Paradox or did I make a mistake?

Steven G said:
Suppose you visit me in my house, I introduce you to my son and I tell you that I have another child.
Then P( my other child is a girl) = 2/3. (sample space is {bb, bg, gb}

Now suppose you visit me in my house, I introduce you to my son who I tell you is my 1st born and I tell you that I have another child.
Then P( my other child is a girl) = 1/2. (Let x,y mean my 1st child is sex x and my 2nd born is sex y. So sample space is {bb, bg})

Similarly, suppose you visit me in my house, I introduce you to my son who I tell you is my 2nd born and I tell you that I have another child.
Then P( my other child is a girl) = 1/2. (sample space is {gb, bb)

So the last two probabilities are the same.

Now suppose suppose you visit me in my house, I introduce you to my son who I tell you is my ??? born and I tell you that I have another child. The ??? means that you did not hear what I said.
Then P( my other child is a girl) = 1/2 because it does matter what I said.

But wait a minute! If you did not hear me then this is equivalent to the very 1st scenario which has P (other child is a girl) =2/3.

Combining this we conclude that P( other child is a girl) equals both 2/3 and 1/2.

So did I make a mistake or is this some kind of a paradox?
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I'm new to these forums. I wouldn't normally resurrect an old question, except that this one did not get a satisfactory, or even correct, answer.

Yes, you do hit on a kind of paradox. It doesn't break math, it says that you did something wrong to get one of the answer. In fact, it is a paradox that is almost 150 years old, and was published to warn against solution methods that would break math. A warning that has not been well heeded.

It even has a name: Bertrand's Box Paradox. In 1889, that was not the name of the problem, it was how Joseph Bertrand showed, using the same paradox you found, that the 2/3 answer you found cannot be right. It creates a paradox, so something is wrong. He used three boxes with two coins each, not four family combinations of two children, but the principle is the same. It also applies to the Monty Hall Problem, or the equivalent Three Prisoner's Problem.

Martin Gardner posed a slightly different version with two children in 1959. He originally gave the equivalent of the the answer you did (he actually said the probability that the other child was a boy is 1/3). But he retracted that answer, because his problem statement was ill-formed, and ambiguous.

Here is what he said:
Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.
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But your problem statement has no such ambiguity. We were not told that "at least one is a boy," we were shown that a specific child was a boy. When you are told about (or shown) a specific child, it does not matter how that child was selected as long as it is not a biased selection, as in Gardner's first case here. It could be the older, the younger, the first one we meet, or the one who sits to Mother's right at the dinner table. The chances that the other child is a boy, or a girl, are both 50%.

+++++

But let's try an actual example of Bertrand's Box Paradox, reconfigured to match the two-child format.
  1. I have four identical bags of marbles. One has a number of blue marbles. One has the same number of green marbles. The other two each have the same number of marbles, but in an unspecified mix of blue and green.
  2. I let you pick a bag at random, but I don't let you look inside of it.
  3. Question #1: What is the probability that this is a mixed bag?
  4. Now I peek in, look at all of the marbles, and pick a blue one which I show to you.
  5. Question #2: To the best of your ability to say, what is the probability that it is a mixed bag?
For Q1, the sample space originally is {Blue, Mix1, Mix2, Green}. The bags are identical, so the chance that this one is mixed is 2/4=1/2.

When I show you a blue marble, the sample space is reduced to {Blue, Mix1, Mix2}. The answer to Q2 seems to be 2/3. But what if I had shown you a green marble? If this logic is correct, the answer is, again, 2/3.

And now for the paradox. Suppose I look in the bag, and take out a marble without showing it to you. What is the probability that this is a mixed bag? Well, if I were to show it to you, no matter what color it is, you would say 2/3. Since that is the inevitable answer regardless of the actual color, you don't need to see it. The answer is 2/3.

But the answer can't be 2/3, since you have no more information than you did in Q1. It is 1/2.

Finally, the reason: You have no idea why I would pick a blue, or a green, marble from a mixed bag. When you have no information about equivalent choices, you can only assume they are made randomly. Being random, the sample space has to distinguish which color was chosen. Before the marble is seen, it is {Blue, Mix1-Blue, Mix1-Green, Mix2-Blue, Mix2-Green, Green} with probabilities {1/4, 1/8, 1/8, 1/8, 1/8, 1/4}. Once I show you a blue marble, it is {Blue, Mix1-Blue, Mix2-Blue} with prior probabilities {1/4, 1/8, 1/8} and updated probabilities {1/2, 1/4, 1/4}.
 
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