Quick Logic Problem

Again, it doesn't matter when the known quantity is placed.

There is only one unknown and two possibilities.

Your math is simply wrong.

Just as it would be with my coin flip scenario.
 
Damocles said:
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.
Click to expand...

Which is unknown? Are you telling me that the first child is known or the second child is known? They are both unknown. All you know is one is a boy, so there are 3 possible combinations left and you have a 1 in 3 shot of it being a boy-boy.
 
If Jane and Joan had both flipped coins twice and we know that each got heads once, what is the probability they both got heads twice?

25%. It will always be 25% regardless even if we know that Joan's first toss was heads.
 
KingCondanomation said:
Which is unknown? Are you telling me that the first child is known or the second child is known? They are both unknown. All you know is one is a boy, so there are 3 possible combinations left and you have a 1 in 3 shot of it being a boy-boy.
Click to expand...
I am telling you that only one child is unknown in each case.

Therefore there is always a 50% chance that the unknown child is a boy.

When taken together it is a 25% probability that they both had the boy-boy combo regardless of the time we know that Joan's first child was a boy.
 
Anyway, the moment in time the child might have been born has no bearing on the probability of its sex.

1 child is unknown, it is a 50% chance the child is a girl. Every time, regardless of birth order.
 
Suppose I am trying NOT to get 2 tails with 2 coin flips. I will keep flipping both coins until I get any combination other than 2 tails.
Logically, I know that the outcome will have to have at least one of the coins with heads.

There is a 1 in 3 chance of me getting the first coin as heads and the second as tails.
There is a 1 in 3 chance of me getting the first coin as tails and the second as heads.
There is a 1 in 3 chance of me getting the first coin as heads and the second as heads.
 
KingCondanomation said:
Suppose I am trying NOT to get 2 tails with 2 coin flips. I will keep flipping both coins until I get any combination other than 2 tails.
Logically, I know that the outcome will have to have at least one of the coins with heads.

There is a 1 in 3 chance of me getting the first coin as heads and the second as tails.
There is a 1 in 3 chance of me getting the first coin as tails and the second as heads.
There is a 1 in 3 chance of me getting the first coin as heads and the second as heads.
Click to expand...
Again,

If you have two people flipping coins twice each, and they both got heads on one of the flips there is a 25% chance that they both got heads on the unknown flip.

It will be that way no matter how many charts you try to create because it doesn't matter the order, the unknown flip will always have a 50% chance of being heads, when two are taken together you multiply them.
 
Damocles said:
Anyway, the moment in time the child might have been born has no bearing on the probability of its sex.

1 child is unknown, it is a 50% chance the child is a girl. Every time, regardless of birth order.
Click to expand...

1 child's sex is known but BOTH children's relation to Jane are not known.

Try this:
If everyone on this board had 2 kids and I asked all those who have at least 1 boy, how many of you have 2 boys? The response would be about a third.

Just like if everyone on this board had 2 kids and I asked all those who have at least 1 girl, how many of you have 2 girls? The response would also be about a third.
 
KingCondanomation said:
1 child's sex is known but BOTH children's relation to Jane are not known.

Try this:
If everyone on this board had 2 kids and I asked all those who have at least 1 boy, how many of you have 2 boys? The response would be about a third.

Just like if everyone on this board had 2 kids and I asked all those who have at least 1 girl, how many of you have 2 girls? The response would also be about a third.
Click to expand...
Each woman had 2 kids.

One was a boy, the chances the other are a boy is 50% regardless of birth order in relation to the mother. The mother is not a variable, nor is the boy and birth order is superfluous.

Hint: It's like the missing dollar, it isn't really missing the math was done incorrectly....

Anyway, there is only one variable regardless of birth order with two choices for each unknown child.
 
Damocles said:
Each woman had 2 kids.

One was a boy, the chances the other are a boy is 50% regardless of birth order in relation to the mother. The mother is not a variable, nor is the boy and birth order is superfluous.

Hint: It's like the missing dollar, it isn't really missing the math was done incorrectly....

Anyway, there is only one variable regardless of birth order with two choices for each unknown child.
Click to expand...
So following your logic, say I were to ask all those parents in the world who have 2 kids and that at least 1 of them is a boy, if both children they have are boys? You are saying that roughly half the respondents would say they had both boys, and the other half would say they had one boy and one girl?
 
You must log in or register to reply here.
Back
Top