Quick Logic Problem

LMAO...this is the don't get laid much thread..or don't have many kids thread...just saying, I gave the correct answer and ....lol.., go figure!
 
Jane has a 1 in 3 chance of having a boy and then a girl, a 1 in 3 chance of having a girl and then a boy and a 1 in 3 chance of having a boy and then a boy.

Joan has a 1 in 2 chance of having a boy and then a boy and a 1 in 2 chance of having a boy and then a girl.
Joan's oldest child is a boy so those are the only 2 logical combinations left.

Thus combined there is a 1 in 6 chance that they have all boys ( 1/3 * 1/2).
 
KingCondanomation said:
Did you know that without reading any of the other posts?
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I saw the other posts, but not your last two posts. I thought it might be a Bayesian, but the second outcome is not conditional on the first, so that wouldn't work. I thought it might be Binomial, but the events are not independent.

So, I drew up the logic table. It turned out to be a simple counting answer.
 
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Thorn said:
Not because it's been established that each already has one boy. So we're calculating the probability only that the second child of each is a boy. The knowledge that each already has one boy is actually irrelevant and is one of those nasty little tricks that such questions often include just to mess with you.
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Yup. There is a 50% chance they would have a child that is a boy, regardless of what type of child they had before.

It's like the coin flip.
 
Damocles said:
Yup. There is a 50% chance they would have a child that is a boy, regardless of what type of child they had before.

It's like the coin flip.
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You are thinking of it from the point of view of Jane and her husband about to be having another child when an older brother already exists. Both children already exist and you don't know that the older child is a boy - just that one is a boy.
 
KingCondanomation said:
You are thinking of it from the point of view of Jane and her husband about to be having another child when an older brother already exists. Both children already exist and you don't know that the older child is a boy - just that one is a boy.
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It doesn't matter. You have fallen for "common sense" rather than reality.

You already know you got heads on a coin flip, what are the odds you will get heads on the next one?

They are exactly the same for every flip regardless of how the coin landed before. Your odds would be 50%.

If you have two people flipping coins and both get heads on the first flip, it would be 25% probability that they would both get heads on the next flip, and the next... and the next. Why? Because it is a 50% probability for each to land heads, .5 x .5 = .25...
 
Damocles said:
There are not.

It does not matter how many children they have, the unknown quantity is still a 50% probability of being a boy.
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Ah but which unknown quantity? The unknown younger sibling or the unknown older sibling? You don't know as they are both unknown.
Thus you have 3 different possibilities (for Jane):
A boy was born and then another boy.
A boy was born and then a girl.
A girl was born and then a boy.
 
Damocles said:
It doesn't matter. You have fallen for "common sense" rather than reality.

You already know you got heads on a coin flip, what are the odds you will get heads on the next one?

They are exactly the same for every flip regardless of how the coin landed before. Your odds would be 50%.

If you have two people flipping coins and both get heads on the first flip, it would be 25% probability that they would both get heads on the next flip, and the next... and the next.
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Damo, there is no next one, you only know you got heads on one of the coin flips - could be the first, could be the second, could be both.
 
KingCondanomation said:
Ah but which unknown quantity? The unknown younger sibling or the unknown older sibling? You don't know as they are both unknown.
Thus you have 3 different possibilities (for Jane):
A boy was born and then another boy.
A boy was born and then a girl.
A girl was born and then a boy.
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It doesn't matter which is unknown.

There is one unknown child for each, for each the probability is 50% that it will be a boy, therefore I do the math.

You are making it unnecessarily "complicated" and making a mistake.

They each had one child that we know was male.

There are only two possibilities.

Boy child - Girl Child
Boy Child - Boy child.
 
Let me make it clearer.

Name each of the boys that we know...

Bob and John

Jane had John
Joan had Bob

Now put it into the same scenario you have.

John - Girl .25%
John - Boy .25%
Girl - John .25%
Boy - John .25%

So you add the times that the other child was a boy together to find that probability... 50%...

It doesn't matter where it is placed there are only two scenarios that can be posibble, either the unknown child is a boy or a girl.
 
Damocles said:
It doesn't matter which is unknown.

There is one unknown child for each, for each the probability is 50% that it will be a boy, therefore I do the math.

You are making it unnecessarily "complicated" and making a mistake.

They each had one child that we know was male.

There are only two possibilities.

Boy child - Girl Child
Boy Child - Boy child.
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No there are 3 possibilities (for Jane):
Boy child - Girl Child
Girl child - Boy Child
Boy Child - Boy child.

We know there is no Girl Child - Girl Child as Jane definitely has one son, that is the only possibility we can eliminate.
 
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