i am again, bored as hell ...

DO A BARREL ROLL!!!!!

Another thinking challenge:

Say that a woman and a man (who are unrelated) each has two children. We know that at least one of the woman's children is a boy and that the man's oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys?

My algebra teacher insists that the probability is greater that the man has two boys, but I think the chances may be the same. What do you think?

The man could have impregnated more than one women and doesn't have to wait 9 month after that, so theoretically, during his lifetime he can make more children (and more boys) than the woman.

no, as stated, he has two children, not more

I hope this makes enough sense.

I've circled the possibilities your scenario constrains us to, and you can see how many of those fulfill the condition that both children are boys.

Jibb

Sorry, but this does not make too much sense. At least not the circles and your conclusion.

If I see that right the first branching is for the first-born child and the second for the second-born.

In this case, the "boy"-case in the first branching has a chance of 100%, because we already know it's a boy. So we don't have to consider the "girl" case, because its propability is 0%. On the next-branching you have a 50%-chance that the second child is also a boy. You have to multiply it with the overall propability of this branching, which is, as I said 100%, so the resulting propability is 50%.

On the woman tree, we have at the first branching a propability of each 50%. If we follow the boy case, we have again a 50% propability for each case. So the propability of having two boys is 50%*50% = 25%.
If the first-born child is a girl, the propability is 0%, so we don't have to consider this branch. So the overall propability is 50%.

BUT WAIT!

We didn't consider that we know that one of the woman's children is a boy. So we should not talk about "first-born" and "second-born" children, but just "first" and "second" child, because it doesn't matter in which order they are born. Because the propability of the man having two boys doesn't change if we say that his second-born child is a boy and the first child's gender is uncertain, the propability doesn't change.
So, we can say the same thing about the woman's tree as we said about the man's tree, which will also result to 50%.

So the propability that he/she has two boys is 50% for each of them.

You're over-complicating things.

For each person (the man and the woman), there are four possible outcomes of equal likelihood.

In the man's case, we already know the first child is a boy, so we only consider that branch. There is a 1:2 chance of having another boy, and thus a 1:2 chance of having two boys in total.

In the woman's case, the only outcome we know will not happen is girl-girl. Taking that out, we're left with three outcomes of totally equal likelihood, and only one of those is boy-boy, thus she has a 1:3 chance of having two boys.

Jibb

The problem of your way is that the three possibilities for the woman are no Laplace propabilities.

The propabilities are like this:
B-B: 50%
B-G: 25%
G-B: 25%
G-G: 0%
(which is not easily seen in your way)

What you need are Laplace propabilites, so you can add all cases where you can add all cases that meet the requirement (two boys) and devided them by the total number of possibilities.


And I'm not over complicating things. Only in the last paragraph of my last post I'm explaining MY actual (easy) solution: For both, the man and the woman, we know that one child, which we can define as the "first" child is a boy. So we just have to look at the last branch, and there we have two possibilities: "boy" or "girl", which both have a propability of 50%. Only if the second child is also a boy, we have two boys, so the propability of having two boys is 50% for both.

(Believe me, I'm working on a program that calculates propabilites of Poker hands in certain situations in Texal Hold'em with trees like you tried to make and for 5 and 7 closed cards my results are those that are on Wikipedia.

first child of the man is boy this is certain, second being boy is 1:2 possibility. (man's possibilities BG and BB = %50 BB)

woman's children's genders are unknown. possibility of the first child to be a boy is 1:2, possibility of ALSO the second child being boy is 1:4.(woman's possibilities = BB,BG,GG,GB = %25 BB)

if the one is boy, possibilites are BB BG and GB = 1:3