.

what have we become

people who don't understand statistics, obviously. wtf, this is not a combinatorial analysis. Since we already know that one child is a boy, the gender of the second kid is an independent event. so it's actually like a 49% chance of having a boy, if you understand biology.

Does one of these count?

Does Bob have a Thai gf?

plz tell me more about bob the dirty child molester

you realize that #3 and #4 are exactly the same, right?

I cannot fathom how the answer is anything other than 50%, unless its the tricky wording.

yeah for sure 2 boys arent children

Yeah -- older brother younger sister is exactly the same as older sister younger brother, LDO.

that has nothing to do with the problem; order doesn't matter

I am in the 2/3 crowd. If Bob is assumed to be a random draw from the entire population of people with exactly 2 children, he either has 2 boys or a boy and a girl. Twice as many people have exactly one boy and one girl; therefore, the overall chance of him having a girl is 2/3.

Obv it's not exactly 2/3...slightly higher in most places, significantly lower in China, but you get the drift.

BG = GB im 100% sure its 50/50

If you're a 50/50 person try this. Flip a coin twice 100 times (or however many it takes until you'll believe the results) recording each set of 2 flips. Once you're done look for every set that has at least one head and see how many of them also have a tail. The answer will be roughly 2/3, which is exactly the same scenario as this question. If it was specified whether the boy was youngest or oldest child (eg only look at every set that has a tail first) it would be 50/50 but without that specification the answer is 2/3.

it's 2/3.

at the start of the movie 21 i think they do something like this too in the lecture

Wow. I can see how you're confused, but I am nonetheless amazed. The order doesn't matter. You either know the gender of child one or of child two. So you can say "Child 1 is a boy," leaving as possibilities b,b and b,g OR you can say "Child 2 is a boy," leaving as possibilities b,b and g,b. The fact that we do not know whether the boy is child one or child two does not change the fact that he can only be one of them. You can't have both.

Hahahaha. While it is true that roughly 2/3 of the sets containing one head will also contain one tail, that has absolutely nothing to do with the probability of an independent variable. The more accurate "experiment" would be:
1. Flip one coin. If it is a tail, disregard it. If it is a head,
2. Flip it again.
3. Count the number of heads and tails occurring in #2. I bet you get 50/50.

Why would you disregard the tail(girl) on the first flip? If the second one is a head(boy), he still has two children and at least one is a boy.

It's basically the same as saying "Bob flipped a coin twice. One of the flips came up heads. What are the odds that one of them was tails?"

Since he was twice as likely to have gotten one of each, it is 2/3.

For those that missed it, I think this illustrates the flaw in the 66% thing in an understandable way:
To come at it another way, imagine that we are flipping ten coins. Before we flip any of them, there are a very large number of possible outcomes, all but one of them involving tails. Then the first nine coins come up heads. What is the probability that one of the ten is tails? It's 50/50. It has nothing to do with the very large probability before the first nine flips that one would be tails. The only thing that matters after the first nine flips is the tenth flip. Hopefully, this will make the error in the 66% analysis more clear.

That is, after the first nine flips (or to make it better mesh with the OP - once we know the result of nine flips), we are NOT left with these possible sets:
1. t,h,h,h,h,h,h,h,h,h
2. h,t,h,h,h,h,h,h,h,h
3. h,h,t,h,h,h,h,h,h,h
4. h,h,h,t,h,h,h,h,h,h
5. h,h,h,h,t,h,h,h,h,h
6. h,h,h,h,h,t,h,h,h,h
7. h,h,h,h,h,h,t,h,h,h
8. h,h,h,h,h,h,h,t,h,h
9. h,h,h,h,h,h,h,h,t,h
10. h,h,h,h,h,h,h,h,h,t
11. h,h,h,h,h,h,h,h,h,h

If these were our possible outcomes after nine flips, we would say that the probability of a tail existing in the set of ten is 90.9%, but only two of them are actually possible:

1. t,h,h,h,h,h,h,h,h,h OR
2. h,t,h,h,h,h,h,h,h,h OR
3. h,h,t,h,h,h,h,h,h,h OR
4. h,h,h,t,h,h,h,h,h,h OR
5. h,h,h,h,t,h,h,h,h,h OR
6. h,h,h,h,h,t,h,h,h,h OR
7. h,h,h,h,h,h,t,h,h,h OR
8. h,h,h,h,h,h,h,t,h,h OR
9. h,h,h,h,h,h,h,h,t,h OR
10. h,h,h,h,h,h,h,h,h,t AND
11. h,h,h,h,h,h,h,h,h,h

I don't know how to explain it more clearly than that. After this, I give up.

This is where you mess up. There is no constant since you don't know which child is a boy.

I'm an exactly average poker player who plays in 3-handed games. Exactly half of players are better than me and half are worse. If I sit down at a table, there are four equal possibilities:

1. Both players are better than me.
2. The player on my left is better and the one on my right is worse.
3. The player on my right is better and the one on my left is worse.
4. Both players are worse than me.

Now...I go and sit at a random table. If it is given that I am better than at least one of the players, this leaves possibilities 2, 3 and 4. Each possibility is equally likely. Therefore, there is a 2/3 chance that I am not as good as one of the players.

I understand that if the player on the left is better than me that there is a 50/50 chance that the other player is also since in this situation, we are left with only 1 and 2. However, this is not the question being asked.

OP I hate you for reminding me of my discrete mathematics 2 class

Saying one child is a boy does not preclude the other being a boy. This is not the "I have 2 coins totaling 35 cents, and one of them is not a quarter..." problem.

67%, btw. Before any births, he is not equally likely to have 2 boys or 1 boy. He is twice as likely to have exactly 1 boy.

Bears.

Its 2/3rds. Its been proven with surveys and such.

*Yawn*

My reply is not a level: Your answer would be correct if the FIRST child was a boy, just as your answer would be correct if the first 9 flips were heads. It does not say that.

I'll simplify your example: How many have at least one tails? Given at least 1 tails, what's the probability of a heads as the other possible outcome?

1. hh
2. tt
3. th
4. ht

2 through 4 have at least 1 tail. Of these 3, 2 have a heads. Explain why we should ignore outcome 4, which you are doing? The tails (son) can come second. You are asserting the distribution of [2 boys/2 girls/1 of each] is exactly the same.

Actually, the order does matter. That "OR" changes the answer. It may not make sense to you intuitively, but it's something you learn in any intro to probability course.

Also, in your example here, the flaw is that you can't count BB as a possibility twice.