one child" policy.
A further consideration might be that 42.35% of statistics are made up; so don't take this site's word for it.
This problem is a simple illustration of how easy it can be to make mistakes in probablilty, and why this area of mathematics lends itself so easily and frequently to paradoxical fallicies.
However, it is intersting to note that if one were only told that Prof Ravenscroft had exactly two children, and one then ASKED whether at least one was a boy, the problem would then be answerable.
If the response was that he did have at least one boy, the probability of him having two boys would then be 1/3. The reason is that in this case all the boy/girl combinations would lead to a possitive response, and only a girl/girl combination would give a negative response, so it would not matter from which sample population the Professor and his children were taken.