Math Challenge #3.1 - Solution posted in thread

[sciencewhiz]

Kranky, your reasoning is wrong.

Lets say that the second child happens to be a girl also. Does this mean that the third child has to be a boy? Of course not.

Punnett squares are only valid for determining the probability of an event once. The probability gets reset each time there is another child. It doesn't matter how many times a couple has a child, there is always a 50/50 chance that the next one will be a boy.

 

[kranky]

I'm not claiming Punnett squares are involved in the answer. This question is similar to the one in the other thread (Monty Hall paradox) in that it involves conditional probability. Conditional probability involves the probability of some event occurring given that some other event has definitely occurred.

It really is as simple as the explanation above. No tricks. It just doesn't SEEM logical.

And, to make it even more bizarre, if the question was "given that the OLDEST child is a girl, what are the odds that the other child is a boy?" the answer would indeed be 1/2. (possible combinations when OLDEST child is a girl: GB and GG. 1/2 chance that second child is a boy)

Some links that explain this reasoning:
Princeton Univ. (search for "problem 3&quot
Swarthmore Univ.

 

[Specialist]

1 to 2 And that is my final answer!

 

[konichiwa]

Nice post, Kranky, that really had me thinking. I see how you're right now.

 

[konichiwa]

We need some more of these Math Challenges, I like them a lot. They expand my mind.

 

[Xede]

Kranky, you worded your problem wrong. The answer to the question that you originally posed IS 1/2.

Original question: "You are in a chat room one night and start talking to a girl you have never met. She tells you she has one sibling. What are the odds that the other child in her family is a boy?"

A = event that the person you're talking to is a girl
B = event that the sibiling is a girl

P(sibling is a boy) = 1 - P(B)

P(B|A) is the probability that the sibling is a girl, given that the speaker is a girl

P(B|A) = P(B and A)/P(A)

A and B are independent events. The sex of one child is independent of the sex of the other.

since A and B are independent, P(B and A) = P(B)*P(A)

so P(B|A) = P(B)*P(A)/P(A) = P(B)

and P(B) = 1/2 since any child will be born a girl with 1/2 probability

and P(sibling is a boy) = 1 - P(B) = 1 - 1/2 = 1/2

The answer to the question you posed is 1/2.

If, instead, you do not know the gender of the speaker, and the speaker says "At least one of us is a girl"
the probability that the sibling (not the speaker) is a girl is 2/3, or 1/3 probability that the sibling is a boy.

B = event that the sibiling is a girl
C = event that at least one of the two children is a girl

P(B|C) = probability that the sibling is a girl, given that at least one of the two is a girl

P(B|C) = P(B and C) / P(C)

Here, B and C are NOT independent, as they were in the originally stated problem.

1st column is the speaker, 2nd column is the sibling

BB
BG (B and C is true here)
GB
GG (B and C is true here)

P(B and C) = 1/2

BB
BG (C is true here)
GB (C is true here)
GG (C is true here)

P(C) = 3/4

P(B|C) = P(B and C) / P(C) = .5 / .75 = 2/3

The problem with your wording is that as soon as you specify that a certain one of the children is a girl (doesn't matter whether you say the "oldest" is a girl, or the "speaker" is a girl), you eliminate the conditional probability and the sex of the other child is an independent event with probability 1/2.

 

[Wangel]

Good Answer - I submit :Q