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# One Girl - One Boy

### #71

Posted 05 August 2007 - 02:54 AM

So if the girl is the first child, the possibilities are:

GG

GB

or 1/2 the time the second child is also a girl

If the girl is the second child, the possibilities are:

GG

BG

or 1/2 the time the first child is also a girl

since the first scenario results in 1/2 and the second scenario results in 1/2, we have:

(50% * 1/2) + (50% * 1/2) = 1/4 + 1/4 = 1/2

Thus, like many have said - the odds are clearly 1/2 that the other child is also a girl.

Doesn't work that way. Why on Earth are you multiplying 50% by 1/2 twice and adding them together?

If you think your answer is logical and correct, PLEASE do the experiment with pennies that I laid out several times. When all TT combinations are removed, about 1/3 of the pennies will be HH, 1/3 will be HT, and 1/3 will be TH.

### #72

Posted 05 August 2007 - 09:11 PM

### #73

Posted 06 August 2007 - 06:27 AM

Q. If the first child is a girl, the probability of the other child being a girl/boy is 1/2.

If the second child is a girl, the probability of the other child being a girl/boy is 1/2.

Ifanyof the children is a girl, how can the probability of the other being a girl be less(i.e., 1/3)? Aren't we merging both cases?

**A**. Yes we are merging two cases,

*creating duplicate members of the set*.

In the first we have 1 in 2 - GB, GG; in the second we have 1 in 2 - BG, GG.

Merging them we have a duplicate member GG which is occuring twice. So 1

*in*2 and 1

*in*2 do not form 2

*in*4 but rather 1

*in*3 due to overlapping members.

Q.Does already having a girl decreases the chance of having another girl?

**A.**Having a girl increases the chance of having a subsequent boy, thus decreasing the chance of having another girl. Just like if we get several heads in succession on tossing a coin, the chance of getting a subsequent tail increases.

### #74

Posted 06 August 2007 - 06:36 AM

Q.Does already having a girl decreases the chance of having another girl?

A.Having a girl increases the chance of having a subsequent boy, thus decreasing the chance of having another girl. Just like if we get several heads in succession on tossing a coin, the chance of getting a subsequent tail increases.

I guess you think if you are at a roulette table and have just seen red come up 6 times in a row you should bet on black? Please, enough inaccurate information has been given in this thread!

### #75

Posted 06 August 2007 - 06:16 PM

### #76

Posted 06 August 2007 - 09:33 PM

The reason assigning which member of the group meets the condition changes the probability is because it changes the possibilities. If you know the oldest child is a girl or the brown bear is a male, then there are only two possible results with one meeting the criteria, which results in the 50% answer. Without assigning either member specifically, then there are three possible results with only one meeting the criteria.

The brown bear/black bear scenario might actually make the whole thing easier to understand, since we get away from the sequencing issue that many people are hung up on. Without any conditions, there are four possible results; brown male/black male, brown male/black female, brown female/black male and brown female/black female. Once you stipulate the criteria that one of the bears be male, you eliminate only the instance where both are female. Leaving three possibilities with only one meeting the criteria.

I hope that adds something to the discussion, but it is really just rewording of what has already been said many times.

### #77

Posted 07 August 2007 - 12:50 AM

A lot of the initial confusion in this riddle is that it was worded differently the first few days. Initially, it was implied that a couple was going to have 2 children (future tense), and the question was, if the first child was a girl, what would the probability be for the second to also be a girl.

I do see with the wording now, the actual intent of the OP. It is indeed 1/3 the way it is worded now, just as it is 1/4 the way it was worded originally.

Not only did the OP never change the wording of the riddle, if it was worded the way you claim, the answer to the problem you wrote above is a probability of 1/2, not 1/4.

This is why in roulette black and red are 2:1 payout

1:1 payout.

### #78

Posted 07 August 2007 - 01:04 AM

It does come down to the wording.

"One of them is a girl"

versus

"One of which is a girl"

The statement in the riddle is not "One of which" though.

"One of them is a girl" means the same exact thing as "One of which is a girl"!

Nor is it "At least one of them is a girl"

You're saying that adding "at least" to "One of them is a girl" changes the meaning? It doesn't! Did you think that "One of them is a girl" didn't mean at least one? Did you think it meant both?

It's a matter of assigning. As soon as you say "one of them" you've assigned, say, the first born, or the second born to be a girl.

Oh really? So which one did the OP assign, the first or the second? The answer is 'neither'. "One of them" means

**either**one of them. Not the first and not the second.

But then again, I may just be "one of them" guys.

Yup!

### #79

Posted 07 August 2007 - 01:12 AM

ANYONE who thinks it is 1/2, PLEASE come forward and state your case. Those who do not we will assume agree with us that it is 1/3. To anyone that still thinks it is 1/2, we will correct the flaw in your arguement.

### #80

Posted 07 August 2007 - 07:22 PM

You're saying that adding "at least" to "One of them is a girl" changes the meaning? It doesn't! Did you think that "One of them is a girl" didn't mean at least one? Did you think it meant both?

Actually, in Propsguy's defense, these two statements can be interpreted differently. "At least one" clearly allows for more than one. However, "One of them" can be interpreted to mean "one and only one". Of course, if this is the interpretation you take, then the answer to the puzzle is "0%", as some clever individual has already posted (Sorry, I forget who it was).

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