Your solution curtails the point of the article. Information affects probability. To be concise, you made indistinguishable children distinguishable. We cannot solve the problem, in the way it is posed, by explicitly selecting one child and moving that child through the outcomes.
By way of demonstration, consider the case in which there are two boys. If the children are distinguishable, then we are at liberty to give each boy his own identity. Let's name one boy Peter. Two outcomes follow:
Peter, Other Boy
Other Boy, Peter
If the children are indistinguishable, then we are prohibited from naming one boy Peter because do not know which boy is the first and which boy is the second. Thus, only one outcome follows:
Boy, Boy
In your solution, you counted the same case twice (Peter-Boy and Boy-Peter), so you mistook two out of four cases as being the answer instead of one out of three.
Slashdot Mirror
Your solution curtails the point of the article. Information affects probability. To be concise, you made indistinguishable children distinguishable. We cannot solve the problem, in the way it is posed, by explicitly selecting one child and moving that child through the outcomes.
By way of demonstration, consider the case in which there are two boys. If the children are distinguishable, then we are at liberty to give each boy his own identity. Let's name one boy Peter. Two outcomes follow:
Peter, Other Boy
Other Boy, Peter
If the children are indistinguishable, then we are prohibited from naming one boy Peter because do not know which boy is the first and which boy is the second. Thus, only one outcome follows:
Boy, Boy
In your solution, you counted the same case twice (Peter-Boy and Boy-Peter), so you mistook two out of four cases as being the answer instead of one out of three.