Funny when you think of it as cards from a finite set, it seems easier to see it's not the same and something would be amiss. Draw two cards, one is red, even money if the other is red.

Quote:JeepsterWhy was it a mistake, it's just a scenario to set out the question.

It gave the info that one child is a boy and the sex of the other is unknown

It did not alter the answer, it's still 1 in 3 that the other is a boy.

Now that I've sobered up, I think I can explain this better: that's not the only information you gave me. You gave me the information that the child you were out with was a boy. Assuming you wouldn't keep a girl locked up, I've essentially taken a random sample and found a boy, which would be impossible if you had two girls, but would be certain if you had two boys, and possible but uncertain if you had one of each. Because it would be less likely to have happened in that world, the fact that it has happened makes that world less likely.

Here's the question you meant to ask: "Given I have at least one boy, what's the chance I have two?" And we get the chance you'd have at least one boy if you had two, 1, times the probability you'd have two boys, 1/4, over the probability you'd have at least one, 3/4, yielding 1/3. But that's wrong.

It's wrong because here's the question you actually asked: "Given this child was a boy, what's the chance I have two?" So what we get is the chance a randomly selected child if yours would be a boy if you had two, 1, times the probability you'd have two, 1/4, divided by the probability a random child would be a boy, 1/2, and that gives 1/2.

Actually, if I know your second child is just born, the answer is 50%. I f I know nothing, the answer is 33.33%Quote:michael99000In other words, if I tell you I have a son and a pregnant wife, the odds of that second kid being a boy are 50/50.

An hour later the kid is born. I tell you I have a son and another kid at home. Now the odds that other kid is a boy is 33% ?

Doesn't make sense, the only thing that changed is the physical location of the second kid .

Similarly, if you tell me "this boy is my elder child", the answer is 50%.

Probability is not about describing the real world. It is about describing information. In the original problem, I have less information about which child happened to wander with its father than in the 'pregnant' or 'elder' situations. That is why probabilities change. In the latter, the lack of info is just about the sex of one child (the absent one, that is forcibly absent, either because it is just born or because it is the younger one). In the original problem, the lack of info is also about which child I have been presented.

Quote:kubikulannActually, if I know your second child is just born, the answer is 50%. I f I know nothing, the answer is 33.33%

Similarly, if you tell me "this boy is my elder child", the answer is 50%.

Probability is not about describing the real world. It is about describing information. In the original problem, I have less information about which child happened to wander with its father than in the 'pregnant' or 'elder' situations. That is why probabilities change. In the latter, the lack of info is just about the sex of one child (the absent one, that is forcibly absent, either because it is just born or because it is the younger one). In the original problem, the lack of info is also about which child I have been presented.

And let me guess - the long run doesn't matter in a real casino, right?

What's so special about birth order? Let's say someone throws two dice, and you're told one of the numbers was a six. There's then a 1/11 chance that they're both sixes. But say you see one die, and it's a six. There's a 1/6 chance that they're both sixes. What's the difference? In the first case, you didn't know which die was the six, so the five cases where only die A was a six and those where only die B was were equally probable. "But," you say, "I don't know which die I saw!" Sure you do. You saw the die the saw. You didn't see the die you didn't see. Only the possibilities where the die you saw was a six are still possible; you don't have to consider that it might have been the other die. If we call that die "die A," we can see that only those six are possible. What if it's die B? Then we've made a mistake in naming them. But you can't do that if you've just been told that one was a six, because then if you pick the wrong name for the die that was a six, you may have made a mistake in naming them, or may have made a mistake in the actual die. Ultimately, if you don't know which die you saw, but have no reason to think one more likely than the other, you can say that all the one-die probabilities have gone down by half, for the 50% chance that that was the die you saw.

In the same way, I saw the child I saw, and didn't see the child I didn't see. It doesn't matter which was born first, because the possibility that the child I saw is a girl and the one I didn't see was a boy is eliminated, whether in birth order that's GB or BG. I don't know which, but they're equally likely, so if you prefer, I halve the probability of each.

(I'm starting to wonder how many ways I can come up with to explain this.)

Quote:24BingoIn the same way, I saw the child I saw, and didn't see the child I didn't see. It doesn't matter which was born first, because the possibility that the child I saw is a girl and the one I didn't see was a boy is eliminated, whether in birth order that's GB or BG. I don't know which, but they're equally likely, so if you prefer, I halve the probability of each.

(I'm starting to wonder how many ways I can come up with to explain this.)

Imagine a coin flip game.

Version 1: Your friend is flipping a penny and a quarter. He flips both coins, and keeps flipping if they both show up tails. That way he's guaranteed to have at least one head. Once he arrives at a valid outcome, which you can't see, your friend slides forward a coin showing heads. If only one of the penny and quarter is heads, that's the one he slides forward. If both the penny and quarter are heads, he picks one at random. Then he says "This coin is a head. I have flipped another coin. What are the chances of both being heads?" It's easy enough to see that the odds of both coins being heads are 1/3 under the circumstances: of the four possible outcomes in flipping 2 heads, TT has been eliminated leaving (with equal probability) HH, HT, and TH.

In Version 1, the two coins were differentiable -- you knew beforehand how to identify them. In Version 2 they're not. Version 2 is exactly the same but with two pennies instead. How could the odds possibly have changed?

(I think you're agreeing with me, but I'm not sure.)

Quote:24BingoOf course, the difference there is that your friend would under no circumstances have shown you a coin that came up tails, whereas there's no reason to think I'd be unlikely to see this man with his daughter if he has one.

(I think you're agreeing with me, but I'm not sure.)

But you didn't see the man with his daughter. You saw him with his son. It's not possible for the man in the story to have two girls, just as it's not possible in the case where you see him with his daughter for him to have two boys.

Let's ask it another way. If you go around to every parent you see in a park with a single child, and find out which of those parents have exactly two children, what is the probability of the child at home of the two-children parents being the same gender as the child in the park?