The other can also be born on Tuesday. (generated id = 1 in rnd(7)) And the other can also have a generated id = 456783123 in rnd(1billion). But as the bound of the rnd is bigger and bigger, the probability of that approaches to zero. So as we know more and more information of one of the children, we are closer and closer to simply uniquely identifying him, so the probability of the other being a boy approaches 2.

If both can be born on Tuesday then telling me one was born on Tuesday does not uniquely identify him! It's not unique if both can do it.

And since it's not unique the rest of your analysis is based on a faulty assumption.

(BTW I did not downmod you, in case you were wondering.)

It is about specifying a child, and then saying it is a boy.

The one extreme is: one of my children is a boy. in this case the other is a boy with 1/3 probability.

The other extreme is: One of my children has a national unique id=... He is a boy. The other is a boy with 1/2 probability.

And other cases are in between. The more information you provide on the first child (the less chance there is that the other can have the same property, the more close you are to 1/2.

Of course in the rnd(7) example we are in-between:

1/3 < 13/27 < 1/2

"the less chance there is that the other can have the same property"

No! That is not true! The two events are independent, specifying information on one has zero impact on the other.

They are independent, but you get information about both ("One of them is a boy."), therefore p=1/3.

If you say "the older is a boy" then you have a statement about one son and no information whatsoever about the younger one, therefore p=1/2.

"One of them is a boy." is a completely different scenario and no one is arguing about it. (BG, GB, BB, GG - by saying one is a boy you exclude GG leaving 1/3.)

In this case we are asking if the other is a boy NOT if the other was born on Tuesday.

Ah, sorry, I misunderstood your question.

You are asking, why the information "one is born on Tuesday" says anything about the gender or birthday of the other, right?

More specificaly, this:

> Let us first assume that it is the older child who was a son born on a Tuesday. In this case the second child could be either of two sexes, and could have been born on any of seven days of the week, for a total of 14 possibilities.

> Now let's suppose it is the younger child who was a son born on a Tuesday. Then the older child could, again, be either of two sexes and could have been born on any of seven days of the week, again providing 14 possibilities. Added to our original 14 that would seem to give 28 possibilities.

> But be careful! One possibility got counted twice. Specifically, the one where both children are boys born on Tuesdays. So really there are only 27 possibilities. And since 13 of them involve the second child being a boy, the probability would be 13/27.

Maybe it helps, if you first imagine all 196 (2 * 2 * 7 * 7, think of four 7-by-7 tables [draw them, it helps a lot ;)]) possibilities. Eliminate 49 of them where both are girls (one of the four tables) and you have 147 possibilities/cells left. Then, in each of the two BG and GB tables leave only one row/column, eg. eliminating another 2 * 6 * 7=84 possibilities.

Now comes the interesting/tricky part: The last case/table, where both are boys contains only 13 possible cases (instead of 2 * 7)! Imagine a 7-by-7 table, where each row and column corresponds to a day. Mark all the cells which are not in a Tuesday-row or a Tuesday-column, eg. eliminate all posibilities where none of the two boys are born on a Tuesday. This eliminates another 49-13=36 cases.

So, we have a total of 196-49-84-36=27 cases, 13 of which "the other is a boy" and 3/27=1/9 cases where "the other is born on a tuesday".

I hope this makes somewhat sense. Once I've drawn all the possibilities and marked all the impossible ones, it became a lot clearer.

ADDENDUM: The tables:

       A: Boy                     A: Boy
       M T W T F S S              M T W T F S S
                           
 B M   . 1 . . . . .        B M   . 1 . . . . .
   T   1 1 1 1 1 1 1          T   . 1 . . . . .
 B W   . 1 . . . . .        G W   . 1 . . . . .
 o T   . 1 . . . . .        i T   . 1 . . . . .
 y F   . 1 . . . . .        r F   . 1 . . . . .
   S   . 1 . . . . .        l S   . 1 . . . . .
   S   . 1 . . . . .          S   . 1 . . . . .
  
  
       A: Girl                    A: Girl
       M T W T F S S              M T W T F S S
     
 B M   . . . . . . .        B M   . . . . . . .
   T   1 1 1 1 1 1 1          T   . . . . . . .
 B W   . . . . . . .        G W   . . . . . . .
 o T   . . . . . . .        i T   . . . . . . .
 y F   . . . . . . .        r F   . . . . . . .
   S   . . . . . . .        l S   . . . . . . .
   S   . . . . . . .          S   . . . . . . .

 total: 196
 of which are possible ('1'): 27
 of which 'the other is a boy': 13
   (all the ones from the upper left table)
 of which 'the other child is born on a tuesday': 3
   (the T/T-cell of each table)

If the older child is the boy born on Tuesday, you have 7 chances the younger child is a boy. If the younger child is a boy born on Tueday, you only have 6 chances the older child is a boy. This is because the 7th chance (older is a boy born on Tueday) was covered in the first scenario.

Summarized from: http://www.sciencenews.org/view/generic/id/60598/title/Math_...