Initially, there are four possibilities, each with probability 1/4:
A) Boy, Boy
B) Boy, Girl
C) Girl, Boy
D) Girl, Girl
If you learn that one of them is a boy, then that eliminates option D, leaving the other three options (A, B, C) with 1⁄3 probability each. So the probability that both are boys given that at least one is a boy (ie., Pr[A] given A-or-B-or-C) is 1⁄3.
On the other hand, if you learn that the first child is a boy, that eliminates options D and C. You’ve ruled out more possibilities—whereas before ‘Girl, Boy’ (C) was an option, now the only options are ‘Boy, Boy’ (A) and ‘Boy, Girl’ (B). So there’s now a 1⁄2 chance that both are boys (i.e., Pr[A] given A-or-B). And the same calculation holds if you learned instead that the second child is a boy, only with B eliminated in place of C.
Thank you for the reply, RobbBB. As I mentioned in my reply to shinoteki (at 03 December 2012 01:48:47AM ), I followed my original post (to which you have just responded) with a post in which there is no reference to birth order. As I also said to shinoteki, that does not mean I see that birth order bears on this. It means simply that I was anticipating the response you, RobbBB, have just posted.
At 06 December 2012 10:18:40AM, as you may see, William Kasper posted a reply to my said second post (the one without reference to birth order). After I post the present comment, I will reply to Mr. Kasper. Thank you again.
Initially, there are four possibilities, each with probability 1/4:
A) Boy, Boy
B) Boy, Girl
C) Girl, Boy
D) Girl, Girl
If you learn that one of them is a boy, then that eliminates option D, leaving the other three options (A, B, C) with 1⁄3 probability each. So the probability that both are boys given that at least one is a boy (ie., Pr[A] given A-or-B-or-C) is 1⁄3.
On the other hand, if you learn that the first child is a boy, that eliminates options D and C. You’ve ruled out more possibilities—whereas before ‘Girl, Boy’ (C) was an option, now the only options are ‘Boy, Boy’ (A) and ‘Boy, Girl’ (B). So there’s now a 1⁄2 chance that both are boys (i.e., Pr[A] given A-or-B). And the same calculation holds if you learned instead that the second child is a boy, only with B eliminated in place of C.
Thank you for the reply, RobbBB. As I mentioned in my reply to shinoteki (at 03 December 2012 01:48:47AM ), I followed my original post (to which you have just responded) with a post in which there is no reference to birth order. As I also said to shinoteki, that does not mean I see that birth order bears on this. It means simply that I was anticipating the response you, RobbBB, have just posted.
At 06 December 2012 10:18:40AM, as you may see, William Kasper posted a reply to my said second post (the one without reference to birth order). After I post the present comment, I will reply to Mr. Kasper. Thank you again.