Wonderfully logical illogical probability problem

[Jokester]

Look at the link to the girl-boy paradox from wikipedia.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

He's right

The example of the daughter being younger is analogous to that of question 1 where the answer is 50%.

 

[[TR]SweetSpotOz]

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[PurpleDinosaur]

But you're given some information (i.e. that one of the children is a girl) so this narrows down the options.
At this point you know that it's impossible for him to have 2 boys.

Think of it this way, you have 2 yellow balls and 2 red balls.
2 of these are put into a box, but you don't know which ones.
You pull out the first ball and it's yellow.
What is the probability of the second ball being red?
This is exactly the same problem, the answer is 2/3.

There are twice as many ways to arrange a yellow ball and a red ball (YR and RY) that there are to arrange two yellows (YY) thus the situation of this occurring happens twice as often as getting another yellow (so probability of this is 1/3)

Sorry but this situation is not to do with probability, it's to do with whether the man saying "this is my daughter" means his other child is male or not, which will vary from person to person. It doesn't rule out his other child being either male or female.

 

[Jokester]

Since people seem to be having a problem with it from a statistics point of view,

You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy.

What is the probability that any one child is male. It's 50%. This is not affected by the sex of other children. The fact one child is female is a moot point.

Would it make a difference if he had said "I have two children, this is my youngest child, Lisa"? (You may assume a 50% chance of any one birth being a boy or a girl.

Again the fact the younger child is female has no bearing on the sex of the other child, it's 50/50.

 

[Jokester]

He said that it is 50% in either situation. It isn't?

This only becomes question 2 in the Wiki link when you remove the constraint that the daughter is younger.

Edit: By stating the daughter is younger reduces the possible combinations to either female + male or female + female, there is no male + female case to give you the 2/3

 

[[TR]SweetSpotOz]

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[[TR]SweetSpotOz]

x

 

[Haircut]

I now feel annoyed as I was sure it was 50/50 but then went with Haircut as the wiki link said the same thing.

You were right to go with me as I posted the correct answer

For the first part, where you don't know which child is oldest, it's definitely 2/3 chance of the other one being a boy.

 

[banja]

Reminds me of the roulette thing where the odds of getting 13 Reds in a row is massive, where as the odds of getting a black after 12 reds is still 1 in 2 (ignoring the zero).

It's still 50% anyway in my opinion, I don't care how many maths paradox problems are spouted.

 

[[TR]SweetSpotOz]

Z

 

[Haircut]

Reminds me of the roulette thing where the odds of getting 13 Reds in a row is massive, where as the odds of getting a black after 12 reds is still 1 in 2 (ignoring the zero).

It's still 50% anyway in my opinion, I don't care how many maths paradox problems are spouted.

But similarly the odds of getting a 13th red after already having had 12 is also 1 in 2.

 

[Jokester]

And replace it by saying that the older child is a boy.

If the older child is a boy then the younger child 50/50 boy girl, or if you still state the younger child is female, then the younger child is 100% female, and the older child 100% male.



A probabilty of 2/3 is not possible if you state that the younger child is definitely female as you've reduced the problem to what is the sex of one child, which is always 50/50.

 

[Jokester]

A-ha:

http://mathforum.org/library/drmath/view/52186.html

Depends on the sample space. ie both are right depending on how it is defined. However in the OP it was defined as being 2/3

Where you went wrong was in saying b:b was as likely as b:g or g:b,
when of course in b:b if we now call the boy we know about x, we have
x:b and b:x, so we have the sets x:g, g:x, b:x, and x:b; and thus a
probability of 1/2.

And this is effectively the case the OP poster makes, except it's further reduced to x:g and x:b because he also states the girl is the younger.

 

[[TR]SweetSpotOz]

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[banja]

But similarly the odds of getting a 13th red after already having had 12 is also 1 in 2.

Correct. That actually proves the point a little better too

 

[[TR]SweetSpotOz]

And this is effectively the case the OP poster makes, except it's further reduced to x:g and x:b because he also states the girl is the younger.

Again, it depends on how it is defined.

 

[Jokester]

Read my link. I'll admit I didn't get what you were saying about the wiki link.

I have read your link, it also demonstrates that in both cases the probability is 50/50 because the OP messed up the question

Again, it depends on how it is defined.

Yes, and as defined by the OP, it is most definitely 50/50

 

[[TR]SweetSpotOz]

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[Haircut]

I don't think the OP did mess up the question.
Just taking the first part:

You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy.

The answer is 2/3

Obviously ignoring any linguistics things where he wouldn't say 'this is my daughter' if he had two daughters.

 

[[TR]SweetSpotOz]

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