Logic riddle:

[Fort Fun Indiana]

Meathead

The gambler's fallacy does not apply to this. The trials have already been completed before the riddle begins. Both coins have already been flipped.

 

[Meathead]

No. Incorrect.

The sex of one child is completely irrelevant to the sex of a sibling.

Yours is a classic gamblers fallacy. You can obfuscate all you want, but in binary sex world it would be a lesson in futility.

Are you black?

 

[Fort Fun Indiana]

The sex of one child is completely irrelevant to the sex of a sibling

Correct! But what you are not getting is that both siblings' sexes have already been chosen.

 

[Fort Fun Indiana]

The sex of one child is completely irrelevant to the sex of a sibling.

Yours is a classic gamblers fallacy. You can obfuscate all you want, but in binary sex world it would be a lesson in futility.

Are you black?

You should save your outbursts for the Trump threads. They are not helping you, here.

Would you like me to try to explain the solution to you again?

 

[Meathead]

You should save your outbursts for the Trump threads. They are not helping you, here.

Would you like me to try to explain the solution to you again?

OK, I get it. So it would be like trying to explain ratio to IM2.

Others can knock on a dead man's door forever. I am not so inclined.

 

[Fort Fun Indiana]

OK, I get it. So it would be like trying to explain ratio to IM2.

Others can knock on a dead man's door forever. I am not so inclined.

No, you don't get it.

You do not have the correct answer.

 

[Dadoalex]

Correct, inasmuch as her sex being chosen at fertilization. But the two independent trials have already occurred. Both coins have already been flipped.

While one coin flip does not affect the other, the sample space is made up of the permutations of the two coin flips.

4 equally occuring permutations:

HH
HT
TH
TT

If you flip the two coins and record the results of both flips (a permutation) the occurrence of each permutation will trend to 25% of all permutations.

For simplicity, you can reduce the sampe space to four permutations, each occuring once.

{HH, HT, TH, TT}

Now, consider ONLY the coin flip pairs where at least one coin is tails.

{HT, TH, TT}

Now choose at random from this sample set. What is the probability you choose the TT permutation?

1/3.

You are describin DEPENDENT probabilities.
The question, as posed, is about INDEPENDENT probabilities.

Flip a coin and it comes up head 999 times in a row
The probability of a head on the 1000th flip is still 50/50.

The probability of the sibling being male or female is 50/50 because the gender of the sibling is independent of the sex of the girl.

 

[Fort Fun Indiana]

You are describin DEPENDENT probabilities.
The question, as posed, is about INDEPENDENT probabilities.

False.

The two "independent trials" -- the fertilization of each sibling -- have already occurred and, yes, they were independent of each other.

The probability of each possible permutation is very much dependent on learning the gender of one of the siblings.

 

[Dadoalex]

False.

The two "independent trials" -- the fertilization of each sibling -- have already occurred and, yes, they were independent of each other.

The probability of each possible permutation is very much dependent on learning the gender of one of the siblings.

You have 1 female.
You have a sibling whose sex is unknown.
The sex of the sibling is independent of the sex of the girl.

The OP asked about the probability of the sibling being female.

It did not ask what the permutations were nor did it ask the probability of a certain permutation.
Permutations are dependent probabilities.
This is an independent probability because of the nature of the question.

 

[mamooth]

33.3% is the correct answer. As simply as I can put it:

Nope, 50% is the correct answer.

There are four equally.possible permutations of two siblings:

That's incorrect. There are two sets of two possibilities. That is _not_ the same thing as 4 possibilities.

If she's the older sibling, the possibilities are GB or GG.

If she's the younger sibling, the possibilities are BG or GG.

We don't know which case it is, but it doesn't matter. In either case, the odds of another girl are 50%.

 

[Grumblenuts]

If she's the older sibling, the possibilities are GB or GG.

If she's the younger sibling, the possibilities are BG or GG.

We don't know which case it is,

So the odds would be 50% if we did.

Same as given the carny didn't know where the money ended up.

Knowledge is power. Rah!!

 

[Fort Fun Indiana]

The sex of the sibling is independent of the sex of the girl

*was, when it was chosen. The revealed gender of one of the siblings affects the possible permutations in the remaining sample space. Not the gender of the other sibling.

The OP asked about the probability of the sibling being female

Yep! 1/3.

 

[Fort Fun Indiana]

That's incorrect. There are two sets of two possibilities. That is _not_ the same thing as 4 possibilities.

4 equally possible permutations.

This is why 50% of the possible combinations consist of one boy, one girl.

 

[Fort Fun Indiana]

If she's the older sibling, the possibilities are GB or GG.

If she's the younger sibling, the possibilities are BG or GG.

We don't know which case it is, but it doesn't matter. In either case, the odds of another girl are 50%.

*in each case where you KNOW if she is the older or younger sibling. Yes, that would be 50%.

When you don't know, it is 33.3%

 

[mamooth]

4 equally possible permutations.

No. One has a 1.0 possibility, two have an 0.5 possibility, and one has a 0 possibility. You inflated the 0.5's to 1.0's, so you got the wrong answer.

 

[Fort Fun Indiana]

No. One has a 1.0 possibility, two have an 0.5 possibility, and one has a 0 possibility. You inflated the 0.5's to 1.0's, so you got the wrong answer.

No.

 

[Fort Fun Indiana]

No.

There are two siblings.

At least one is female.

What are the odds the other is also female?

1/3