1) Suppose I tell you that I have 2 children and one of them is a boy, what is the probability that I have 2 boys?
The correct answer is not 1/2 but 1/3. How can that be? Well there are 4 possible combinations, BB,GG,BG and GB but but at least one is a boy so you can get rid of GG. So all that's left is BB,BG and GB; and in only one of those 3 possibilities do I have two boys.
2) Now I tell you that I have 2 children and one of them is a boy born on a Tuesday. What is the probability that I have 2 boys?
You may think that Tuesday is not useful information in this matter so the answer would be the same as the previous example, but you would be wrong. The correct answer is 13/27. How can that be?
Well there are 14 possibilities for EACH kid: B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su
But I told you the one of my kids (the first or the second) was a boy born on a Tuesday so that narrows down the field of possibilities to:
First child: B-Tu, second child: B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
Second child: B-Tu, first child: B-Mo, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
No need to put B-Tu in the second row because it's already accounted for in the first row. So now just count them out, 14+13= 27 possibilities. How many result in 2 boys? Count them out again 7+6=13. So 13 out of 27 possibilities give you 2 boys.
The way the puzzle is posed is ambiguous, as it does not state exactly how the trial is carried on. It may be reworded in (to my understanding) 2 different reasonnable ways, leading to 2 different answers.
1) Out of the set of 4 boys and 4 girls, that is 4 families representing the 4 combinations BB,BG,GB,GG, you choose at random a boy (rejecting the trial if you get a girl). Then the probability of the other member of the family being a boy is 1/2. This is the same as Ariskatsaris scenario B) below.
2) Out of the 4 families, you chose at random one that has at least one boy (rejecting the family with 2 girls), then the probabilty is is 1/3.
The way the puzzle is worded appears to me closer to scenario #2 than #1, but this my biased interpretation.
There are 2 probability puzzles that I like:
1) Suppose I tell you that I have 2 children and one of them is a boy, what is the probability that I have 2 boys?
The correct answer is not 1/2 but 1/3. How can that be? Well there are 4 possible combinations, BB,GG,BG and GB but but at least one is a boy so you can get rid of GG. So all that's left is BB,BG and GB; and in only one of those 3 possibilities do I have two boys.
2) Now I tell you that I have 2 children and one of them is a boy born on a Tuesday. What is the probability that I have 2 boys?
You may think that Tuesday is not useful information in this matter so the answer would be the same as the previous example, but you would be wrong. The correct answer is 13/27. How can that be?
Well there are 14 possibilities for EACH kid:
B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su
G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su
But I told you the one of my kids (the first or the second) was a boy born on a Tuesday so that narrows down the field of possibilities to:
First child: B-Tu, second child: B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
Second child: B-Tu, first child: B-Mo, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
No need to put B-Tu in the second row because it's already accounted for in the first row.
So now just count them out, 14+13= 27 possibilities. How many result in 2 boys? Count them out again 7+6=13. So 13 out of 27 possibilities give you 2 boys.
John K Clark